Electrostatics

Class 11 Physics, unit 10, book pages 184–196

Practise 72 MCQs Take the unit quiz Review 23 cards

Electrostatics is the study of electric charges at rest and of the forces between them. This unit gives Coulomb's law for the force between two point charges, shows how several forces are added by superposition, and introduces the electric field, its field lines and the potential gradient that links field to potential. It ends with the motion of a charged particle in a uniform field, ferrofluids and the Faraday cage.

Learning outcomes (9)

State that an electric field is an example of a field of force.
Define and calculate electric field strength [use $F = q E$ for the force on a charge in an electric field; use $E = Δ V /Δ r$ to calculate the field strength of the uniform field between charged parallel plates].
Represent an electric field by means of field lines.
Describe the effect of a uniform electric field on the motion of charged particles.
State that, for a point outside a spherical conductor, the charge on the sphere may be considered to be a point charge at its centre.
Explain how a Faraday cage works [by inducing internal electric fields that work to shield the inside from the influence of external electric fields].
State and apply Coulomb's law [ $F = k \frac{q _{1} q _{2}}{r ^{2}}$ for the force between two point charges in free space, where $k = \frac{1}{4 π ε _{0}}$ ].
Use $E = k Q / r^{2}$ for the electric field strength due to a point charge in free space.
Describe how ferrofluids work [they make use of temporary soft magnetic materials suspended in liquids to develop fluids that react to the poles of a magnet and have many applications in fields such as electronics].

Electric charge and electrostatics

All matter is made of atoms, and atoms are made of electrons, protons and neutrons. Each proton carries one unit of positive charge and each electron carries one unit of negative charge. A neutron has no charge. Protons and neutrons are packed tightly in the nucleus, while the electrons can be pictured as small charged clouds around the nucleus.

The three particles of an atom.
Particle	Charge	Where it is in the atom
Proton	one unit of positive charge	in the nucleus
Neutron	no charge	in the nucleus
Electron	one unit of negative charge	in a cloud around the nucleus

An atom normally has as many electrons as protons, so it is electrically neutral. If an electron is removed from an atom, the atom is left with a net positive charge. If an extra electron is added, it acquires a net negative charge.

The cover picture of the unit shows this at work. When a comb is rubbed in the hair it becomes charged, and when it is brought near small pieces of paper the pieces are attracted to it because of induction.

Electric charge: A property of matter that causes a force when it is near another charge. Charge comes in two forms: positive and negative.
Electrostatics: The study of electric charges at rest and the forces between them.

In this unit we study the electric forces between charges (Coulomb's law), the electric field a charge sets up and how it is drawn with field lines, the potential gradient, the motion of a charged particle in a field, ferrofluids and the Faraday cage.

10.1Coulomb's lawOften examined

The French physicist Charles Coulomb (1736–1806) performed experiments on the electrostatic force and proposed a formula to calculate it. The law that gives the force between two charges is named after him.

Coulomb's law: The force of attraction or repulsion between two charged bodies is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Take two point charges $q_{1}$ and $q_{2}$ separated by a distance $r$ , as in Fig. 10.1. According to Coulomb's law the magnitude of the force $F$ between them obeys two proportionalities.

Figure 10.1: Two similar charges

q_{1}

and

q_{2}

separated by a distance

r

F \propto q_{1} q_{2}

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

Combining the two relations gives one proportionality, and putting in a constant of proportionality $k$ turns it into an equation.

F \propto \frac{q _{1} q _{2}}{r ^{2}}

F = k \frac{q _{1} q _{2}}{r ^{2}}

(10.1)

Here $k$ is the constant of proportionality, known as Coulomb's constant. It is usually written in terms of the permittivity of free space $ε_{0}$ .

k = \frac{1}{4 π ε _{0}}, ε_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}

Putting this in Eq. (10.1) gives the mathematical form of Coulomb's law. It represents the force between the two charges when there is a vacuum between them.

F = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}}

(10.2)

Coulomb's constant: The constant of proportionality $k$ in Coulomb's law. Its value depends on the system of units and on the medium between the charges. In the SI system, for a vacuum, $k = 9 \times 1 0^{9} N m^{2} C^{- 2}$ .

How the Coulomb force falls with distance

r / r_{0}

F / F_{0}

(1, 1)

(2, \frac{1}{4})

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

For fixed charges,

F \propto 1/ r^{2}

F_{0}

is the force at a distance

r_{0}

. At twice the distance the force is one quarter.

