Electric potential and capacitor

Class 12 Physics, unit 19, book pages 86–105

This unit connects electric fields to energy. Electric potential is the potential energy per unit charge at a point, it falls as $1/ r$ around a point charge, and the electric field is the negative of its gradient. Capacitors use these ideas to store charge and energy: their capacitance depends on plate area, separation and dielectric, they can be combined in series and parallel, and they charge and discharge exponentially with time constant $R C$ . The unit ends with bioelectricity: how cell membranes, nerves and animals such as the electric eel use potential differences.

Learning outcomes (13)

Define and calculate electric potential at a point as the work done per unit positive charge in bringing a small test charge from infinity to the point, and use $V = \frac{q}{4 π ε _{0} r}$ for the potential in the field of a point charge.
Use the fact that the electric field at a point is equal to the negative of the potential gradient at that point.
State how the concept of electric potential leads to the electric potential energy of two point charges, and use $E_{p} = \frac{q Q}{4 π ε _{0} r}$ .
Define and calculate capacitance, as applied to both isolated spherical conductors and parallel plate capacitors.
Derive and apply formulae for the combined capacitance of capacitors in series and in parallel.
Use the capacitance formulae for capacitors in series and in parallel.
Determine the electric potential energy stored in a capacitor from the area under the potential to charge graph, and use $W = \frac{1}{2} Q V = \frac{1}{2} C V^{2}$ to solve problems.
Analyse graphs of the variation with time of potential difference, charge and current for a capacitor discharging through a resistor, using $τ = R C$ for the time constant.
Use equations of the form $x = x_{0} e^{- t / R C}$ , where $x$ could represent current, charge or potential difference for a capacitor discharging through a resistor.
List the uses of capacitors in household appliances, such as flash guns, refrigerators, electric fans and rectification circuits.
Illustrate how bioelectricity is generated in animals: ion channels in the cell membrane control the flow of specific ions; when a cell is stimulated, positive ions enter through open channels and trigger electrical pulses called action potentials; patterns of action potentials start the correct movements, thoughts and behaviours.
State that several species of aquatic life, such as Electrophorus electricus, can naturally generate external electric shocks through internal biological mechanisms that act as batteries.
Explain, with examples of animals with this ability, that electroreception is the ability to detect weak naturally occurring electrostatic fields in the environment.

19.1Electric potential energyOften examined

Electric circuits take electrical energy from a source and change it into heat, light or motion. A battery stores this energy, power lines carry it across the country, and a capacitor can hold it for later use. The energy a charge has because of its position in an electric field is called electric potential energy.

Energy and voltage are different quantities. A small motorcycle battery and a large car battery both have a voltage of $12 V$ , but the car battery can deliver much more energy because it can move much more charge (see Example 19.1). This unit explains how voltage and energy are related.

A charge between two charged plates

Place a small positive test charge $q_{0}$ between two oppositely charged parallel plates (Fig. 19.1). The uniform electric field $E$ points from the positive plate to the negative plate, and the charge feels a force

F = q_{0} E

$F$ : electric force on the charge (N), $q_{0}$ : test charge (C), $E$ : electric field strength ( $N C^{- 1}$ ).

If the charge is free, the field pushes it from the positive plate towards the negative plate and it gains kinetic energy. To move it the other way, against the field, an external agent must do work. This work is not lost: it is stored as electric potential energy of the charge.

Fig. 19.1: A test charge

+ q_{0}

between oppositely charged plates. The field pushes it towards the negative plate, so moving it from B to A needs work, which is stored as potential energy.

Electric potential energy: The energy that is needed to move a charge against an electric field to its position. It is the energy a charge has because of its position in the field. SI unit: joule (J).

Electric potential energy depends on two factors: (1) the size of the charge itself, and (2) its position relative to the other charges that create the field. More energy is needed to move a larger charge, to move it farther, or to move it through a stronger field.

Change in potential energy

Only the change in potential energy can be measured. When the charge $q_{0}$ is carried from point B to point A against the field, slowly so that it stays in equilibrium (it gains no kinetic energy), the increase in its potential energy equals the work done by the applied force:

Δ U = W_{B A} or U_{A} - U_{B} = W_{B A} (19.1)

$U_{A}$ , $U_{B}$ : potential energies at A and B (J); $W_{B A}$ : work done by the applied force in carrying $q_{0}$ from B to A (J).

19.1Electric potential and potential differenceOften examined

The potential energy of a charge depends on how big the charge is, so it does not describe the place in the field on its own. A test charge twice as big has twice the potential energy at the same point. If we divide the potential energy by the charge, we get a quantity that depends only on the location. This quantity is the electric potential.

Electric potential: The electric potential energy per unit positive charge at a point, $V = U / q_{0}$ . Equivalently, it is the work done in bringing a unit positive charge from infinity to that point while keeping it in equilibrium. It is a scalar. SI unit: volt ( $1 V = 1 J C^{- 1}$ ).

V = \frac{U}{q _{0}}

$V$ : electric potential (V), $U$ : potential energy of the test charge at the point (J), $q_{0}$ : test charge (C).

Example from the book: if the potential at a point is $12 J C^{- 1}$ , then every coulomb placed there has $12 J$ of potential energy. The potential energy changes with the charge, but the potential does not:

Electric potential is a property of the location, not of the charge placed there.
Charge placed at the point	Potential energy $U$	Potential $V = U / q$
$0.5 C$	$6 J$	$12 V$
$1 C$	$12 J$	$12 V$
$2 C$	$24 J$	$12 V$

A positive test charge is at a high potential when it is held close to a positive source charge and at a lower potential when it is held farther away. Positive charges, when free, move from high potential to low potential; negative charges (such as electrons) move from low potential to high potential.

Potential difference

Potential difference: The work done in moving a unit positive charge from one point to another while keeping the charge in electrostatic equilibrium. SI unit: volt (V).

If $V_{A}$ and $V_{B}$ are the potentials at points A and B, the potential difference between them is

Δ V = V_{A} - V_{B} = \frac{W _{B A}}{q _{0}} = \frac{Δ U}{q _{0}} (19.2)

Rearranging gives the change in potential energy of any charge $q_{0}$ that moves through a potential difference $Δ V$ :

Δ U = q_{0} Δ V (19.3)

Unit of electric potential: the volt

The unit of electric potential (and of potential difference) is the joule per coulomb, called the volt (V). The potential difference between two points is one volt if one joule of work is done in moving one coulomb of charge from one point to the other.

1 volt = \frac{1 joule}{1 coulomb} or 1 V = 1 J C^{- 1}

Multiples and sub-multiples of the volt.
Unit	Symbol	Value
millivolt	$mV$	$1 0^{- 3} V$
microvolt	$μ V$	$1 0^{- 6} V$
kilovolt	$kV$	$1 0^{3} V$
megavolt	$MV$	$1 0^{6} V$
gigavolt	$GV$	$1 0^{9} V$

Example 19.1: Energy delivered by two batteries

You have a

12.0 V

motorcycle battery that can move

5000 C

of charge, and a

12.0 V

car battery that can move

60000 C

of charge. How much energy does each deliver?

Given

$q_{m} = 5000 C$ charge moved by the motorcycle battery
$q_{c} = 60000 C$ charge moved by the car battery
$Δ V = 12.0 V$ same for both

To find

$Δ U_{m}$
$Δ U_{c}$

Energy delivered equals charge times potential difference, Eq. (19.3).
$Δ U = q Δ V$
Motorcycle battery:
$Δ U_{m} = (5000 C) (12.0 V) = 6.00 \times 1 0^{4} J$
Car battery:
$Δ U_{c} = (60000 C) (12.0 V) = 7.20 \times 1 0^{5} J$

Δ U_{m} = 6.00 \times 1 0^{4} J, Δ U_{c} = 7.20 \times 1 0^{5} J

Assignment 19.1: Proton accelerated between two plates

A proton is injected at a speed of

1.0 \times 1 0^{6} m s^{- 1}

between two plates

5.0 cm

apart. The proton accelerates across the gap and exits through the opening. (a) What must the electric potential difference be if the exit speed is to be

3.0 \times 1 0^{6} m s^{- 1}

? (b) What is the magnitude of the electric field intensity between the plates, if it is assumed constant?

The field between the plates speeds the proton up from

u

v

Given

$u = 1.0 \times 1 0^{6} m s^{- 1}$
$v = 3.0 \times 1 0^{6} m s^{- 1}$
$d = 5.0 cm = 0.050 m$
$m_{p} = 1.67 \times 1 0^{- 27} kg$ mass of proton
$e = 1.6 \times 1 0^{- 19} C$ charge of proton

To find

(a) $Δ V$
(b) $E$

(a) The loss of electric potential energy equals the gain in kinetic energy.
$e Δ V = \frac{1}{2} m_{p} v^{2} - \frac{1}{2} m_{p} u^{2} \Rightarrow Δ V = \frac{m _{p} ( v ^{2} - u ^{2} )}{2 e}$
Substitute the values.
$Δ V = \frac{( 1.67 \times 1 0 ^{- 27} ) ( 9.0 \times 1 0 ^{12} - 1.0 \times 1 0 ^{12} )}{2 ( 1.6 \times 1 0 ^{- 19} )} = \frac{1.336 \times 1 0 ^{- 14}}{3.2 \times 1 0 ^{- 19}}$
Evaluate.
$Δ V = 4.2 \times 1 0^{4} V$
(b) For a uniform field the field strength is the potential difference per metre.
$E = \frac{Δ V}{d} = \frac{4.175 \times 1 0 ^{4}}{0.050} = 8.4 \times 1 0^{5} V m^{- 1}$

Δ V \approx 4.2 \times 1 0^{4} V, E \approx 8.4 \times 1 0^{5} V m^{- 1}

19.2Electric potential due to a point chargeOften examined

In a uniform field (for example between parallel plates) the potential changes steadily with distance: $Δ V = - E Δ r$ . Around a point charge the field is not uniform: $E = \frac{1}{4 π ε _{0}} \frac{Q}{r ^{2}}$ becomes weaker with distance. So $Δ V = - E Δ r$ cannot be used over a long path in one go.

The trick is to split the path into many tiny steps. In each step the field hardly changes, so the work in that step is easy to find. Adding the work of all the steps gives the total work, and from it the potential.

Derivation Potential $V$ at distance $r$ from a point charge $Q$

Aim. Find the electric potential at a point 1, a distance $r$ from an isolated point charge $+ Q$ , by bringing a test charge $q$ from infinity to that point.

Assumptions.

$+ Q$ is an isolated point charge fixed at the origin, in vacuum or air.
The test charge $q$ is small and positive, and it is moved slowly so that it stays in equilibrium (no kinetic energy is gained).
The path from a far point N to point 1 is divided into small steps of length $Δ r$ ; in each step the field $E$ is almost constant. $r_{1}, r_{2}, \dots, r_{N}$ are the distances of points $1, 2, \dots, N$ from $Q$ .
The potential is taken as zero at infinity.

Fig. 19.2: The test charge

q

is moved from a far point N to point 1 against the field

E

+ Q

, in small steps

Δ r

The field of $+ Q$ points away from $Q$ . To move $q$ towards $Q$ slowly (in equilibrium), we apply a force $F$ equal and opposite to the electric force. So $F = q E$ and it points towards $Q$ , in the same direction as the displacement.
$F = q E = \frac{1}{4 π ε _{0}} \frac{Qq}{r ^{2}}$
Take the first step, from N to N−1. Its length is $Δ r = r_{N} - r_{N - 1}$ . The force and the displacement are in the same direction ( $θ = 0^{\circ}$ ), so the work done by the applied force is
$Δ W_{N \to N - 1} = F Δ r cos 0^{\circ} = \frac{1}{4 π ε _{0}} \frac{Qq}{r ^{2}} (r_{N} - r_{N - 1})$
Inside this tiny step the distance lies between $r_{N - 1}$ and $r_{N}$ , which are very close. So we take $r^{2}$ as the product of the two end distances (a very good average for a small step).
$r^{2} \approx r_{N - 1} r_{N}$
Substitute and split the fraction into two terms.
$Δ W_{N \to N - 1} = \frac{Qq}{4 π ε _{0}} \frac{r _{N} - r _{N - 1}}{r _{N - 1} r _{N}} = \frac{Qq}{4 π ε _{0}} (\frac{1}{r _{N - 1}} - \frac{1}{r _{N}}) (i)$
The same working holds for every other step.
$Δ W_{N - 1 \to N - 2} Δ W_{2 \to 1} = \frac{Qq}{4 π ε _{0}} (\frac{1}{r _{N - 2}} - \frac{1}{r _{N - 1}}) ⋮ = \frac{Qq}{4 π ε _{0}} (\frac{1}{r _{1}} - \frac{1}{r _{2}}) (ii) (iii)$
Add the work of all the steps. Each middle term such as $1/ r_{N - 1}$ appears once with a plus sign and once with a minus sign, so it cancels. Only the first and the last terms are left.
$Δ W_{N \to 1} = \frac{Qq}{4 π ε _{0}} [(\frac{1}{r _{N - 1}} - \frac{1}{r _{N}}) + \dots + (\frac{1}{r _{1}} - \frac{1}{r _{2}})] = \frac{Qq}{4 π ε _{0}} (\frac{1}{r _{1}} - \frac{1}{r _{N}}) (iv)$
Now let the starting point N be at infinity: $r_{N} \to \infty$ , so $1/ r_{N} \to 0$ . Write $r$ for $r_{1}$ . This work is stored as the potential energy $U$ of $q$ at distance $r$ from $Q$ .
$U = Δ W_{\infty \to 1} = \frac{1}{4 π ε _{0}} \frac{Qq}{r} (19.4)$
Electric potential is potential energy per unit charge, so divide by $q$ .
$V = \frac{U}{q} = \frac{1}{4 π ε _{0}} \frac{Qq}{r} \div q = \frac{1}{4 π ε _{0}} \frac{Q}{r} (19.5)$

Result

V = \frac{1}{4 π ε _{0}} \frac{Q}{r} = \frac{k Q}{r}, k = \frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}

The potential of a point charge falls off as $1/ r$ , more slowly than the field ( $1/ r^{2}$ ). It has the same value at all points equidistant from $Q$ , so the equipotential surfaces around a point charge are spheres centred on $Q$ . $V$ is positive around a positive charge and negative around a negative charge, and it is zero at infinity.

Potential is a scalar. The potential due to several point charges is the algebraic sum of their separate potentials, with the sign of each charge included:

V = \frac{1}{4 π ε _{0}} (\frac{Q _{1}}{r _{1}} + \frac{Q _{2}}{r _{2}} + \frac{Q _{3}}{r _{3}} + \dots)

Potential and field around a point charge

r / r_{0}

V, E

(1, 1)

V : \frac{1}{2}

E : \frac{1}{4}

V \propto \frac{1}{r}

E \propto \frac{1}{r ^{2}}

Values are relative to those at a distance

r_{0}

. Doubling the distance halves the potential but reduces the field to one quarter.

Example: Potential and energy near a point charge

Find the electric potential at a distance of

30 cm

from a point charge of

+ 3.0 μ C

. What is the potential energy of a

+ 2.0 nC

charge placed there?