Worked example: Coulomb force and how it scales

Two point charges

q_{1} = + 3.0 μ C

and

q_{2} = - 5.0 μ C

are

0.30 m

apart in vacuum. (a) Find the force between them and say whether it attracts or repels. (b) What is the force if the distance is halved? (c) What is the force if, instead, both charges are doubled and the distance is doubled?

Given

$q_{1} = 3.0 μ C = 3.0 \times 1 0^{- 6} C$
$q_{2} = - 5.0 μ C = - 5.0 \times 1 0^{- 6} C$
$r = 0.30 m$
$k = 9 \times 1 0^{9} N m^{2} C^{- 2}$

To find

(a) $F$ and its kind
(b) $F$ at $r /2$
(c) $F$ for $2 q_{1}$ , $2 q_{2}$ , $2 r$

(a) Convert microcoulombs to coulombs, then use the magnitudes of the charges in Coulomb's law.
$F = k \frac{∣ q _{1} ∣∣ q _{2} ∣}{r ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 3.0 \times 1 0 ^{- 6} ) ( 5.0 \times 1 0 ^{- 6} )}{( 0.30 ) ^{2}} = \frac{0.135}{0.090} = 1.5 N$
The charges have opposite signs, so the force is attractive.
(b) The force goes as $1/ r^{2}$ . Halving $r$ makes $r^{2}$ four times smaller, so $F$ becomes four times larger.
$F' = 4 F = 4 \times 1.5 = 6.0 N (check: \frac{0.135}{( 0.15 ) ^{2}} = 6.0)$
(c) The force goes as $q_{1} q_{2}$ and as $1/ r^{2}$ . Doubling both charges gives a factor $2 \times 2 = 4$ , and doubling the distance gives a factor $1/ 2^{2} = 1/4$ .
$F'' = F \times \frac{2 \times 2}{2 ^{2}} = F = 1.5 N$

F = 1.5 N (attractive), F' = 6.0 N, F'' = 1.5 N

When does Coulomb's law apply?

The value of $k = 9 \times 1 0^{9} N m^{2} C^{- 2}$ is applicable when there is a vacuum between the charges. Coulomb's law is applicable to point charges and to a uniform spherical charge distribution.

Point charge: An imaginary charge located at a single point in space. An electron can be considered a point charge.

A charged sphere acts as a point charge

A charged sphere also behaves as a point charge for a point outside the spherical conductor. The reason is that the electric field at any point outside a spherical charge distribution is the same as the field that would be produced by a point charge located at its centre. So for such a point the sphere's charge may be taken as a point charge at its centre, and $r$ is measured from the centre.

This is true only if the spherical conductor is isolated. If another charged object is brought near the spherical conductor, it induces surface charges on the sphere that are not spherically symmetric, and then we can no longer treat the sphere as a point charge.

10.1Vector form of Coulomb's law

Force is a vector quantity, so when we use Coulomb's law we must treat it as one. The force exerted on charge $q_{1}$ by a second charge $q_{2}$ is written in vector form as

F_{12} = k \frac{q _{1} q _{2}}{r ^{2}} \overset{r}{^}_{12}

(10.3)

where $\overset{r}{^}_{12}$ is a unit vector in the direction of $F_{12}$ . In the same way, the force exerted on $q_{2}$ by $q_{1}$ is

F_{21} = k \frac{q _{1} q _{2}}{r ^{2}} \overset{r}{^}_{21}

(10.4)

where $\overset{r}{^}_{21}$ is a unit vector in the direction of $F_{21}$ . The two unit vectors point in opposite directions, $\overset{r}{^}_{12} = - \overset{r}{^}_{21}$ . Comparing Eqs. (10.3) and (10.4) gives

F_{21} = - F_{12}

(10.5)

Equation (10.5) shows that the two forces are equal in magnitude but opposite in direction, as illustrated in Fig. 10.2.

Figure 10.2: (a) Charges with the same sign repel each other. (b) Charges with opposite signs attract each other.

The sign of the product of the charges decides the kind of force.
Charges	Product $q_{1} q_{2}$	Force
Like (both positive or both negative)	positive	repulsion
Unlike (one positive, one negative)	negative	attraction

Like other forces, electric forces obey Newton's third law: the forces $F_{12}$ and $F_{21}$ are equal in magnitude but opposite in direction, however different the sizes of $q_{1}$ and $q_{2}$ may be.

Coulomb's force and gravitational force

Compare the Coulomb force with the gravitational force. Both act at a distance without direct contact; such a force is called a field force. Both are inversely proportional to the square of the distance, and both act along the line connecting the two bodies. The mathematical form is the same: the masses $m_{1}$ and $m_{2}$ of Newton's law are replaced by the charges $q_{1}$ and $q_{2}$ , and Newton's constant $G$ is replaced by Coulomb's constant $k$ .