Given

$Q = 3.0 μ C = 3.0 \times 1 0^{- 6} C$
$r = 30 cm = 0.30 m$
$q = 2.0 nC = 2.0 \times 1 0^{- 9} C$

To find

Use Eq. (19.5).
$V = \frac{k Q}{r} = \frac{( 9 \times 1 0 ^{9} ) ( 3.0 \times 1 0 ^{- 6} )}{0.30} = 9.0 \times 1 0^{4} V$
Potential energy of the second charge: $U = q V$ .
$U = (2.0 \times 1 0^{- 9}) (9.0 \times 1 0^{4}) = 1.8 \times 1 0^{- 4} J$

V = 9.0 \times 1 0^{4} V, U = 1.8 \times 1 0^{- 4} J

19.2Electric potential energy of two point charges

Eq. (19.4) also gives the potential energy of a pair of point charges. The charge Q sets up a potential V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r} at distance r. Placing a second charge q there gives it potential energy U=qV:

U=qV=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{qQ}{r}

The SLO box writes this as E_p=\dfrac{qQ}{4\pi\varepsilon_0 r}. U (or E_p): potential energy of the pair (J), q, Q: the charges (C), r: their separation (m).

This energy equals the work done in bringing one charge from infinity to a distance r from the other. It belongs to the pair of charges (the system), not to one charge alone, and it is a scalar.

What the sign of the potential energy tells you.
Charges	Sign of U	Meaning
Like charges (both + or both −)	positive	They repel. Work had to be done to push them together; if released, they fly apart and the stored energy becomes kinetic energy.
Unlike charges (one + and one −)	negative	They attract. Energy must be supplied to pull them apart to infinity (where U=0).

Example: Potential energy of a pair of charges

Charges of +2.0\,\mu\mathrm{C} and -3.0\,\mu\mathrm{C} are 30\,\mathrm{cm} apart. Find their electric potential energy and explain its sign.

Use U=\dfrac{kqQ}{r} and keep the signs of both charges.
U=\dfrac{(9\times10^{9})(2.0\times10^{-6})(-3.0\times10^{-6})}{0.30}
Evaluate.
U=-0.18\,\mathrm{J}
The negative sign shows that the charges attract. To separate them completely (to infinity) an external agent must do 0.18\,\mathrm{J} of work.

U=-0.18\,\mathrm{J}

19.2Potential gradient and the electric fieldOften examined

The electric field and the electric potential describe the same field in two ways. The field is the force per unit charge (a vector). The potential is the energy per unit charge (a scalar). They are linked by how fast the potential changes with distance.

Potential gradient: The rate of change of electric potential with displacement, \Delta V/\Delta r. SI unit: \mathrm{V\,m^{-1}}. It is largest along a field line.

Derivation E=-\dfrac{\Delta V}{\Delta r}

Aim. Show that the electric field at a point equals the negative of the potential gradient at that point.

Assumptions.

A and B lie on the same field line, a small distance \Delta r apart, so the field E between them is almost uniform.
A small positive test charge q_0 is moved slowly from A to B, so it stays in equilibrium.

A positive test charge is moved a small distance \Delta r along the field, from A to B. The applied force F=-q_0E is opposite to the displacement.

The field pushes +q_0 with a force q_0\vec E along the field. To move the charge slowly (in equilibrium), we apply an equal and opposite force.
\vec F=-q_0\vec E
The displacement \Delta r from A to B is along \vec E, so the applied force is opposite to the displacement (\theta=180^\circ). The work done by the applied force is
W_{AB}=F\,\Delta r\cos180^\circ=-q_0E\,\Delta r
Potential difference is the work done per unit positive charge.
\Delta V=V_B-V_A=\dfrac{W_{AB}}{q_0}=\dfrac{-q_0E\,\Delta r}{q_0}=-E\,\Delta r
Divide by \Delta r to get the field.
E=-\dfrac{\Delta V}{\Delta r}
For a very small step the ratio becomes a derivative (calculus form).
E=-\dfrac{dV}{dr}

ResultE=-\dfrac{\Delta V}{\Delta r}

The electric field at a point equals the negative of the potential gradient there. The minus sign means that the field points in the direction in which the potential decreases (from high potential to low potential). Where the potential changes quickly with distance, the field is strong.

The unit of potential gradient is the same as the unit of electric field:

1\,\mathrm{V\,m^{-1}}=1\,\dfrac{\mathrm{J}}{\mathrm{C\,m}}=1\,\dfrac{\mathrm{N\,m}}{\mathrm{C\,m}}=1\,\mathrm{N\,C^{-1}}

Uniform field between parallel plates

Between two parallel plates with potential difference V and separation d, the potential falls steadily from the positive plate to the negative plate. The gradient is the same everywhere, so the field is uniform:

E=\dfrac{V}{d}

E: field between the plates (\mathrm{V\,m^{-1}}), V: potential difference (V), d: plate separation (m).

Potential between parallel plates

Distance x from the positive plate in cm. Between plates 5\,\mathrm{cm} apart with 100\,\mathrm{V} across them, V falls linearly. Slope =-20\,\mathrm{V\,cm^{-1}}, so E=2000\,\mathrm{V\,m^{-1}}.

Field from the slope of the V to r graph

The field at P equals minus the slope of the tangent there. The curve is steeper close to the charge, so the field is stronger there.

19.3Capacitors and capacitanceOften examined

A capacitor is a device that stores electric charge, and with it electrical energy. Capacitors are found in almost every electronic circuit. The common form is the parallel plate capacitor: two identical parallel conducting plates, a distance d apart, with an insulator between them. The insulating medium (air, paper, mica, ceramic or plastic) is called the dielectric.

Fig. 19.3: A parallel plate capacitor connected to a battery. The plates get equal and opposite charges +Q and -Q.

How a capacitor is charged

When the battery is connected, it pulls electrons off the plate joined to its positive terminal and pushes the same number onto the plate joined to its negative terminal.
One plate becomes +Q and the other -Q. The charge on one plate induces an equal and opposite charge on the inner surface of the other plate (electrostatic induction).
The flow stops when the potential difference between the plates equals the battery voltage V.
The attraction between +Q and -Q holds the charges on the inner surfaces, so the capacitor stays charged even after the battery is removed.

Capacitance

Experiments show that the charge stored is directly proportional to the potential difference across the plates:

Q\propto V\ \Rightarrow\ Q=CV\qquad\text{or}\qquad C=\dfrac{Q}{V}\qquad(19.6)

Q: charge on either plate (C), V: potential difference between the plates (V), C: capacitance (F).

Capacitance: The capability of a capacitor to store charge. It is the charge stored per unit potential difference, C=Q/V. SI unit: farad (F).
Farad: The capacitance of a capacitor that stores a charge of 1 coulomb when the potential difference between its plates is 1 volt: 1\,\mathrm{F}=1\,\mathrm{C\,V^{-1}}. Named after Michael Faraday (1791–1867).

One farad is a very large capacitance, because one coulomb is a very large charge. Practical capacitors range from picofarads to millifarads:

Unit	Symbol	Value
millifarad	\mathrm{mF}	10^{-3}\,\mathrm{F}
microfarad	\mu\mathrm{F}	10^{-6}\,\mathrm{F}
nanofarad	\mathrm{nF}	10^{-9}\,\mathrm{F}
picofarad	\mathrm{pF}	10^{-12}\,\mathrm{F}

Charge against potential difference

V in volts. Q against V is a straight line through the origin. Its slope is the capacitance: the steeper line belongs to the larger capacitor.

19.3Capacitance of a parallel plate capacitorOften examined

The capacitance of a parallel plate capacitor depends only on its construction. To find it, we write the field between the plates in two ways: once from the potential difference and once from the charge, and then equate them.

Derivation C=\dfrac{A\varepsilon_0}{d} for a parallel plate capacitor

Aim. Find the capacitance of a parallel plate capacitor with plate area A and separation d, first with vacuum (or air) and then with a dielectric between the plates.

Assumptions.

Each plate has area A; the separation d is small compared with the size of the plates, so the field between them is uniform and the edge effects can be ignored.
The plates carry charges +Q and -Q and are at potentials V_1 and V_2.
At first the space between the plates is vacuum (or air). \varepsilon_0=8.85\times10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}} is the permittivity of free space.

Plates of area A and separation d carry +Q and -Q. The field E between them is uniform.

The field between the plates is uniform, so it equals minus the potential gradient. The positive plate is at V_1 and the negative plate at V_2, a distance d apart.
E=-\dfrac{\Delta V}{\Delta r}=-\dfrac{V_2-V_1}{d}=\dfrac{V_1-V_2}{d}=\dfrac{V}{d}\quad\text{(i)}\qquad(V_1-V_2=V)
The charge Q is spread evenly over the plate area A, so the surface charge density is
\sigma=\dfrac{Q}{A}
Gauss's law gives the field between two oppositely charged plates in terms of \sigma.
E=\dfrac{\sigma}{\varepsilon_0}\quad\text{(ii)}
Substitute \sigma=Q/A in (ii).
E=\dfrac{Q}{A\varepsilon_0}\quad\text{(iii)}
Both (i) and (iii) give the same field, so equate them and rearrange for Q/V.
\dfrac{Q}{A\varepsilon_0}=\dfrac{V}{d}\ \Rightarrow\ \dfrac{Q}{V}=\dfrac{A\varepsilon_0}{d}
Q/V is the capacitance. With vacuum (or air) between the plates we call it C_{vac}.
C_{vac}=\dfrac{A\varepsilon_0}{d}\qquad(19.7\,a)
When a dielectric of relative permittivity \varepsilon_r fills the space, the capacitance increases by the factor \varepsilon_r.
C_{med}=\dfrac{A\varepsilon_0\varepsilon_r}{d}\qquad(19.7\,b)

ResultC_{vac}=\dfrac{A\varepsilon_0}{d},\qquad C_{med}=\dfrac{A\varepsilon_0\varepsilon_r}{d}

Capacitance is directly proportional to the plate area and to the relative permittivity of the dielectric, and inversely proportional to the separation of the plates. It does not depend on the charge or the voltage.

Relative permittivity

Dividing Eq. (19.7 b) by Eq. (19.7 a) cancels A, \varepsilon_0 and d:

\varepsilon_r=\dfrac{C_{med}}{C_{vac}}\qquad(19.8)

Relative permittivity (dielectric constant): The ratio of the capacitance of a capacitor with a dielectric between its plates to the capacitance of the same capacitor with vacuum between the plates, \varepsilon_r=C_{med}/C_{vac}. It has no unit. For vacuum \varepsilon_r=1 and for air \varepsilon_r\approx1.

Typical values for comparison (not from the book).
Dielectric	Typical \varepsilon_r
Vacuum	1 (exactly)
Air	1.0006
Paper	about 3.7
Glass	about 5 to 10
Mica	about 6
Water	about 80

In an electric field the molecules of a dielectric are polarised. Their charges partly cancel the field inside the dielectric.

Why a dielectric increases capacitance. The field between the plates polarises the dielectric: each molecule becomes a small dipole, with its negative end towards the positive plate. The induced charges on the faces of the dielectric oppose the field of the plates, so the field inside falls by the factor \varepsilon_r. For the same charge Q, the potential difference V=Ed falls, so C=Q/V rises by \varepsilon_r. If the battery stays connected, V stays fixed and the capacitor takes \varepsilon_r times more charge.

How to increase the capacitance

The equation C_{med}=\dfrac{A\varepsilon_0\varepsilon_r}{d} shows that the capacitance depends on three things:

**The area of the plates, A.** Larger plates give more capacitance (C\propto A).
**The distance between the plates, d.** Bringing the plates closer gives more capacitance (C\propto 1/d).
**The medium (dielectric) between the plates, \varepsilon_r.** A dielectric with a high relative permittivity gives more capacitance (C\propto\varepsilon_r).

Example 19.2: Capacitance and charge of a large capacitor

What is the capacitance of a parallel plate capacitor with metal plates, each of area 1.00\,\mathrm{m^2}, separated by 1.00\,\mathrm{mm}? What charge is stored in this capacitor if a voltage of 3.00\times10^{3}\,\mathrm{V} is applied to it?

Given

A = 1.00\,\mathrm{m^2}
d = 1.00\,\mathrm{mm}=1.00\times10^{-3}\,\mathrm{m}
\Delta V = 3.00\times10^{3}\,\mathrm{V}

To find

(a) C
(b) Q

(a) Air between the plates, so use Eq. (19.7 a).
C=\dfrac{A\varepsilon_0}{d}=\dfrac{(1.00)(8.85\times10^{-12})}{1.00\times10^{-3}}=8.85\times10^{-9}\,\mathrm{F}=8.85\,\mathrm{nF}
(b) Charge stored, from Q=CV.
Q=(8.85\times10^{-9})(3.00\times10^{3})=2.66\times10^{-5}\,\mathrm{C}=26.6\,\mu\mathrm{C}

C=8.85\,\mathrm{nF},\qquad Q=26.6\,\mu\mathrm{C}

Assignment 19.2: Capacitance of a Van de Graaff terminal

What is the capacitance of a large Van de Graaff generator's terminal, given that it stores 8.00\,\mathrm{mC} of charge at a voltage of 12.0\,\mathrm{MV}?

Use the definition of capacitance with Q=8.00\times10^{-3}\,\mathrm{C} and V=12.0\times10^{6}\,\mathrm{V}.
C=\dfrac{Q}{V}=\dfrac{8.00\times10^{-3}}{12.0\times10^{6}}=6.67\times10^{-10}\,\mathrm{F}=667\,\mathrm{pF}

C\approx6.67\times10^{-10}\,\mathrm{F}=667\,\mathrm{pF}

19.3Capacitance of an isolated spherical conductor

The SLO box asks for the capacitance of an isolated spherical conductor as well as of a parallel plate capacitor. A single conductor can store charge too: the “second plate” is the surroundings at zero potential (infinity or the Earth). The book text does not derive this result, so it is given here.

Derivation C=4\pi\varepsilon_0R for an isolated sphere

Aim. Find the capacitance of an isolated conducting sphere of radius R.

Assumptions.

The sphere is a conductor of radius R carrying charge Q, far from other objects, in vacuum or air.
The potential is zero at infinity.

Charge Q on an isolated sphere of radius R spreads over its surface.

The charge Q spreads evenly over the outer surface. Outside a uniformly charged sphere, the field and potential are the same as if all the charge were a point charge at the centre O.
So the potential of the sphere (at its surface, r=R) follows from Eq. (19.5).
V=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{R}
Capacitance is charge per unit potential.
C=\dfrac{Q}{V}=\dfrac{Q}{\dfrac{Q}{4\pi\varepsilon_0R}}
Q cancels, which confirms that C does not depend on the charge.
C=4\pi\varepsilon_0R

ResultC=4\pi\varepsilon_0R

The capacitance of an isolated sphere is proportional to its radius: a bigger sphere holds more charge at the same potential. In a medium of relative permittivity \varepsilon_r it becomes C=4\pi\varepsilon_0\varepsilon_rR.

Example: Capacitance of the Earth

Treat the Earth as an isolated conducting sphere of radius 6.4\times10^{6}\,\mathrm{m}. Find its capacitance.

Use C=4\pi\varepsilon_0R.
C=4\pi(8.85\times10^{-12})(6.4\times10^{6})
Evaluate.
C=7.1\times10^{-4}\,\mathrm{F}\approx711\,\mu\mathrm{F}

C\approx7.1\times10^{-4}\,\mathrm{F}

19.4.1Series combination of capacitorsOften examined

Several capacitors can be connected together. The whole group then behaves like one equivalent capacitor. Its capacitance depends on the individual capacitances and on how they are connected. The two simple types of connection are series and parallel.

Equivalent capacitance: The capacitance of the single capacitor that would store the same charge as the whole combination when connected to the same potential difference.
Series combination: Capacitors connected plate to plate, one after another (the right plate of one joined to the left plate of the next), so that there is only one path between the end points A and B.

Fig. 19.4: Series combination of three capacitors between A and B. Every capacitor carries the same charge Q; the battery voltage is shared.

Derivation Equivalent capacitance of capacitors in series

Aim. Show that for capacitors in series \dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\cdots

Assumptions.

Capacitors C_1, C_2 and C_3 are initially uncharged and are connected in series between A and B across a battery of voltage V.
V_1, V_2, V_3 are the potential differences across C_1, C_2, C_3.

The battery puts +Q on the left plate of C_1 and -Q on the right plate of C_3. Each inner pair of connected plates was neutral, so by electrostatic induction every capacitor gets the same charge.
Q_1=Q_2=Q_3=Q
There is only one path, so the battery voltage is shared among the capacitors (the energy given to each coulomb is used up across the three capacitors in turn).
V=V_1+V_2+V_3\qquad(19.9)
For each capacitor V=Q/C. Substitute.
V=\dfrac{Q_1}{C_1}+\dfrac{Q_2}{C_2}+\dfrac{Q_3}{C_3}=Q\left(\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\right)\quad\text{(i)}
Divide both sides by Q.
\dfrac{V}{Q}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}
The equivalent capacitor stores the same Q at the same V, so V/Q=1/C_e.
\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}
The same steps work for any number n of capacitors.
\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\cdots+\dfrac{1}{C_n}

Result\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\cdots

The reciprocal of the equivalent capacitance is the sum of the reciprocals. So the equivalent capacitance of capacitors in series is always less than the smallest individual capacitance. Joining capacitors in series is like increasing the separation of the plates.

Useful results for series

\text{Two capacitors: }\ C_e=\dfrac{C_1C_2}{C_1+C_2}\qquad\qquad n\text{ identical capacitors: }\ C_e=\dfrac{C}{n}

The voltage divides in the inverse ratio of the capacitances, because V_1=Q/C_1, V_2=Q/C_2 and Q is the same. The smallest capacitor gets the largest share of the voltage.

Example 19.3 (a): Two capacitors in series

Find the equivalent capacitance of two capacitors of capacitance 4\,\mu\mathrm{F} and 8\,\mu\mathrm{F} when connected in series.

Two capacitors in series: use the product over sum.
\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}=\dfrac{C_1+C_2}{C_1C_2}\ \Rightarrow\ C_e=\dfrac{C_1C_2}{C_1+C_2}
Substitute. Microfarads can be kept as they are, since the unit is the same throughout.
C_e=\dfrac{(4)(8)}{4+8}=\dfrac{32}{12}=2.67\,\mu\mathrm{F}

C_e=2.67\,\mu\mathrm{F}

Assignment 19.3: Three capacitors in series

Find the equivalent capacitance of three capacitors of capacitance 4\,\mu\mathrm{F}, 6\,\mu\mathrm{F} and 8\,\mu\mathrm{F}, connected in series.

Add the reciprocals using the common denominator 24.
\dfrac{1}{C_e}=\dfrac14+\dfrac16+\dfrac18=\dfrac{6+4+3}{24}=\dfrac{13}{24}\,\mu\mathrm{F^{-1}}
Invert.
C_e=\dfrac{24}{13}=1.85\,\mu\mathrm{F}

C_e\approx1.85\,\mu\mathrm{F}

19.4.2Parallel combination of capacitorsOften examined

Parallel combination: Two or more capacitors connected between the same two points, so that one plate of each is joined to point A and the other plate of each to point B.

Fig. 19.5: Parallel combination of three capacitors between A and B. Each capacitor has the full battery voltage V; the charge is shared.

Derivation Equivalent capacitance of capacitors in parallel

Aim. Show that for capacitors in parallel C_e=C_1+C_2+C_3+\cdots

Assumptions.

Capacitors C_1, C_2 and C_3 are connected in parallel between A and B across a battery of voltage V.
Q_1, Q_2, Q_3 are the charges on C_1, C_2, C_3.

Each capacitor is connected directly across A and B, so each has the same potential difference as the battery.
V_1=V_2=V_3=V
The battery supplies a total charge Q, which is shared among the capacitors.
Q=Q_1+Q_2+Q_3\quad\text{(i)}
Use Q=CV for each capacitor, with the same V.
Q=C_1V+C_2V+C_3V=V\,(C_1+C_2+C_3)
Divide both sides by V.
\dfrac{Q}{V}=C_1+C_2+C_3\quad\text{(ii)}
The equivalent capacitor stores the same Q at the same V, so Q/V=C_e.
C_e=C_1+C_2+C_3
For n capacitors in parallel:
C_e=C_1+C_2+C_3+\cdots+C_n\qquad(19.10)

ResultC_e=C_1+C_2+C_3+\cdots

The equivalent capacitance of capacitors in parallel is the sum of the individual capacitances, so it is always larger than the largest one. Joining capacitors in parallel is like increasing the area of the plates.

n\text{ identical capacitors in parallel: }\ C_e=nC

The charge divides in the direct ratio of the capacitances, because Q_1=C_1V, Q_2=C_2V and V is the same. The largest capacitor takes the most charge.

Example 19.3 (b): Two capacitors in parallel

Find the equivalent capacitance of two capacitors of capacitance 4\,\mu\mathrm{F} and 8\,\mu\mathrm{F} when connected in parallel.

In parallel the capacitances simply add.
C_e=C_1+C_2=4\,\mu\mathrm{F}+8\,\mu\mathrm{F}=12\,\mu\mathrm{F}

C_e=12\,\mu\mathrm{F}

Series and parallel combinations compared (the rules are the opposite of those for resistors).
Feature	Series	Parallel
How connected	One after another, single path	Across the same two points
Same for every capacitor	Charge Q	Potential difference V
Shared among capacitors	Voltage: V=V_1+V_2+V_3	Charge: Q=Q_1+Q_2+Q_3
Equivalent capacitance	\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots	C_e=C_1+C_2+\cdots
Size of C_e	Less than the smallest	Greater than the largest
n identical capacitors C	C/n	nC

19.4Solving capacitor networksOften examined

Many exam questions give a network that is partly series and partly parallel. Reduce it step by step:

Find a group that is clearly in parallel (both ends joined to the same two junctions) or clearly in series (one after another with no branch between them).
Replace that group by its equivalent capacitance.
Redraw the simpler circuit and repeat until one capacitor is left between A and B.
Check the answer: a series group is smaller than its smallest member, and a parallel group is larger than its largest member.

Example: A mixed network

C_1=3\,\mu\mathrm{F} is connected in series with a parallel pair C_2=2\,\mu\mathrm{F} and C_3=4\,\mu\mathrm{F}, as shown. Find the equivalent capacitance between A and B. If 12\,\mathrm{V} is applied between A and B, find the charge on each capacitor.

Given

C_1 = 3\,\mu\mathrm{F}
C_2 = 2\,\mu\mathrm{F}
C_3 = 4\,\mu\mathrm{F}
V = 12\,\mathrm{V}

To find

C_e
Q_1, Q_2, Q_3

C_2 and C_3 are in parallel: add them.
C_{23}=C_2+C_3=2+4=6\,\mu\mathrm{F}
C_1 is in series with C_{23}: product over sum.
C_e=\dfrac{C_1C_{23}}{C_1+C_{23}}=\dfrac{(3)(6)}{3+6}=2\,\mu\mathrm{F}
The total charge from the supply passes through C_1 (series), so Q_1 is the total charge.
Q_1=Q=C_eV=(2\,\mu\mathrm{F})(12\,\mathrm{V})=24\,\mu\mathrm{C}
Voltage across C_1 and across the parallel pair.
V_1=\dfrac{Q_1}{C_1}=\dfrac{24}{3}=8\,\mathrm{V},\qquad V_{23}=12-8=4\,\mathrm{V}
Charges on the parallel capacitors (same voltage, Q=CV).
Q_2=(2)(4)=8\,\mu\mathrm{C},\qquad Q_3=(4)(4)=16\,\mu\mathrm{C}
Check: the parallel charges add up to the series charge, 8+16=24\,\mu\mathrm{C}.

C_e=2\,\mu\mathrm{F};\quad Q_1=24\,\mu\mathrm{C},\ Q_2=8\,\mu\mathrm{C},\ Q_3=16\,\mu\mathrm{C}

19.5Energy stored in a capacitorOften examined

To charge a capacitor, the battery must do work to move charge from one plate to the other against the growing potential difference. This work is stored as electric potential energy in the electric field between the plates. It depends on the charge Q and the voltage V.

The capacitor starts with zero voltage and gradually comes up to its full voltage. The first charge moved sees \Delta V=0, so it needs almost no work. The last charge moved sees the full voltage V. Since Q=CV, the voltage rises in direct proportion to the charge, so the graph of V against Q is a straight line through the origin.

Derivation Energy stored from the V to Q graph

Aim. Find the energy E stored in a capacitor of capacitance C that holds charge Q at potential difference V.

Assumptions.

The capacitor starts uncharged and is charged in many small steps.
The potential difference is proportional to the charge at every moment, V=Q/C.

Potential difference against charge

Fig. 19.6: The graph of V against Q is a straight line through the origin. The shaded triangle OAB is the energy stored; the average p.d. while charging is V/2.

To move a small charge \Delta Q across a potential difference V, the work needed is \Delta W=V\Delta Q (from \Delta U=q\Delta V). On the graph this is the area of a thin strip of height V and width \Delta Q.
\Delta W=V\,\Delta Q
Adding all the strips from zero to the final charge gives the total work, which is the area under the V to Q graph. The graph is a straight line, so the area is the right-angled triangle OAB.
E=\text{area of triangle }OAB=\tfrac12\times\text{base}\times\text{height}
The base OB is the final charge Q and the height AB is the final potential difference V.
E=\tfrac12\,QV\qquad(19.11\,a)
Substitute Q=CV to write the energy in terms of C and V.
E=\tfrac12\,(CV)\times V=\tfrac12\,CV^{2}\qquad(19.11\,b)
Substitute V=Q/C to write it in terms of Q and C.
E=\tfrac12\,Q\times\dfrac{Q}{C}=\dfrac{Q^{2}}{2C}\qquad(19.11\,c)

ResultE=\tfrac12\,QV=\tfrac12\,CV^{2}=\dfrac{Q^{2}}{2C}

The energy is half of QV, because on average the charge was moved across only half the final voltage (V/2). For a given capacitor the energy grows as V^{2}: doubling the voltage stores four times the energy.

Example 19.4: Energy in a large capacitor

The capacitance of a capacitor is 50\,\mathrm{F}. It is charged to a potential of 100\,\mathrm{V}. Calculate the energy stored in it.

C and V are given, so use Eq. (19.11 b).
E=\tfrac12CV^{2}=\tfrac12(50)(100)^{2}
Evaluate.
E=250\times10^{3}\,\mathrm{J}=250\,\mathrm{kJ}

E=2.5\times10^{5}\,\mathrm{J}=250\,\mathrm{kJ}

Assignment 19.4: Energy from charge and voltage

Calculate the energy stored in a capacitor having 5\,\mathrm{C} of charge and a potential difference of 15\,\mathrm{V}.

Q and V are given, so use Eq. (19.11 a).
E=\tfrac12QV=\tfrac12(5)(15)=37.5\,\mathrm{J}

E=37.5\,\mathrm{J}

19.5Uses of capacitorsOften examined

Capacitors are very important circuit elements in electronics. They store charge and energy, release it quickly when needed, block steady (direct) current while letting changing current pass, and together with resistors they set time delays. Common uses in daily life:

Uses of capacitors in daily life.
Appliance or circuit	What the capacitor does
Radio and TV sets	Filters out unwanted frequencies. A variable capacitor in the tuning circuit selects the station.
Camera flash (flash gun)	The small battery charges the capacitor slowly; the capacitor then releases a large amount of energy in about a millisecond through the flash tube, giving a bright flash.
Refrigerator (and air conditioner)	A start or run capacitor with the compressor motor gives the extra push (starting torque) needed to start the compressor and keeps it running smoothly.
Electric (ceiling) fan	A capacitor in the fan's second winding shifts the phase of the current, which produces a rotating magnetic field. This starts the fan and sets its speed; the book notes it is used to increase the speed of the fan.
Rectification circuit (power supply, mobile charger)	After an a.c. supply is rectified, a capacitor acts as a filter: it charges at each voltage peak and discharges between peaks, reducing the ripple voltage and giving a smoother d.c.
Heart defibrillator	A capacitor charged to a high voltage delivers a large pulse of energy to the heart in a very short time.
Computers	Tiny capacitors store the bits of memory (charged or uncharged).

19.6Charging a capacitor through a resistorOften examined

A circuit with a resistor R and a capacitor C in series is called an RC circuit. When the switch is set at point 1 (Fig. 19.7 (a)), a battery of e.m.f. \varepsilon charges the capacitor through the resistor, and the charge builds up gradually to a maximum value Q_0.

Fig. 19.7 (a): Charging circuit. With the switch at 1, the battery, R and C form one loop.

What happens during charging

At t=0 the capacitor is uncharged (Q=0), so the voltage across it is zero and the whole e.m.f. is across R. The current is at its maximum: I_0=\varepsilon/R.
Electrons flow onto one plate and off the other. As charge builds up, it repels the charge that is arriving, and the capacitor voltage V_C=Q/C rises. The voltage left across R is \varepsilon-V_C, so the current falls.
The capacitor voltage rises quickly at first (large current) and then more and more slowly.
Charging stops when the potential difference across the plates equals the e.m.f. of the battery. The current is then zero and the charge is maximum.

Q_0=C\varepsilon\qquad(\text{maximum charge}=\text{capacitance}\times\text{e.m.f. of battery})

Experiments show that charging is not linear but exponential:

Q=Q_0\left(1-e^{-t/RC}\right)\qquad(19.12)

Q: charge at time t (C), Q_0: maximum charge (C), R: resistance (\Omega), C: capacitance (F), e=2.718: base of natural logarithms.

Charging: charge rises, current falls

Fig. 19.7 (b) with the current added. After one time constant the charge is 63.2\% of Q_0 and the current has fallen to 36.8\% of I_0.

Time constant

Time constant: The product \tau=RC of an RC circuit. It is the time in which the charge on a charging capacitor reaches 63.2\% of its maximum value (or, when discharging, falls to 36.8\% of its initial value). SI unit: second.