Gravitational and electric forces compared.
	Gravitational force	Coulomb (electric) force
Acts at a distance without contact	yes (field force)	yes (field force)
Dependence on distance	inversely proportional to $r^{2}$	inversely proportional to $r^{2}$
Direction	along the line joining the bodies	along the line joining the charges
Law	$F = G \frac{m _{1} m _{2}}{r ^{2}}$	$F = k \frac{q _{1} q _{2}}{r ^{2}}$
Attractive or repulsive	always attractive	attractive or repulsive
Between charged elementary particles	very weak	far stronger

10.1Superposition principle of forceOften examined

Superposition principle of force: When a number of separate charges act on the charge of interest, each exerts an electric force on it. These forces can all be computed separately, one at a time, and then added as vectors.

In a region with several charges, each charge experiences a net force due to all the other charges placed at suitable distances. Each pair is handled with Coulomb's law as if the others were absent, and the results are added as vectors.

Figure 10.2 (c): Four charges q_1, q_2, q_3 and q_4 placed near each other. The forces on q_1 are drawn.

For example, if four charges q_1, q_2, q_3 and q_4 are placed near each other as in Fig. 10.2 (c), the resultant force exerted by q_2, q_3 and q_4 on q_1 is

\vec F=\vec F_{12}+\vec F_{13}+\vec F_{14}

The following example illustrates this procedure in two dimensions.

Example 10.1: Resultant force on q_3

Three point charges are located at the corners of a right triangle, as shown in Fig. 10.3, where q_1=q_3=5.00\,\mu\mathrm{C}, q_2=-2.00\,\mu\mathrm{C} and a=0.100\,\mathrm{m}. Find the resultant force exerted on q_3.

Figure 10.3: Three point charges located at the corners of a right triangle. F_{32} and F_{31} are the forces on q_3.

Given

q_1=q_3 = 5.00\,\mu\mathrm{C}=5.00\times10^{-6}\,\mathrm{C} two positive charges
q_2 = -2.00\,\mu\mathrm{C}=-2.00\times10^{-6}\,\mathrm{C} negative charge
a = 0.100\,\mathrm{m} side of the triangle

To find

\vec F_3 (resultant force on q_3)

Place the charges as in Fig. 10.3: q_1 at the origin, q_2 at (0,a) and q_3 at (a,a). The distance from q_2 to q_3 is a and the distance from q_1 to q_3 is the hypotenuse, \sqrt2\,a. Because q_3 is near two other charges it feels two forces, so we use the superposition principle.
Directions. q_2 and q_3 have opposite signs, so \vec F_{32} is attractive: it points towards q_2, to the left (the negative x-direction). q_1 and q_3 are both positive, so \vec F_{31} is repulsive: it points away from q_1 along the hypotenuse, at 45^\circ to the x-axis.
Magnitude of F_{32} (distance a).
F_{32}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_2q_3}{a^{2}}=9\times10^{9}\,\dfrac{(2\times10^{-6})(5\times10^{-6})}{(0.100)^{2}}=8.99\,\mathrm{N}
Magnitude of F_{31} (distance \sqrt2\,a, so r^2=2a^2).
F_{31}=9\times10^{9}\,\dfrac{(5\times10^{-6})(5\times10^{-6})}{(\sqrt2\,a)^{2}}=9\times10^{9}\,\dfrac{25\times10^{-12}}{2(0.100)^{2}}=11.2\,\mathrm{N}
Resolve \vec F_{31} into components, using 45^\circ.
F_{31x}=F_{31}\cos45^\circ=7.94\,\mathrm{N},\qquad F_{31y}=F_{31}\sin45^\circ=7.94\,\mathrm{N}
\vec F_{32} lies along the x-axis only: F_{32x}=-8.99\,\mathrm{N} and F_{32y}=0. Add the components.
F_{3x}=F_{31x}+F_{32x}=7.94\,\mathrm{N}+(-8.99\,\mathrm{N})=-1.05\,\mathrm{N}
The same for the y-components.
F_{3y}=F_{31y}+F_{32y}=7.94\,\mathrm{N}+0=7.94\,\mathrm{N}
In vector form the resultant force on q_3 is
\vec F_3=(-1.04\,\hat i+7.94\,\hat j)\,\mathrm{N}
Its magnitude and direction (an extra check, not asked in the book).
\begin{aligned}F_3&=\sqrt{(1.04)^2+(7.94)^2}\approx8.0\,\mathrm{N}\\ \theta&=180^\circ-\tan^{-1}\dfrac{7.94}{1.04}\approx97.5^\circ\ \text{from the }+x\text{-axis}\end{aligned}

\vec F_3=(-1.04\,\hat i+7.94\,\hat j)\,\mathrm{N}

Assignment 10.1: Iron nucleus and its innermost electron

What is the magnitude of the force of attraction between an iron nucleus bearing charge q=26e and its innermost electron, if the distance between them is 1\times10^{-12}\,\mathrm{m}?