Derivation Charge after one time constant

Aim. Show that after a time t=RC a charging capacitor holds 63.2\% of its maximum charge.

Put t=RC in Eq. (19.12). The exponent becomes -1.
Q=Q_0\left(1-e^{-RC/RC}\right)=Q_0\left(1-e^{-1}\right)
Evaluate e^{-1}.
e^{-1}=\dfrac{1}{2.718}=0.368
Substitute.
Q=Q_0(1-0.368)=0.632\,Q_0\qquad\text{or}\qquad\dfrac{Q}{Q_0}=63.2\%

ResultQ=0.632\,Q_0\quad\text{at}\quad t=\tau=RC

In one time constant the capacitor gains 63.2\% of its final charge. In each further time constant it gains 63.2\% of what is still missing, so after about 5RC it is more than 99\% charged and is taken as fully charged.

The time constant really is a time: an ohm times a farad is a second.

\Omega\times\mathrm{F}=\dfrac{\mathrm{V}}{\mathrm{A}}\times\dfrac{\mathrm{C}}{\mathrm{V}}=\dfrac{\mathrm{C}}{\mathrm{A}}=\dfrac{\mathrm{A\,s}}{\mathrm{A}}=\mathrm{s}

Fraction of the maximum charge after whole numbers of time constants.
Time t	Q/Q_0 while charging	Q/Q_0 while discharging
RC	0.632	0.368
2RC	0.865	0.135
3RC	0.950	0.050
4RC	0.982	0.018
5RC	0.993	0.007

The capacitor voltage follows the same curve as the charge, and the current falls exponentially:

V_C=\varepsilon\left(1-e^{-t/RC}\right),\qquad I=I_0\,e^{-t/RC},\qquad I_0=\dfrac{\varepsilon}{R}

Derivation The charging equation (for understanding, using calculus)

Aim. Show where Q=Q_0\left(1-e^{-t/RC}\right) comes from. The book states this result without proof.

At any instant the e.m.f. is shared between the resistor and the capacitor.
\varepsilon=IR+\dfrac{Q}{C}
The current is the rate at which charge arrives, I=dQ/dt. Multiply by C and use Q_0=C\varepsilon.
RC\,\dfrac{dQ}{dt}=C\varepsilon-Q=Q_0-Q
Separate the variables and integrate from Q=0 at t=0 to Q at time t.
\int_0^{Q}\dfrac{dQ}{Q_0-Q}=\int_0^{t}\dfrac{dt}{RC}\ \Rightarrow\ -\ln\dfrac{Q_0-Q}{Q_0}=\dfrac{t}{RC}
Remove the logarithm.
Q_0-Q=Q_0\,e^{-t/RC}\ \Rightarrow\ Q=Q_0\left(1-e^{-t/RC}\right)

ResultQ=Q_0\left(1-e^{-t/RC}\right)

The rate of charging is proportional to how much charge is still missing, Q_0-Q. That is the signature of an exponential approach: fast at first, then slower and slower.

19.6Discharging a capacitor through a resistorOften examined

When the switch is moved to terminal 2 (Fig. 19.8 (a)), the battery is cut out of the loop and the charged capacitor is connected across the resistor alone. The charge -Q on the negative plate can now flow through the resistor and neutralise the charge +Q on the positive plate.

Fig. 19.8 (a): Discharging circuit. With the switch at 2, the capacitor discharges through R. The current through R is opposite to the charging current.

At the start the capacitor has its full charge Q_0 and voltage V_0, so the current is at its maximum, I_0=V_0/R.
As charge leaves the plates, the voltage V=Q/C falls, so the current I=V/R also falls.
The current reaches zero when the capacitor is fully discharged and no charge is left on it.

Assume a fully charged capacitor starts discharging at t=0. The charge left on either plate at time t is

Q=Q_0\,e^{-t/RC}\qquad(19.13)

Derivation Charge left after one time constant

Aim. Show that after a time t=RC a discharging capacitor keeps 36.8\% of its initial charge.

Put t=RC in Eq. (19.13).
Q=Q_0\,e^{-RC/RC}=Q_0\,e^{-1}
Evaluate e^{-1}.
e^{-1}=\dfrac{1}{2.718}=0.368
So
Q=0.368\,Q_0\qquad\text{or}\qquad\dfrac{Q}{Q_0}=36.8\%

ResultQ=0.368\,Q_0\quad\text{at}\quad t=\tau=RC

In every time constant the charge falls to 36.8\% of the value it had at the start of that interval. A smaller time constant means a more rapid discharge.

Discharging: exponential decay

Fig. 19.8 (b). Charge, voltage and current all decay along the same curve. The dashed tangent at t=0 meets the time axis at t=\tau.

General forms and graphs

The book writes the charging and discharging equations in a general form:

\begin{aligned}\text{Charging: }\ x&=x_0\left(1-e^{-t/RC}\right)&&(19.14)\\ \text{Discharging: }\ x&=x_0\,e^{-t/RC}&&(19.15)\end{aligned}

For a capacitor discharging through a resistor, x can be the charge, the potential difference or the current, because V=Q/C and I=V/R are both proportional to Q:

Q=Q_0e^{-t/RC},\qquad V=V_0e^{-t/RC},\qquad I=I_0e^{-t/RC}

Analysing discharge graphs

All three graphs (Q, V and I against t) have the same shape: they start at their maximum, fall quickly at first and then more and more slowly, and never quite reach zero.
In equal time intervals the value falls by the same fraction. Every time constant it is multiplied by 0.368.
To find \tau from a graph, read the time at which the value has fallen to 37\% of its starting value, or draw the tangent at t=0: it meets the time axis at t=\tau.

Example: Discharging a capacitor

A 100\,\mu\mathrm{F} capacitor is charged to 12\,\mathrm{V} and then discharged through a 50\,\mathrm{k\Omega} resistor. Find (a) the time constant, (b) the initial current, (c) the p.d. across the capacitor after 10\,\mathrm{s}, and (d) the time taken for the p.d. to halve.

Given

C = 100\,\mu\mathrm{F}=1.0\times10^{-4}\,\mathrm{F}
R = 50\,\mathrm{k\Omega}=5.0\times10^{4}\,\Omega
V_0 = 12\,\mathrm{V}

To find

\tau
I_0
V at t=10\,\mathrm{s}
t_{1/2}

(a) Time constant.
\tau=RC=(5.0\times10^{4})(1.0\times10^{-4})=5.0\,\mathrm{s}
(b) At the start the full 12\,\mathrm{V} is across R.
I_0=\dfrac{V_0}{R}=\dfrac{12}{5.0\times10^{4}}=2.4\times10^{-4}\,\mathrm{A}=0.24\,\mathrm{mA}
(c) t=10\,\mathrm{s} is two time constants.
V=V_0e^{-t/RC}=12\,e^{-10/5}=12\,e^{-2}=12(0.135)=1.6\,\mathrm{V}
(d) Put V=V_0/2: e^{-t/RC}=\tfrac12, so t=RC\ln2.
t_{1/2}=0.693\,RC=0.693(5.0)=3.5\,\mathrm{s}

\tau=5.0\,\mathrm{s},\quad I_0=0.24\,\mathrm{mA},\quad V(10\,\mathrm{s})=1.6\,\mathrm{V},\quad t_{1/2}=3.5\,\mathrm{s}

Derivation The discharging equation (for understanding, using calculus)

Aim. Show where Q=Q_0e^{-t/RC} comes from. The book states this result without proof.

With the battery removed, the capacitor voltage drives the current through R, so the voltages around the loop add to zero.
IR+\dfrac{Q}{C}=0
The current is the rate of change of the charge on the plates, I=dQ/dt (negative, because Q is decreasing).
\dfrac{dQ}{dt}=-\dfrac{Q}{RC}
Separate the variables and integrate from Q_0 at t=0 to Q at time t.
\int_{Q_0}^{Q}\dfrac{dQ}{Q}=-\int_0^{t}\dfrac{dt}{RC}\ \Rightarrow\ \ln\dfrac{Q}{Q_0}=-\dfrac{t}{RC}
Remove the logarithm.
Q=Q_0\,e^{-t/RC}

ResultQ=Q_0\,e^{-t/RC}

The charge leaves at a rate proportional to the charge that is left. This gives exponential decay, the same law as radioactive decay.

19.7Bioelectricity

Bioelectricity: The generation or action of electric currents or voltages in biological processes.

Electrical activity in living tissue is a cellular process that depends on the cell membrane. The membrane is a thin insulating layer with conducting salt solutions on both sides. Charged ions collect on its two faces, positive on one side and negative on the other, so the cell membrane acts like a capacitor that stores energy as charged ions on opposite sides.

How a potential difference arises

In most solutions every ion is accompanied by ions of the opposite charge, so the solution has no net charge. Now suppose two solutions of different concentrations are separated by a membrane that lets one kind of ion pass but blocks the other. The ions that can pass diffuse from the concentrated side to the dilute side and carry their charge with them, while their partners are left behind. Equal and opposite net charges build up on the two sides, and this imbalance gives rise to an electric potential difference across the membrane.

A living cell has two such solutions: the fluid inside the cell and the fluid outside it. The cell membrane between them is semipermeable: it lets certain ions through and blocks others. Nerve-cell and muscle-cell membranes are slightly permeable to potassium ions (\mathrm{K^+}). These diffuse out of the cell and leave a net negative charge inside.

Fig. 19.9: A channel protein forms a tunnel that lets a specific ion cross the membrane, down its concentration gradient. The membrane has positive charge outside and negative inside, like a charged capacitor.

Ion channel: A protein that spans the cell membrane and makes a water-filled (hydrophilic) tunnel through it, allowing specific ions such as \mathrm{Na^+}, \mathrm{K^+}, \mathrm{Ca^{2+}} or \mathrm{Cl^-} to diffuse across. Channels are very selective: each accepts only one type of ion (or a few closely related ones).
Resting potential: The potential difference across the membrane of a cell at rest. The book gives about 50\,\mathrm{mV}; for a nerve cell it is typically about 70\,\mathrm{mV}, with the inside negative.
Action potential: A brief electrical pulse produced when a stimulated cell lets positive ions in through open ion channels, so that the inside of the cell becomes positive for a moment.

Action potentials

At rest, the inside of the cell is negative compared with the outside (the resting potential).
Stimulus. When the cell is stimulated, ion channels open and positive charges (mainly \mathrm{Na^+} ions) enter the cell. The inside becomes more positive, even positive for a moment.
Pulse. This change triggers the channels next to it, so further electrical currents flow and an electrical pulse, the action potential, travels along the cell (for example along a nerve fibre).
Recovery. Potassium channels open and \mathrm{K^+} ions flow out, restoring the negative inside. Ion pumps then restore the original concentrations.

An action potential (simplified)

Time t in ms. Typical values for a nerve cell. The inside is negative at rest; when \mathrm{Na^+} channels open it becomes positive for about a millisecond, then \mathrm{K^+} flowing out restores it.

The bodies of many organisms use certain patterns of action potentials to start the correct movements, thoughts and behaviours. Bioelectric phenomena include fast signalling in nerves and the triggering of physical processes in muscles or glands.

19.7Electrogenic and electroreceptive animals

Several species of aquatic life, such as Electrophorus electricus (the electric eel), can naturally generate external electric shocks through internal biological mechanisms that act as batteries. Such species are called electrogenic animals. Electric eels are the best known: they stun their prey with shocks of up to 860\,\mathrm{V}.

Some types of bacteria, yeast and fish are also electrogenic.

Electrogenic animal: An animal that can produce electricity (electric fields or shocks) through internal biological mechanisms that act like batteries, for example the electric eel.
Electroreception: The ability of some animals to detect weak, naturally occurring electric (electrostatic) fields in their environment. Such animals are called electroreceptive.

Electroreceptive animals include echidnas, platypuses, bees, spiders, dolphins, sharks and rays.

Most electrogenic animals are also electroreceptive. But there are many electroreceptive animals that are not electrogenic.

Electrogenic and electroreceptive animals compared.
	Electrogenic	Electroreceptive
Ability	Produces electric fields or shocks	Detects weak electric fields
Examples	Electric eel (Electrophorus electricus), some fish, bacteria and yeast	Sharks, rays, platypus, echidna, dolphins, bees, spiders
Used for	Stunning prey, defence, communication	Finding prey, navigation, sensing surroundings

Formula sheet

Force on a charge in a field

F=q_0E

q_0: charge, E: electric field strength

Unit: \mathrm{N}

Change in potential energy

U_A-U_B=W_{BA}

W_{BA}: work done against the field in carrying the charge from B to A

Unit: \mathrm{J}

Electric potential

V=\dfrac{U}{q_0}

U: potential energy of test charge q_0 at the point

Unit: \mathrm{V}=\mathrm{J\,C^{-1}}

Potential difference

\Delta V=V_A-V_B=\dfrac{W_{BA}}{q_0}=\dfrac{\Delta U}{q_0}

W_{BA}: work done in moving q_0 from B to A

Unit: \mathrm{V}

Energy change of a charge

\Delta U=q_0\,\Delta V

q_0: charge moved through potential difference \Delta V

Units: \mathrm{J} (or eV)

Kinetic energy gained from rest

\tfrac12mv^{2}=q\,\Delta V

m: mass, v: final speed, q: charge of the particle

Unit: \mathrm{J}

Potential energy of two point charges

U=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{qQ}{r}

q, Q: charges (with signs), r: separation

Unit: \mathrm{J}

Potential of a point charge

V=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{r}=\dfrac{kQ}{r}

Q: source charge, r: distance from it, k=9\times10^{9}\,\mathrm{N\,m^2\,C^{-2}}

Unit: \mathrm{V}

Field and potential gradient

E=-\dfrac{\Delta V}{\Delta r}

\Delta V/\Delta r: potential gradient along the field

Unit: \mathrm{V\,m^{-1}}=\mathrm{N\,C^{-1}}

Uniform field between plates

E=\dfrac{V}{d}

V: p.d. between the plates, d: separation

Unit: \mathrm{V\,m^{-1}}

Capacitance

Q=CV\qquad C=\dfrac{Q}{V}

Q: charge on either plate, V: p.d. between the plates

Unit: \mathrm{F}=\mathrm{C\,V^{-1}}

Surface charge density

\sigma=\dfrac{Q}{A}

A: area of a plate

Unit: \mathrm{C\,m^{-2}}

Field between charged plates (Gauss's law)

E=\dfrac{\sigma}{\varepsilon_0}

\varepsilon_0=8.85\times10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}}

Unit: \mathrm{N\,C^{-1}}

Parallel plate capacitor in vacuum or air

C_{vac}=\dfrac{A\varepsilon_0}{d}

A: plate area, d: plate separation

Unit: \mathrm{F}

Parallel plate capacitor with a dielectric

C_{med}=\dfrac{A\varepsilon_0\varepsilon_r}{d}

\varepsilon_r: relative permittivity of the dielectric

Unit: \mathrm{F}

Relative permittivity

\varepsilon_r=\dfrac{C_{med}}{C_{vac}}

ratio of capacitances with and without the dielectric

no unit

Isolated spherical conductor

C=4\pi\varepsilon_0R

R: radius of the sphere

Unit: \mathrm{F}

Capacitors in series

\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\cdots

C_e: equivalent capacitance; same charge on each

Unit: \mathrm{F}

Two capacitors in series

C_e=\dfrac{C_1C_2}{C_1+C_2}

product over sum

Unit: \mathrm{F}

Capacitors in parallel

C_e=C_1+C_2+C_3+\cdots

same p.d. across each

Unit: \mathrm{F}

Identical capacitors

\text{series: }C_e=\dfrac{C}{n}\qquad\text{parallel: }C_e=nC

n: number of identical capacitors of capacitance C

Unit: \mathrm{F}

Energy stored in a capacitor

E=\tfrac12QV=\tfrac12CV^{2}=\dfrac{Q^{2}}{2C}

E: energy (the book's symbol), Q: charge, V: p.d., C: capacitance

Unit: \mathrm{J}

Time constant

\tau=RC

R: resistance, C: capacitance

Unit: \mathrm{s} (\Omega\times\mathrm{F}=\mathrm{s})

Maximum charge

Q_0=C\varepsilon

\varepsilon: e.m.f. of the charging battery

Unit: \mathrm{C}

Charging

Q=Q_0\left(1-e^{-t/RC}\right),\qquad I=I_0e^{-t/RC},\ \ I_0=\dfrac{\varepsilon}{R}

Q: charge at time t; the capacitor voltage follows the same curve as Q

Units: \mathrm{C}, \mathrm{A}

Discharging

Q=Q_0e^{-t/RC},\qquad x=x_0e^{-t/RC}

x can be the charge, p.d. or current of a discharging capacitor

Unit: same as x

Values after one time constant

t=RC:\quad Q=0.632\,Q_0\ \text{(charging)},\quad Q=0.368\,Q_0\ \text{(discharging)}

e^{-1}=0.368

no unit (fractions)

Key terms

Electric potential energy: The energy a charge has because of its position in an electric field; the work needed to move it against the field to that position. Unit: joule.
Electric potential: The potential energy per unit positive charge at a point; the work done in bringing a unit positive charge from infinity to the point in equilibrium. Scalar; unit: volt.
Potential difference: The work done in moving a unit positive charge from one point to another while keeping it in electrostatic equilibrium. Unit: volt.
Volt: The potential difference between two points when 1\,\mathrm{J} of work is done in moving 1\,\mathrm{C} of charge between them: 1\,\mathrm{V}=1\,\mathrm{J\,C^{-1}}.
Potential gradient: The rate of change of potential with displacement, \Delta V/\Delta r. The electric field equals the negative of the potential gradient, E=-\Delta V/\Delta r. Unit: \mathrm{V\,m^{-1}}.
Capacitor: A device that stores charge and electrical energy, usually two conducting plates separated by an insulator (dielectric).
Dielectric: The insulating medium between the plates of a capacitor, such as air, paper, mica, glass, ceramic or plastic.
Capacitance: The capability of a capacitor to store charge; the charge stored per unit potential difference, C=Q/V.
Farad: The capacitance of a capacitor that stores 1\,\mathrm{C} of charge when the p.d. between its plates is 1\,\mathrm{V}: 1\,\mathrm{F}=1\,\mathrm{C\,V^{-1}}.
Permittivity of free space: The constant \varepsilon_0=8.85\times10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}} that appears in Coulomb's law and in C=A\varepsilon_0/d.
Relative permittivity (dielectric constant): The ratio of the capacitance with a dielectric between the plates to the capacitance with vacuum, \varepsilon_r=C_{med}/C_{vac}. No unit.
Surface charge density: Charge per unit area of a surface, \sigma=Q/A. Unit: \mathrm{C\,m^{-2}}.
Equivalent capacitance: The capacitance of a single capacitor that stores the same charge as a combination at the same potential difference.
Series combination: Capacitors joined one after another, plate to plate, in a single path; each carries the same charge and the voltages add.
Parallel combination: Capacitors connected between the same two points; each has the same p.d. and the charges add.
RC circuit: A circuit containing a resistor and a capacitor in series, used to charge or discharge the capacitor gradually.
Time constant: \tau=RC: the time in which a charging capacitor reaches 63.2\% of its maximum charge, or a discharging capacitor falls to 36.8\% of its initial charge.
Bioelectricity: The generation or action of electric currents or voltages in biological processes.
Ion channel: A protein that spans the cell membrane and forms a tunnel through which specific ions can cross it.
Resting potential: The potential difference across the membrane of a cell at rest; the book gives about 50\,\mathrm{mV} (inside negative).
Action potential: A short electrical pulse produced when a stimulated cell lets positive ions in through open ion channels; it carries signals along nerves and triggers muscles.
Electrogenic animal: An animal, such as the electric eel Electrophorus electricus, that can produce electric shocks through internal biological mechanisms that act like batteries.
Electroreception: The ability of some animals, such as sharks, rays and the platypus, to detect weak naturally occurring electric fields in their environment.

Multiple choice questionsp. 102–103

Which of the following is equivalent to 1.0 volt?

Anewton/second
Bjoule/second
Cjoule/coulombAnswer
Dcoulomb/joule

Why

Potential difference is work done per unit charge, V=W/q, so 1\,\mathrm{V}=1\,\mathrm{J\,C^{-1}}. A joule per second (B) is a watt, the unit of power, and a coulomb per joule (D) is the reciprocal of a volt.

The quantity electric potential is defined as the amount of ______.

Aelectric potential energy
Bforce acting upon a charge
Cpotential energy per unit chargeAnswer
Dforce per unit charge

Why

By definition V=U/q: electric potential is the potential energy per unit (positive) charge. Force per unit charge (D) is the electric field strength, not the potential.

A negatively charged particle is placed in a uniform electric field directed from South to North. In which direction will the particle move after it is released?

AEast
BSouthAnswer
CNorth
DNorth-West

Why

The force on a charge is \vec F=q\vec E. For a negative charge the force is opposite to the field. The field points from South to North, so the force, and the motion from rest, is towards the South.

What is the charge that appears on the plates of the 10\,\mathrm{pF} capacitor if it is connected to a battery of 9\,\mathrm{V}?

A90\,\mathrm{pC}Answer
B90\,\mathrm{mC}
C90\,\mathrm{kC}
D90\,\mu\mathrm{C}

Why

Q=CV=(10\times10^{-12}\,\mathrm{F})(9\,\mathrm{V})=9.0\times10^{-11}\,\mathrm{C}=90\,\mathrm{pC}. The prefix stays pico because only the number is multiplied by 9.

What is the voltage that should be applied to the 5\,\mu\mathrm{F} capacitor to accumulate 1\,\mu\mathrm{C} charge on its plates?

A0.2\,\mathrm{V}Answer
B0.2\,\mathrm{mV}
C0.2\,\mathrm{kV}
D0.2\,\mu\mathrm{V}

Why

V=\dfrac{Q}{C}=\dfrac{1\times10^{-6}\,\mathrm{C}}{5\times10^{-6}\,\mathrm{F}}=0.2\,\mathrm{V}. The micro prefixes cancel, so the answer is in plain volts.

Capacitor is a device used to ______.

Astore electrical energyAnswer
Bvary the resistance
Cstore magnetic energy
Ddissipate energy

Why

A capacitor stores electrical energy in the electric field between its plates, E=\tfrac12CV^{2}. Magnetic energy (C) is stored by an inductor, and a resistor dissipates energy (D).

What is the total capacitance when three capacitors, C_1, C_2 and C_3 are connected in parallel?

A\dfrac{C_1}{C_2+C_3}
BC_1+C_2+C_3Answer
C\dfrac{C_2}{C_1+C_3}
D\dfrac{1}{C_1+C_2+C_3}

Why

In parallel each capacitor has the same p.d. and the charges add, so the capacitances add: C_e=C_1+C_2+C_3. None of the fractions has the unit of capacitance except B.

If three capacitors of capacitances 1\,\mathrm{F}, 2\,\mathrm{F} and 10\,\mathrm{F} are connected in parallel then their equivalent capacitance will be:

A10\,\mathrm{F}
B15\,\mathrm{F}
C13\,\mathrm{F}Answer
D20\,\mathrm{F}

Why

Parallel capacitances add: C_e=1+2+10=13\,\mathrm{F}.

When capacitors are connected in parallel, the total capacitance is always ______ the individual capacitance values.

AGreater thanAnswer
BLess than
CEqual to
DCannot be determined

Why

C_e=C_1+C_2+\cdots is a sum of positive terms, so it is greater than every individual capacitance, even the largest. (For series it is less than the smallest.)

What is the capacitance of a capacitor?

AThe ratio of charge to electric potential differenceAnswer
BThe ratio of electric potential difference to charge
CThe product of charge and electric potential difference
DThe ratio of electric field strength to charge density

Why

C=Q/V: the charge on either plate divided by the potential difference between the plates. Option B is the reciprocal, 1/C.

What is the electric potential difference between two points?

AThe work done in moving a unit charge between the two pointsAnswer
BThe force exerted on a unit charge at one of the points
CThe electric field strength between the two points
DThe distance between the two points

Why

\Delta V=W/q: the potential difference is the work done in moving a unit positive charge from one point to the other. Force on a unit charge (B) is the electric field.

A 220\,\Omega resistor is in series with a 2.2\,\mu\mathrm{F} capacitor. The time constant is

A48\,\mu\mathrm{s}
B480\,\mu\mathrm{s}Answer
C2.42\,\mu\mathrm{s}
D24\,\mu\mathrm{s}

Why

\tau=RC=(220\,\Omega)(2.2\times10^{-6}\,\mathrm{F})=4.84\times10^{-4}\,\mathrm{s}=484\,\mu\mathrm{s}\approx480\,\mu\mathrm{s}.

Note. The exact value is 484\,\mu\mathrm{s}; option B gives it rounded to two significant figures.

The formula for electrostatic potential is ______.

AElectrostatic potential = Work done × charge
BElectrostatic potential = Work done/chargeAnswer
CElectrostatic potential = Work done + charge
DElectrostatic potential = Charge/Work done

Why

Potential is work done (or potential energy) per unit charge, V=W/q.

Note. Option D is printed as “Charge Work/done”; it means charge divided by work done, the reciprocal of the correct formula.

Short questionsp. 103–104

13 marks

What is the relationship between electric potential and electric potential energy?

Solution

Electric potential at a point is the electric potential energy per unit positive charge placed at that point:

V=\dfrac{U}{q}\qquad\text{or}\qquad U=qV

So the potential energy of a charge equals its charge multiplied by the potential of the point where it is placed. Potential energy (unit: joule) depends on both the size of the charge and its position, while potential (unit: volt, \mathrm{J\,C^{-1}}) depends only on the position in the field. Both are scalars.

Example: at a point where V=12\,\mathrm{V}, a 2\,\mathrm{C} charge has U=24\,\mathrm{J} and a 0.5\,\mathrm{C} charge has U=6\,\mathrm{J}, but the potential is 12\,\mathrm{V} for both. In the same way, a change of potential changes the energy by \Delta U=q\,\Delta V.

23 marks

If you wish to store a large amount of energy in a capacitor bank, would you connect capacitors in series or parallel? Explain.

Solution

Connect them in parallel.

The energy stored is E=\tfrac12C_eV^{2}. In parallel the equivalent capacitance is the sum, C_e=C_1+C_2+\cdots, larger than any single capacitor, and every capacitor gets the full supply voltage. In series \dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots, so C_e is smaller than the smallest capacitor and much less energy is stored at the same voltage.

Example: two 10\,\mu\mathrm{F} capacitors on a 100\,\mathrm{V} supply:

\begin{aligned}\text{parallel: } E&=\tfrac12(20\times10^{-6})(100)^{2}=0.10\,\mathrm{J}\\ \text{series: } E&=\tfrac12(5\times10^{-6})(100)^{2}=0.025\,\mathrm{J}\end{aligned}

The parallel bank stores four times as much energy. (Series connection is useful only when the supply voltage is too high for one capacitor to withstand.)

33 marks

What are the units of (a) electric potential difference (b) electric potential energy (c) capacitance?

Solution

(a) Electric potential difference: volt (V). From V=W/q, 1\,\mathrm{V}=1\,\mathrm{J\,C^{-1}}; in base units \mathrm{kg\,m^{2}\,s^{-3}\,A^{-1}}.
(b) Electric potential energy: joule (J). From U=qV, 1\,\mathrm{J}=1\,\mathrm{C\,V}=1\,\mathrm{kg\,m^{2}\,s^{-2}}. For single particles the electronvolt is also used: 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}.
(c) Capacitance: farad (F). From C=Q/V, 1\,\mathrm{F}=1\,\mathrm{C\,V^{-1}}; in base units \mathrm{A^{2}\,s^{4}\,kg^{-1}\,m^{-2}}. The farad is very large, so \mu\mathrm{F}, \mathrm{nF} and \mathrm{pF} are used in practice.

Each unit follows from the defining equation: potential difference is energy per unit charge, potential energy is charge times potential, and capacitance is charge per unit potential difference.

43 marks

What is the net amount of charge on a charged capacitor?

Solution

The net charge on a charged capacitor is zero.

When a capacitor is charged, the battery does not create any charge. It only moves electrons from one plate to the other. The plate that loses electrons gets a charge +Q and the plate that gains them gets an equal charge -Q, so

Q_{net}=(+Q)+(-Q)=0

When we say that a capacitor “stores a charge Q”, we mean the magnitude of the charge on either plate; this is the Q in Q=CV. The equal and opposite charges attract each other and stay on the inner surfaces of the plates, and the energy is stored in the electric field between them.

52026 Annual I, Set D, Q.2 (iii) OR3 marks

Write some applications of capacitors in real life.

Solution

Capacitors store charge and energy, release it quickly, block steady d.c. while passing changing currents, and set time delays with resistors. Some applications:

Camera flash: the battery charges a capacitor slowly; the capacitor then dumps its energy through the flash tube in about a millisecond.
Radio and TV sets: capacitors filter unwanted frequencies, and a variable capacitor tunes the set to a station.
Power supplies and chargers (rectification): a smoothing capacitor reduces the ripple in rectified d.c.
Refrigerators and air conditioners: a start or run capacitor helps the compressor motor start and keep running.
Ceiling fans: a capacitor shifts the phase of the current in one winding to start the fan and control its speed.
Heart defibrillators: a charged capacitor gives the heart a large, short pulse of energy.
Computers and touch screens: memory cells store bits as charge on tiny capacitors; a finger changes the capacitance of a touch screen.

63 marks

Would you place the plates of a parallel-plate capacitor closer together or farther apart to increase their capacitance?

Solution

Closer together. The capacitance of a parallel plate capacitor is

C=\dfrac{\varepsilon_0\varepsilon_rA}{d}\qquad\Rightarrow\qquad C\propto\dfrac{1}{d}

so halving the separation doubles the capacitance. Physically, for the same voltage V, a smaller gap gives a stronger field E=V/d between the plates, and a stronger field needs more charge on the plates (\sigma=\varepsilon_0E). So more charge is stored per volt.

The plates must not touch, and if they are very close the dielectric may break down and spark. In practice a thin but strong dielectric sheet is placed between them.

73 marks

What is meant by electroreception?

Solution

Electroreception is the ability of some animals to detect weak, naturally occurring electric (electrostatic) fields in their environment. Such animals are called electroreceptive.

They have special sense organs that respond to tiny potential differences. Sharks and rays use jelly-filled pores (ampullae of Lorenzini) to sense the weak fields produced by the muscles of prey hidden in sand. The platypus finds food in muddy water with its eyes closed, using receptors in its bill. Other examples are echidnas, dolphins, bees (which sense the electric field of flowers) and spiders.

Most electrogenic animals, such as the electric eel, are also electroreceptive, but many electroreceptive animals cannot produce electricity themselves.

83 marks

If you were asked to design a capacitor in which small size and large capacitance were required, what would be the two most important factors in your design?

Solution

From C=\dfrac{\varepsilon_0\varepsilon_rA}{d}, the two most important factors are:

**A dielectric of very high relative permittivity \varepsilon_r** (for example a special ceramic). It multiplies the capacitance by \varepsilon_r without making the capacitor any bigger.
**A very small plate separation d**, that is, a very thin dielectric layer. Since C\propto1/d, a thin layer gives a large capacitance.

The dielectric must also have a high dielectric strength so that the thin layer does not break down at the working voltage. To fit a large plate area A into a small volume, long thin foils are rolled up or many thin layers are stacked, as in electrolytic and multilayer ceramic capacitors.

93 marks

If a capacitor is fully charged and then left for discharging, how much charge will be left on the plate of the capacitor after time equal to one time constant?

Solution

About 37% of the original charge (more exactly 36.8\%) is left.

During discharge the charge falls exponentially, Q=Q_0e^{-t/RC}. Put t=RC (one time constant):

Q=Q_0\,e^{-1}=\dfrac{Q_0}{2.718}=0.368\,Q_0

So after one time constant the capacitor has lost 63.2\% of its charge and keeps 36.8\% (the book rounds this down to 36.7\%). After each further time constant the remaining charge is again multiplied by 0.368, for example 0.135\,Q_0 after 2RC.

Comprehensive questionsp. 104

Each answer is written for a Section C question of about 7 marks: definition, explanation, derivation or diagram, then the result.

12026 Annual I, Set B, Q.4 OR7 marks

Define and explain the term electric potential. Derive an expression for electric potential at a field point due to a source charge.

Solution

Definition

Electric potential at a point in an electric field is the work done in bringing a unit positive charge from infinity to that point, keeping the charge in equilibrium. Equivalently, it is the electric potential energy per unit positive charge at that point:

V=\dfrac{W}{q_0}=\dfrac{U}{q_0}

It is a scalar quantity. Its SI unit is the volt: 1\,\mathrm{V}=1\,\mathrm{J\,C^{-1}}. The potential at a point is one volt if one joule of work is needed to bring one coulomb of positive charge from infinity to that point.

Explanation

The potential energy of a charge depends on its size: a charge twice as large has twice the energy at the same point. Dividing by the charge removes this dependence, so the potential depends only on the location in the field. For example, if the potential at a point is 12\,\mathrm{V}, a 1\,\mathrm{C} charge there has 12\,\mathrm{J} of energy, a 2\,\mathrm{C} charge 24\,\mathrm{J} and a 0.5\,\mathrm{C} charge 6\,\mathrm{J}. A positive test charge is at high potential near a positive source charge and at lower potential farther away. In practice we measure potential differences: \Delta V=\Delta U/q_0, the work per unit positive charge in moving it from one point to another.

Derivation

Derivation Potential V at distance r from a point charge Q

Aim. Find the electric potential at a point 1, a distance r from an isolated point charge +Q, by bringing a test charge q from infinity to that point.

Assumptions.

+Q is an isolated point charge fixed at the origin, in vacuum or air.
The test charge q is small and positive, and it is moved slowly so that it stays in equilibrium (no kinetic energy is gained).
The path from a far point N to point 1 is divided into small steps of length \Delta r; in each step the field E is almost constant. r_1, r_2,\ldots,r_N are the distances of points 1, 2,\ldots,N from Q.
The potential is taken as zero at infinity.

Fig. 19.2: The test charge q is moved from a far point N to point 1 against the field E of +Q, in small steps \Delta r.

The field of +Q points away from Q. To move q towards Q slowly (in equilibrium), we apply a force F equal and opposite to the electric force. So F=qE and it points towards Q, in the same direction as the displacement.
F=qE=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r^{2}}
Take the first step, from N to N−1. Its length is \Delta r=r_N-r_{N-1}. The force and the displacement are in the same direction (\theta=0^\circ), so the work done by the applied force is
\Delta W_{N\to N-1}=F\,\Delta r\cos0^\circ=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r^{2}}\,(r_N-r_{N-1})
Inside this tiny step the distance lies between r_{N-1} and r_N, which are very close. So we take r^2 as the product of the two end distances (a very good average for a small step).
r^{2}\approx r_{N-1}\,r_N
Substitute and split the fraction into two terms.
\Delta W_{N\to N-1}=\dfrac{Qq}{4\pi\varepsilon_0}\,\dfrac{r_N-r_{N-1}}{r_{N-1}\,r_N}=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_{N-1}}-\dfrac{1}{r_N}\right)\quad\text{(i)}
The same working holds for every other step.
\begin{aligned}\Delta W_{N-1\to N-2}&=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_{N-2}}-\dfrac{1}{r_{N-1}}\right)&&\text{(ii)}\\ &\ \ \vdots\\ \Delta W_{2\to1}&=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_{1}}-\dfrac{1}{r_{2}}\right)&&\text{(iii)}\end{aligned}
Add the work of all the steps. Each middle term such as 1/r_{N-1} appears once with a plus sign and once with a minus sign, so it cancels. Only the first and the last terms are left.
\Delta W_{N\to1}=\dfrac{Qq}{4\pi\varepsilon_0}\left[\left(\dfrac{1}{r_{N-1}}-\dfrac{1}{r_N}\right)+\cdots+\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\right]=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_1}-\dfrac{1}{r_N}\right)\quad\text{(iv)}
Now let the starting point N be at infinity: r_N\to\infty, so 1/r_N\to0. Write r for r_1. This work is stored as the potential energy U of q at distance r from Q.
U=\Delta W_{\infty\to1}=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r}\qquad(19.4)
Electric potential is potential energy per unit charge, so divide by q.
V=\dfrac{U}{q}=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r}\div q=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{r}\qquad(19.5)

ResultV=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{r}=\dfrac{kQ}{r},\qquad k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

The potential of a point charge falls off as 1/r, more slowly than the field (1/r^2). It has the same value at all points equidistant from Q, so the equipotential surfaces around a point charge are spheres centred on Q. V is positive around a positive charge and negative around a negative charge, and it is zero at infinity.

Result

V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}. The potential is inversely proportional to the distance from the source charge. It is the same at all points at the same distance (spherical equipotential surfaces), positive near a positive charge, negative near a negative charge, and zero at infinity.

Potential against distance

Arbitrary units. V\propto1/r: positive around +Q, negative around -Q, and approaching zero far away.

22026 Annual I, Old syllabus (Curriculum 2006), Q.2 (v) OR, old syllabus7 marks

What is meant by potential gradient? Explain.

Solution

Definition

The potential gradient is the rate of change of electric potential with displacement along a field line:

\text{potential gradient}=\dfrac{\Delta V}{\Delta r}

Its SI unit is \mathrm{V\,m^{-1}}, which is the same as \mathrm{N\,C^{-1}}, the unit of electric field.

Relation with the electric field

The electric field at a point equals the negative of the potential gradient at that point, E=-\Delta V/\Delta r. Proof:

Derivation E=-\dfrac{\Delta V}{\Delta r}

Aim. Show that the electric field at a point equals the negative of the potential gradient at that point.

Assumptions.

A and B lie on the same field line, a small distance \Delta r apart, so the field E between them is almost uniform.
A small positive test charge q_0 is moved slowly from A to B, so it stays in equilibrium.

A positive test charge is moved a small distance \Delta r along the field, from A to B. The applied force F=-q_0E is opposite to the displacement.

The field pushes +q_0 with a force q_0\vec E along the field. To move the charge slowly (in equilibrium), we apply an equal and opposite force.
\vec F=-q_0\vec E
The displacement \Delta r from A to B is along \vec E, so the applied force is opposite to the displacement (\theta=180^\circ). The work done by the applied force is
W_{AB}=F\,\Delta r\cos180^\circ=-q_0E\,\Delta r
Potential difference is the work done per unit positive charge.
\Delta V=V_B-V_A=\dfrac{W_{AB}}{q_0}=\dfrac{-q_0E\,\Delta r}{q_0}=-E\,\Delta r
Divide by \Delta r to get the field.
E=-\dfrac{\Delta V}{\Delta r}
For a very small step the ratio becomes a derivative (calculus form).
E=-\dfrac{dV}{dr}

ResultE=-\dfrac{\Delta V}{\Delta r}

Explanation

The minus sign shows that the field points in the direction in which the potential decreases. Moving along E the potential falls; moving against E it rises.
The strength of the field equals how quickly the potential changes with distance. Where equipotential surfaces are close together the gradient, and so the field, is large.
Along an equipotential surface \Delta V=0, so the field has no component along it: the field is always perpendicular to equipotential surfaces.
Units check: 1\,\mathrm{V\,m^{-1}}=1\,\mathrm{J\,C^{-1}\,m^{-1}}=1\,\mathrm{N\,m\,C^{-1}\,m^{-1}}=1\,\mathrm{N\,C^{-1}}.

Examples

Uniform field. Between parallel plates with p.d. V and separation d the potential falls steadily, so E=V/d. With 100\,\mathrm{V} across 5\,\mathrm{cm} the gradient is -100/0.05=-2000\,\mathrm{V\,m^{-1}}, so E=2000\,\mathrm{V\,m^{-1}} directed from the positive to the negative plate.

Potential between parallel plates

Distance x from the positive plate in cm. Between plates 5\,\mathrm{cm} apart with 100\,\mathrm{V} across them, V falls linearly. Slope =-20\,\mathrm{V\,cm^{-1}}, so E=2000\,\mathrm{V\,m^{-1}}.

Point charge. V=kQ/r is a curve that is steepest near the charge. Its slope is dV/dr=-kQ/r^{2}, so E=-dV/dr=kQ/r^{2}, which is exactly the Coulomb field of a point charge. This confirms the relation.

32026 Annual I, Set D, Q.4 OR7 marks

By using graph, derive an expression for the energy stored in a capacitor.

Solution

Explanation

When a capacitor is charged, the battery does work to move charge from one plate to the other against the potential difference that builds up between them. This work is stored as electric potential energy in the electric field between the plates.

At the start the capacitor is uncharged, so the first charge moves across \Delta V=0 and needs almost no work. As the charge grows, the p.d. grows in proportion, V=Q/C. The last charge is moved across the full p.d. V. So the graph of p.d. against charge is a straight line through the origin (Fig. 19.6).

Derivation

Derivation Energy stored from the V to Q graph

Aim. Find the energy E stored in a capacitor of capacitance C that holds charge Q at potential difference V.

Assumptions.

The capacitor starts uncharged and is charged in many small steps.
The potential difference is proportional to the charge at every moment, V=Q/C.

Potential difference against charge

Fig. 19.6: The graph of V against Q is a straight line through the origin. The shaded triangle OAB is the energy stored; the average p.d. while charging is V/2.

To move a small charge \Delta Q across a potential difference V, the work needed is \Delta W=V\Delta Q (from \Delta U=q\Delta V). On the graph this is the area of a thin strip of height V and width \Delta Q.
\Delta W=V\,\Delta Q
Adding all the strips from zero to the final charge gives the total work, which is the area under the V to Q graph. The graph is a straight line, so the area is the right-angled triangle OAB.
E=\text{area of triangle }OAB=\tfrac12\times\text{base}\times\text{height}
The base OB is the final charge Q and the height AB is the final potential difference V.
E=\tfrac12\,QV\qquad(19.11\,a)
Substitute Q=CV to write the energy in terms of C and V.
E=\tfrac12\,(CV)\times V=\tfrac12\,CV^{2}\qquad(19.11\,b)
Substitute V=Q/C to write it in terms of Q and C.
E=\tfrac12\,Q\times\dfrac{Q}{C}=\dfrac{Q^{2}}{2C}\qquad(19.11\,c)

ResultE=\tfrac12\,QV=\tfrac12\,CV^{2}=\dfrac{Q^{2}}{2C}

Result

The energy stored in a capacitor is E=\tfrac12QV=\tfrac12CV^{2}=\dfrac{Q^{2}}{2C}, in joules. It is half of QV because on average the charge was moved across only half the final voltage. For a given capacitor the energy is proportional to V^{2}: doubling the voltage stores four times the energy. For example, a 100\,\mu\mathrm{F} capacitor at 12\,\mathrm{V} stores \tfrac12(100\times10^{-6})(12)^{2}=7.2\times10^{-3}\,\mathrm{J}.

47 marks

Derive the expression for the equivalent capacitance for parallel combination of capacitors.

Solution

Parallel combination

When two or more capacitors are connected between the same two points, so that one plate of each is joined to point A and the other plate of each to point B, the combination is called a parallel combination. The equivalent capacitance C_e is the capacitance of the single capacitor that would store the same total charge at the same potential difference.

Fig. 19.5: Parallel combination of three capacitors.

Derivation

Derivation Equivalent capacitance of capacitors in parallel

Aim. Show that for capacitors in parallel C_e=C_1+C_2+C_3+\cdots

Assumptions.

Capacitors C_1, C_2 and C_3 are connected in parallel between A and B across a battery of voltage V.
Q_1, Q_2, Q_3 are the charges on C_1, C_2, C_3.

Each capacitor is connected directly across A and B, so each has the same potential difference as the battery.
V_1=V_2=V_3=V
The battery supplies a total charge Q, which is shared among the capacitors.
Q=Q_1+Q_2+Q_3\quad\text{(i)}
Use Q=CV for each capacitor, with the same V.
Q=C_1V+C_2V+C_3V=V\,(C_1+C_2+C_3)
Divide both sides by V.
\dfrac{Q}{V}=C_1+C_2+C_3\quad\text{(ii)}
The equivalent capacitor stores the same Q at the same V, so Q/V=C_e.
C_e=C_1+C_2+C_3
For n capacitors in parallel:
C_e=C_1+C_2+C_3+\cdots+C_n\qquad(19.10)

ResultC_e=C_1+C_2+C_3+\cdots

Result and features

C_e=C_1+C_2+C_3+\cdots: the equivalent capacitance is the sum of the individual capacitances, so it is greater than the largest capacitance.
Every capacitor has the same p.d. V; the total charge is shared in the direct ratio of the capacitances (Q_1=C_1V, Q_2=C_2V, ...).
For n identical capacitors, C_e=nC.
Parallel connection acts like increasing the plate area, and it is used to obtain a large capacitance or to store a large amount of energy.

Example

4\,\mu\mathrm{F} and 8\,\mu\mathrm{F} in parallel give C_e=12\,\mu\mathrm{F}. On a 10\,\mathrm{V} supply they take 40\,\mu\mathrm{C} and 80\,\mu\mathrm{C}; the total 120\,\mu\mathrm{C}=C_eV, as expected.

57 marks

Derive the formula for the combined capacitance of capacitors in series.

Solution

Series combination

When capacitors are connected plate to plate, the right plate of one joined to the left plate of the next, so that there is only one path between the end points, the combination is called a series combination. The equivalent capacitance C_e is the capacitance of the single capacitor that would store the same charge at the same potential difference.

Fig. 19.4: Series combination of three capacitors.

Derivation

Derivation Equivalent capacitance of capacitors in series

Aim. Show that for capacitors in series \dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\cdots

Assumptions.

Capacitors C_1, C_2 and C_3 are initially uncharged and are connected in series between A and B across a battery of voltage V.
V_1, V_2, V_3 are the potential differences across C_1, C_2, C_3.

The battery puts +Q on the left plate of C_1 and -Q on the right plate of C_3. Each inner pair of connected plates was neutral, so by electrostatic induction every capacitor gets the same charge.
Q_1=Q_2=Q_3=Q
There is only one path, so the battery voltage is shared among the capacitors (the energy given to each coulomb is used up across the three capacitors in turn).
V=V_1+V_2+V_3\qquad(19.9)
For each capacitor V=Q/C. Substitute.
V=\dfrac{Q_1}{C_1}+\dfrac{Q_2}{C_2}+\dfrac{Q_3}{C_3}=Q\left(\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\right)\quad\text{(i)}
Divide both sides by Q.
\dfrac{V}{Q}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}
The equivalent capacitor stores the same Q at the same V, so V/Q=1/C_e.
\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}
The same steps work for any number n of capacitors.
\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\cdots+\dfrac{1}{C_n}

Result\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\cdots

Result and features

\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\cdots: the equivalent capacitance is less than the smallest capacitance.
Every capacitor carries the same charge Q (because of electrostatic induction); the voltage is shared in the inverse ratio of the capacitances, so the smallest capacitor gets the largest voltage.
For two capacitors C_e=\dfrac{C_1C_2}{C_1+C_2}; for n identical capacitors C_e=C/n.
Series connection acts like increasing the plate separation. It is used when the supply voltage is higher than one capacitor can withstand.

Example

4\,\mu\mathrm{F} and 8\,\mu\mathrm{F} in series give C_e=\dfrac{4\times8}{4+8}=2.67\,\mu\mathrm{F}. On a 12\,\mathrm{V} supply, Q=C_eV=32\,\mu\mathrm{C} on each, so V_1=32/4=8\,\mathrm{V} and V_2=32/8=4\,\mathrm{V}, which add up to 12\,\mathrm{V}.

67 marks

Explain the process of charging and discharging of a capacitor.

Solution

RC circuit

A capacitor C is connected in series with a resistor R, a battery of e.m.f. \varepsilon and a two-way switch. With the switch at 1 the capacitor charges through R; with the switch at 2 the battery is removed from the loop and the capacitor discharges through R.

Charging

Fig. 19.7 (a): Charging circuit.

At t=0, Q=0 and the capacitor voltage is zero, so the whole e.m.f. is across R and the current is maximum, I_0=\varepsilon/R.
Electrons flow onto one plate and off the other. The charge already on the plates repels further charge, the capacitor voltage Q/C rises and the current falls.
Charging stops when the capacitor voltage equals \varepsilon; the charge is then maximum, Q_0=C\varepsilon, and the current is zero.

Q=Q_0\left(1-e^{-t/RC}\right)\qquad(19.12)

Charging: charge rises, current falls

Fig. 19.7 (b) with the current added. After one time constant the charge is 63.2\% of Q_0 and the current has fallen to 36.8\% of I_0.

Time constant

The product \tau=RC is the time constant (unit: second, since \Omega\times\mathrm{F}=\mathrm{s}). Putting t=RC in Eq. (19.12) gives Q=Q_0(1-e^{-1})=0.632\,Q_0: the time constant is the time in which the capacitor gains 63.2\% of its maximum charge. A smaller R or C means faster charging and discharging.

Discharging

Fig. 19.8 (a): Discharging circuit.

With the switch at 2, the charge -Q on the negative plate flows through R and neutralises +Q on the positive plate. The current flows through R in the opposite direction to the charging current.
The current is largest at the start, I_0=V_0/R, and decreases as the charge and voltage decrease.
The current becomes zero when no charge is left on the plates.

Q=Q_0\,e^{-t/RC}\qquad(19.13)

At t=RC, Q=Q_0e^{-1}=0.368\,Q_0: after one time constant 36.8\% of the charge remains. The p.d. and the current fall along the same curve, V=V_0e^{-t/RC} and I=I_0e^{-t/RC}.

Discharging: exponential decay

Fig. 19.8 (b). Charge, voltage and current all decay along the same curve. The dashed tangent at t=0 meets the time axis at t=\tau.

Summary of charging and discharging.
Quantity	Charging	Discharging
Charge Q	rises: Q_0(1-e^{-t/RC})	falls: Q_0e^{-t/RC}
Capacitor p.d. V	rises: \varepsilon(1-e^{-t/RC})	falls: V_0e^{-t/RC}
Current I	falls: I_0e^{-t/RC}	falls: I_0e^{-t/RC} (opposite direction)
At t=RC	Q=0.632\,Q_0	Q=0.368\,Q_0

77 marks

How bioelectricity is generated in animals? Explain.

Solution

Bioelectricity

Bioelectricity is the generation or action of electric currents or voltages in biological processes. Electrical activity in living tissue is a cellular process that depends on the cell membrane.

The cell membrane as a capacitor

The cell membrane is a thin insulating lipid layer with conducting salt solutions inside and outside the cell. Charged ions gather on its two faces, positive outside and negative inside, so the membrane acts like a charged capacitor that stores energy.

How the membrane potential arises

In a solution each ion is normally accompanied by an ion of opposite charge, so there is no net charge. When two solutions of different concentrations are separated by a membrane that lets one kind of ion pass but blocks the other, the ions that can pass diffuse from the concentrated side to the dilute side, carrying their charge. Equal and opposite net charges build up on the two sides, producing a potential difference. Nerve-cell and muscle-cell membranes are slightly permeable to \mathrm{K^+} ions; these diffuse out and leave the inside of the cell negative. This resting potential is about 50\,\mathrm{mV} according to the book (about 70\,\mathrm{mV} for a typical nerve cell).

Fig. 19.9: An ion channel lets a specific ion cross the membrane.

Ion channels

Cells control the flow of specific ions across the membrane with proteins that sit in the membrane and open a tunnel for certain ions. These proteins are called ion channels. They are very selective: each lets through only one kind of ion (such as \mathrm{Na^+}, \mathrm{K^+}, \mathrm{Ca^{2+}} or \mathrm{Cl^-}), which diffuses down its concentration gradient.

Action potentials

When a cell is stimulated, its ion channels open and positive ions (mainly \mathrm{Na^+}) enter the cell.
The inside of the cell becomes more positive. This triggers further electrical currents in the neighbouring part of the membrane, so an electrical pulse, the action potential, travels along the cell.
\mathrm{K^+} ions then flow out and restore the negative resting potential.
The body uses particular patterns of action potentials to start the correct movements, thoughts and behaviours; they carry fast signals in nerves and trigger muscles and glands.

Electric animals

Some aquatic animals, such as Electrophorus electricus (the electric eel), are electrogenic: special cells called electrocytes act like tiny batteries. Thousands of them are stacked in series so their small voltages add up, giving shocks of up to 860\,\mathrm{V} that stun prey. Many animals, such as sharks, rays and the platypus, are electroreceptive: they can detect the weak electric fields of other animals.

Numerical problemsp. 104–105

A heart defibrillator delivers 4\times10^{2}\,\mathrm{J} of energy by discharging a capacitor initially at 1\times10^{4}\,\mathrm{V}. What is its capacitance?

Solution

Given

E = 4\times10^{2}\,\mathrm{J} energy delivered
V = 1\times10^{4}\,\mathrm{V} initial p.d.

To find

The energy delivered is the energy that was stored in the capacitor. Energy and voltage are given, so use
E=\tfrac12CV^{2}\ \Rightarrow\ C=\dfrac{2E}{V^{2}}
Substitute.
C=\dfrac{2(4\times10^{2})}{(1\times10^{4})^{2}}=\dfrac{8\times10^{2}}{1\times10^{8}}
Evaluate.
C=8\times10^{-6}\,\mathrm{F}=8\,\mu\mathrm{F}

C=8\times10^{-6}\,\mathrm{F}=8\,\mu\mathrm{F}

Book answer. 8\,\mu\mathrm{F}

Two capacitors of capacitance C_1=6\,\mu\mathrm{F} and C_2=3\,\mu\mathrm{F} are connected in series. Calculate the equivalent capacitance.

Solution

Given

C_1 = 6\,\mu\mathrm{F}
C_2 = 3\,\mu\mathrm{F}

To find

For two capacitors in series:
\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\ \Rightarrow\ C_e=\dfrac{C_1C_2}{C_1+C_2}
Substitute (the unit \mu\mathrm{F} is the same throughout).
C_e=\dfrac{(6)(3)}{6+3}=\dfrac{18}{9}=2\,\mu\mathrm{F}

C_e=2\,\mu\mathrm{F}

Book answer. 2\,\mu\mathrm{F}

A 2200\,\mu\mathrm{F} capacitor is charged up with a 1.5\,\mathrm{V} cell. Calculate the charge and energy stored in the capacitor.

Solution

Given

C = 2200\,\mu\mathrm{F}=2.2\times10^{-3}\,\mathrm{F}
V = 1.5\,\mathrm{V}

To find

Charge stored:
Q=CV=(2.2\times10^{-3})(1.5)=3.3\times10^{-3}\,\mathrm{C}=3.3\,\mathrm{mC}
Energy stored:
E=\tfrac12CV^{2}=\tfrac12(2.2\times10^{-3})(1.5)^{2}=\tfrac12(2.2\times10^{-3})(2.25)
Evaluate. (Check with E=\tfrac12QV=\tfrac12(3.3\times10^{-3})(1.5), which gives the same value.)
E=2.475\times10^{-3}\,\mathrm{J}\approx2.48\times10^{-3}\,\mathrm{J}

Q=3.3\,\mathrm{mC},\qquad E\approx2.48\times10^{-3}\,\mathrm{J}

Book answer. 3.3\,\mathrm{mC}, 0.00248\,\mathrm{J}

Find the equivalent capacitance for the following circuit if C_1=1\,\mathrm{pF}, C_2=0.5\,\mathrm{pF}, C_3=1\,\mathrm{pF}.

Both capacitors of the parallel pair are labelled C_2 in the book.

Solution

Given

C_1 = 1\,\mathrm{pF}
C_2 = 0.5\,\mathrm{pF} each capacitor of the parallel pair
C_3 = 1\,\mathrm{pF}

To find

The two C_2 capacitors are in parallel: add them.
C_p=C_2+C_2=0.5+0.5=1\,\mathrm{pF}
Now C_1, C_p and C_3 are in series.
\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_p}+\dfrac{1}{C_3}=\dfrac11+\dfrac11+\dfrac11=3\,\mathrm{pF^{-1}}
Invert.
C_e=\dfrac13\,\mathrm{pF}\approx0.33\,\mathrm{pF}

C_e=\tfrac13\,\mathrm{pF}\approx0.33\,\mathrm{pF}

Book answer. \tfrac13\,\mathrm{pF}

Suppose electrons in a TV tube are accelerated through a potential difference of 2.0\times10^{4}\,\mathrm{V} from the heated cathode (where they are produced) toward the screen (which also serves as the anode), 25.0\,\mathrm{cm} away. (a) At what speed would the electrons impact the phosphors on the screen? Assume they accelerate from rest and ignore relativistic effects. (b) What is the magnitude of the electric field, if it is assumed constant?

Solution

The electron gains kinetic energy e\Delta V as it crosses from the cathode to the screen.

Given

\Delta V = 2.0\times10^{4}\,\mathrm{V}
d = 25.0\,\mathrm{cm}=0.250\,\mathrm{m}
e = 1.6\times10^{-19}\,\mathrm{C}
m_e = 9.11\times10^{-31}\,\mathrm{kg}
u = 0 starts from rest

To find

(a) v
(b) E

(a) The loss of electric potential energy equals the gain in kinetic energy (the electron starts from rest).
\tfrac12m_ev^{2}=e\,\Delta V\ \Rightarrow\ v=\sqrt{\dfrac{2e\,\Delta V}{m_e}}
Substitute.
v=\sqrt{\dfrac{2(1.6\times10^{-19})(2.0\times10^{4})}{9.11\times10^{-31}}}=\sqrt{7.02\times10^{15}}
Evaluate.
v=8.38\times10^{7}\,\mathrm{m\,s^{-1}}
(b) For a constant field, E is the potential difference per metre.
E=\dfrac{\Delta V}{d}=\dfrac{2.0\times10^{4}}{0.250}=8.0\times10^{4}\,\mathrm{V\,m^{-1}}

(a)\ v=8.38\times10^{7}\,\mathrm{m\,s^{-1}}\qquad(b)\ E=8.0\times10^{4}\,\mathrm{V\,m^{-1}}

Book answer. (a) 8.38\times10^{7}\,\mathrm{m\,s^{-1}} (b) 8.0\times10^{4}\,\mathrm{V\,m^{-1}}

A network of five capacitors of capacitance C is connected to a 100\,\mathrm{V} supply, as shown in the figure below. Determine the equivalent capacitance of the network.

Solution

Top branch: three capacitors C in series (n identical in series give C/n).
C_{top}=\dfrac{C}{3}
Lower branch: two capacitors C in series.
C_{low}=\dfrac{C}{2}
The two branches are both connected between A and B, so they are in parallel: add them.
C_e=\dfrac{C}{3}+\dfrac{C}{2}=\dfrac{2C+3C}{6}=\dfrac{5C}{6}

The 100\,\mathrm{V} is not needed for C_e, but it gives the charges: the supply delivers Q=C_eV=\dfrac{5C}{6}(100). Each branch has the full 100\,\mathrm{V}, so each capacitor of the top branch has 33.3\,\mathrm{V} and each capacitor of the lower branch has 50\,\mathrm{V}.

C_e=\dfrac{5C}{6}

Book answer. 5C/6

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, Q.2 (xiv) OR3 marks

Four capacitors of capacitance ‘C’ are connected as shown. Calculate their equivalent capacitance.

Solution

Each branch has two capacitors C in series.
\dfrac{1}{C_s}=\dfrac{1}{C}+\dfrac{1}{C}=\dfrac{2}{C}\ \Rightarrow\ C_s=\dfrac{C}{2}
The two branches are joined between the same points A and B, so they are in parallel: add them.
C_e=\dfrac{C}{2}+\dfrac{C}{2}
Result.
C_e=C

C_e=C

2026 Annual I, Set B, Q.4 OR1+5+1 = 7 marks

What is meant by electric potential? Derive an expression for electric potential in the field due to a point charge. Make sketch.

Solution

Electric potential (1 mark)

Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, keeping the charge in equilibrium. It is the potential energy per unit charge, V=\dfrac{W}{q_0}=\dfrac{U}{q_0}. It is a scalar, measured in volts (1\,\mathrm{V}=1\,\mathrm{J\,C^{-1}}).

Derivation (5 marks)

The field of a point charge is not uniform, so the path from infinity is divided into small steps in which the field is almost constant. The work in every step is found and the steps are added.

Derivation Potential V at distance r from a point charge Q

Aim. Find the electric potential at a point 1, a distance r from an isolated point charge +Q, by bringing a test charge q from infinity to that point.

Assumptions.

+Q is an isolated point charge fixed at the origin, in vacuum or air.
The test charge q is small and positive, and it is moved slowly so that it stays in equilibrium (no kinetic energy is gained).
The path from a far point N to point 1 is divided into small steps of length \Delta r; in each step the field E is almost constant. r_1, r_2,\ldots,r_N are the distances of points 1, 2,\ldots,N from Q.
The potential is taken as zero at infinity.

The field of +Q points away from Q. To move q towards Q slowly (in equilibrium), we apply a force F equal and opposite to the electric force. So F=qE and it points towards Q, in the same direction as the displacement.
F=qE=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r^{2}}
Take the first step, from N to N−1. Its length is \Delta r=r_N-r_{N-1}. The force and the displacement are in the same direction (\theta=0^\circ), so the work done by the applied force is
\Delta W_{N\to N-1}=F\,\Delta r\cos0^\circ=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r^{2}}\,(r_N-r_{N-1})
Inside this tiny step the distance lies between r_{N-1} and r_N, which are very close. So we take r^2 as the product of the two end distances (a very good average for a small step).
r^{2}\approx r_{N-1}\,r_N
Substitute and split the fraction into two terms.
\Delta W_{N\to N-1}=\dfrac{Qq}{4\pi\varepsilon_0}\,\dfrac{r_N-r_{N-1}}{r_{N-1}\,r_N}=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_{N-1}}-\dfrac{1}{r_N}\right)\quad\text{(i)}
The same working holds for every other step.
\begin{aligned}\Delta W_{N-1\to N-2}&=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_{N-2}}-\dfrac{1}{r_{N-1}}\right)&&\text{(ii)}\\ &\ \ \vdots\\ \Delta W_{2\to1}&=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_{1}}-\dfrac{1}{r_{2}}\right)&&\text{(iii)}\end{aligned}
Add the work of all the steps. Each middle term such as 1/r_{N-1} appears once with a plus sign and once with a minus sign, so it cancels. Only the first and the last terms are left.
\Delta W_{N\to1}=\dfrac{Qq}{4\pi\varepsilon_0}\left[\left(\dfrac{1}{r_{N-1}}-\dfrac{1}{r_N}\right)+\cdots+\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\right]=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_1}-\dfrac{1}{r_N}\right)\quad\text{(iv)}
Now let the starting point N be at infinity: r_N\to\infty, so 1/r_N\to0. Write r for r_1. This work is stored as the potential energy U of q at distance r from Q.
U=\Delta W_{\infty\to1}=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r}\qquad(19.4)
Electric potential is potential energy per unit charge, so divide by q.
V=\dfrac{U}{q}=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r}\div q=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{r}\qquad(19.5)

ResultV=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{r}=\dfrac{kQ}{r},\qquad k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

Sketch (1 mark)

The test charge q is moved from a far point N (at infinity) to point 1 against the field of +Q, in small steps \Delta r.

The potential falls off as 1/r with distance from the charge:

Potential and field around a point charge

Values are relative to those at a distance r_0. Doubling the distance halves the potential but reduces the field to one quarter.

V=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{r}

2026 Annual I, Set D, MCQ 91 mark

Electric potential due to a point charge varies with distance as:

A1/r^2
Br^2
C1/rAnswer
Dr

Why

V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}, so V\propto1/r. The tempting option A, 1/r^{2}, is how the electric field (and the force) varies, not the potential.

2026 Annual I, Set D, MCQ 101 mark

The capacitance of a parallel plate capacitor increases when:

AArea of plates increasesAnswer
BUsing battery of low potential
CDistance between plates increases
DArea of plates decreases

Why

C=\dfrac{\varepsilon_0\varepsilon_rA}{d}, so C is directly proportional to the plate area. Increasing d (C) or decreasing A (D) reduces C, and the battery voltage (B) changes the charge stored, not the capacitance.

2026 Annual I, Set D, Q.2 (iii) OR1x3 = 3 marks

Briefly explain the use of capacitors in any three household appliances.

Solution

One mark for each appliance with the role of its capacitor:

Ceiling fan: a capacitor connected with the fan's auxiliary (starting) winding shifts the phase of its current. This produces a rotating magnetic field, which starts the motor and keeps it turning; a capacitor of suitable value increases the speed of the fan.
Refrigerator: a start (or run) capacitor with the compressor motor gives the extra starting torque the compressor needs and keeps it running smoothly.
Camera flash (flash gun): the small battery charges a capacitor slowly; the capacitor then releases its stored energy in about a millisecond through the flash tube, producing a bright flash.
Also acceptable: in TV and radio sets capacitors filter unwanted frequencies and tune to a station; in a mobile charger or any d.c. power supply a capacitor smooths the rectified voltage by reducing the ripple.

2026 Annual I, Set D, Q.2 (xiv) OR3 marks

Deduce the equivalent capacitance between A and B for the given circuit.

Solution

Given

C_1 = 2\,\mu\mathrm{F}
C_2 = 4\,\mu\mathrm{F}
C_3 = 4\,\mu\mathrm{F}
C_4 = 2\,\mu\mathrm{F}

To find

C_e between A and B

C_2 and C_3 are connected between the same two junctions, so they are in parallel.
C_{23}=C_2+C_3=4+4=8\,\mu\mathrm{F}
C_1, C_{23} and C_4 are now in series.
\dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_{23}}+\dfrac{1}{C_4}=\dfrac12+\dfrac18+\dfrac12=\dfrac{4+1+4}{8}=\dfrac98\,\mu\mathrm{F^{-1}}
Invert.
C_e=\dfrac89\,\mu\mathrm{F}\approx0.89\,\mu\mathrm{F}

C_e=\tfrac89\,\mu\mathrm{F}\approx0.89\,\mu\mathrm{F}

2026 Annual I, Set D, Q.4 OR(3+1)+3 = 7 marks

a. Determine the electrical potential energy stored in a capacitor by using graph. b. Calculate the energy stored, in a capacitor having 5\,\mathrm{C} of charge and potential different of 15\,\mathrm{V}.

Solution

(a)

Explanation

While a capacitor is being charged, the battery does work in moving charge from one plate to the other against the growing potential difference. This work is stored as electric potential energy in the field between the plates. The p.d. rises in proportion to the charge, V=Q/C, from zero for the first charge to V for the last, so the graph of V against Q is a straight line through the origin.

Graph (1 mark)

Potential difference against charge

Fig. 19.6: The graph of V against Q is a straight line through the origin. The shaded triangle OAB is the energy stored; the average p.d. while charging is V/2.

Derivation (3 marks)

Derivation Energy stored from the V to Q graph

Aim. Find the energy E stored in a capacitor of capacitance C that holds charge Q at potential difference V.

Assumptions.

The capacitor starts uncharged and is charged in many small steps.
The potential difference is proportional to the charge at every moment, V=Q/C.

To move a small charge \Delta Q across a potential difference V, the work needed is \Delta W=V\Delta Q (from \Delta U=q\Delta V). On the graph this is the area of a thin strip of height V and width \Delta Q.
\Delta W=V\,\Delta Q
Adding all the strips from zero to the final charge gives the total work, which is the area under the V to Q graph. The graph is a straight line, so the area is the right-angled triangle OAB.
E=\text{area of triangle }OAB=\tfrac12\times\text{base}\times\text{height}
The base OB is the final charge Q and the height AB is the final potential difference V.
E=\tfrac12\,QV\qquad(19.11\,a)
Substitute Q=CV to write the energy in terms of C and V.
E=\tfrac12\,(CV)\times V=\tfrac12\,CV^{2}\qquad(19.11\,b)
Substitute V=Q/C to write it in terms of Q and C.
E=\tfrac12\,Q\times\dfrac{Q}{C}=\dfrac{Q^{2}}{2C}\qquad(19.11\,c)

ResultE=\tfrac12\,QV=\tfrac12\,CV^{2}=\dfrac{Q^{2}}{2C}

E=\tfrac12QV=\tfrac12CV^{2}=\dfrac{Q^{2}}{2C}

(b)

Given

Q = 5\,\mathrm{C}
V = 15\,\mathrm{V}

To find

Charge and p.d. are given, so use
E=\tfrac12QV
Substitute.
E=\tfrac12(5\,\mathrm{C})(15\,\mathrm{V})
Evaluate.
E=37.5\,\mathrm{J}

E=37.5\,\mathrm{J}

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 1, old syllabus1 mark

If two charges experience a coulomb force of 10\,\mathrm{N} when medium is air, what will be the coulomb force if air is replaced by a medium of relative permittivity 2?

A3\,\mathrm{N}
B5\,\mathrm{N}Answer
C10\,\mathrm{N}
D0.5\,\mathrm{N}

Why

In a medium the Coulomb force is divided by the relative permittivity: F_{med}=\dfrac{1}{4\pi\varepsilon_0\varepsilon_r}\dfrac{q_1q_2}{r^{2}}=\dfrac{F_{air}}{\varepsilon_r}=\dfrac{10\,\mathrm{N}}{2}=5\,\mathrm{N}. The medium's polarisation weakens the field between the charges, the same effect that raises the capacitance of a capacitor by \varepsilon_r.

Note. This is Coulomb's law in a medium, which the new book treats with electrostatics; it is answered here because it uses relative permittivity.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 5, old syllabus1 mark

Ohm × Farad =

ASecondAnswer
BWeber
CHenry
DTesla

Why

RC is the time constant, so its unit is the second: \Omega\times\mathrm{F}=\dfrac{\mathrm{V}}{\mathrm{A}}\times\dfrac{\mathrm{C}}{\mathrm{V}}=\dfrac{\mathrm{C}}{\mathrm{A}}=\mathrm{s}. Weber, henry and tesla are units of magnetic flux, inductance and magnetic field.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ii), old syllabus3 marks

Find electric field at a distance of 30\,\mathrm{cm} from a 3\,\mu\mathrm{C} point charge.

Solution

Given

q = 3\,\mu\mathrm{C}=3\times10^{-6}\,\mathrm{C}
r = 30\,\mathrm{cm}=0.30\,\mathrm{m}
k = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

To find

Field of a point charge (Coulomb's law divided by the test charge):
E=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{q}{r^{2}}=\dfrac{kq}{r^{2}}
Substitute.
E=\dfrac{(9\times10^{9})(3\times10^{-6})}{(0.30)^{2}}=\dfrac{2.7\times10^{4}}{0.09}
Evaluate. The field points radially away from the charge (taking it as positive).
E=3.0\times10^{5}\,\mathrm{N\,C^{-1}}

E=3.0\times10^{5}\,\mathrm{N\,C^{-1}}

Note. The field of a point charge belongs to electrostatics; the map assigned it to this unit.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (v) OR, old syllabus3 marks

Prove that E=-\dfrac{\Delta V}{\Delta r}.

Solution

Consider two points A and B on a field line, a small distance \Delta r apart, so the field E between them is uniform. A positive test charge q_0 is moved slowly from A to B, along the field.

The applied force F=-q_0E is opposite to the displacement \Delta r.

To keep q_0 in equilibrium, the applied force must balance the electric force.
\vec F=-q_0\vec E
The force is opposite to the displacement (\theta=180^\circ), so the work done from A to B is
W_{AB}=F\,\Delta r\cos180^\circ=-q_0E\,\Delta r
Potential difference is work done per unit charge.
\Delta V=V_B-V_A=\dfrac{W_{AB}}{q_0}=-E\,\Delta r
Rearrange.
E=-\dfrac{\Delta V}{\Delta r}

\Delta V/\Delta r is the potential gradient. The negative sign shows that the field points in the direction of decreasing potential.

E=-\dfrac{\Delta V}{\Delta r}

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (x) OR, old syllabus3 marks

How can one increase the capacitance of a capacitor?

Solution

The capacitance of a parallel plate capacitor is

C=\dfrac{\varepsilon_0\varepsilon_rA}{d}

So the capacitance can be increased in three ways:

**Increase the area A of the plates** (C\propto A): larger plates hold more charge at the same voltage. Long foils are rolled up to get a large area in a small space.
**Decrease the separation d between the plates** (C\propto1/d): halving the gap doubles the capacitance.
**Insert a dielectric of high relative permittivity \varepsilon_r** between the plates (C\propto\varepsilon_r): for example, mica (\varepsilon_r\approx6) makes the capacitance about six times its value in air. The polarised dielectric weakens the field between the plates, so more charge is stored per volt.

The effective capacitance of a circuit can also be increased by connecting more capacitors in parallel, since C_e=C_1+C_2+\cdots.

Exam focus

Define electric potential and derive V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r} for a point charge, with a sketch. (long)
Came as a 7-mark Section C question in 2026 (Set B Q.4 OR) and is comprehensive question 1 of the book.
Go to answer
Derive the energy stored in a capacitor, E=\tfrac12QV=\tfrac12CV^{2}=\dfrac{Q^{2}}{2C}, from the V to Q graph. (long)
Asked in 2026 (Set D Q.4 OR) with a numerical part; comprehensive question 3 of the book.
Go to answer
Derive the equivalent capacitance of capacitors in series and in parallel. (long)
Two book comprehensive questions (4 and 5); a standard 3 to 5 mark derivation.
Go to answer
Reduce a capacitor network to a single equivalent capacitance. (numerical)
Both 2026 new-syllabus sets had a network in Q.2 (xiv) OR; book numericals 4 and 6.
Go to answer
Define potential gradient and prove E=-\Delta V/\Delta r. (short)
Old-syllabus 2026 Q.2 (v) OR and book comprehensive question 2; the minus sign is often asked.
Go to answer
Explain charging and discharging of a capacitor; define the time constant and show Q=0.632\,Q_0 and 0.368\,Q_0 at t=RC. (long)
Book comprehensive question 6 and short question 9; MCQs ask for \tau=RC values and the unit \Omega\times\mathrm{F}=\mathrm{s} (2026 old paper).
Go to answer
Derive C=\dfrac{A\varepsilon_0\varepsilon_r}{d} and state how to increase the capacitance. (short)
2026 Set D MCQ 10 and old-syllabus Q.2 (x) OR; book short questions 6 and 8.
Go to answer
Explain the use of capacitors in three household appliances. (short)
Asked in 2026 (Set D Q.2 (iii) OR, 1 mark each); book short question 5.
Go to answer
Energy and charge numericals: Q=CV, E=\tfrac12CV^{2}, E=\tfrac12QV. (numerical)
2026 Set D Q.4 OR (b) and book numericals 1 and 3 use these directly.
Go to answer
Explain how bioelectricity is generated (ion channels, resting and action potentials) and what electroreception is. (long)
New NCP topic: book comprehensive question 7 and short question 7.
Go to answer