Given

q_1 = 26e=26\times1.6\times10^{-19}\,\mathrm{C} iron nucleus
q_2 = e=1.6\times10^{-19}\,\mathrm{C} innermost electron (magnitude)
r = 1\times10^{-12}\,\mathrm{m} distance between them

To find

Both are treated as point charges, so use Coulomb's law with the magnitudes of the charges.
F=k\,\dfrac{q_1q_2}{r^{2}}=9\times10^{9}\,\dfrac{(26e)(e)}{r^{2}}
Put in e=1.6\times10^{-19}\,\mathrm{C} and r=1\times10^{-12}\,\mathrm{m}.
F=9\times10^{9}\times\dfrac{26\,(1.6\times10^{-19})^{2}}{(1\times10^{-12})^{2}}=9\times10^{9}\times\dfrac{6.66\times10^{-37}}{10^{-24}}
Evaluate. The charges are opposite, so the force is attractive.
F\approx6.0\times10^{-3}\,\mathrm{N}

F\approx6.0\times10^{-3}\,\mathrm{N}\ \text{(attractive)}

Worked example: resultant force from three equal charges at the corners of a square

Four equal charges of +2.0\,\mu\mathrm{C} are placed at the corners of a square of side a=0.10\,\mathrm{m}. Find the magnitude and direction of the resultant force on the charge at corner D due to the other three.

Given

q = 2.0\times10^{-6}\,\mathrm{C}
a = 0.10\,\mathrm{m}
k = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

To find

magnitude and direction of the resultant force on D

Forces on the charge at D from the charges at A, C and B. F_A and F_C are perpendicular; F_B lies along the diagonal. All three add to the resultant F (accent), which also points along the diagonal. Arrow lengths are drawn to scale.

The charges at A and C are each a distance a from D. Each repels D with the same force.
F_A=F_C=k\dfrac{q^{2}}{a^{2}}=\dfrac{(9\times10^{9})(2.0\times10^{-6})^{2}}{(0.10)^{2}}=\dfrac{0.036}{0.010}=3.6\,\mathrm{N}
The charge at B is on the diagonal, at a distance \sqrt{2}\,a, so r^{2}=2a^{2}=0.020\,\mathrm{m^{2}}.
F_B=k\dfrac{q^{2}}{2a^{2}}=\dfrac{0.036}{0.020}=1.8\,\mathrm{N}
F_A and F_C are at right angles (along the two sides, away from A and C), so combine them with Pythagoras' theorem. Their resultant points along the diagonal, away from B.
F_{AC}=\sqrt{F_A^{2}+F_C^{2}}=\sqrt{3.6^{2}+3.6^{2}}=3.6\sqrt{2}=5.09\,\mathrm{N}
F_B also points along the diagonal away from B, in the same direction as F_{AC}, so the two simply add.
F=F_{AC}+F_B=5.09+1.8=6.9\,\mathrm{N}
Direction: along the diagonal BD, away from B, at 45^\circ to each side of the square.

F\approx6.9\,\mathrm{N}\text{ along the diagonal, away from the opposite corner}

10.2Electric field intensityOften examined

An electric field, like other fields (for example gravitational or magnetic), is a vector field that surrounds a charged object. Electric fields are found around electric charges and help us determine the direction and magnitude of the force the charge exerts on a nearby charged particle. The concept of a field was developed by Michael Faraday (1791–1867) in the context of electric forces.

Electric field: The region around a charge in which a test charge can feel an electric force.

An electric field is said to exist in the region of space around a charged object, the source charge. When another charged object, called the test charge, enters this field, an electric force acts on it. As an example, Fig. 10.4 shows a small positive test charge q placed near a second object carrying a much greater positive charge Q.

Figure 10.4: A small positive test charge q placed near an object carrying a much larger positive charge Q experiences an electric field \vec E established by the source charge Q. The test charge is assumed so small that the field of the source charge is unaffected by its presence.

If the test charge q is moved away from the source charge Q, the force on it decreases, and beyond a certain distance it becomes practically zero. This is because q is then no longer in the field of Q. We always assume that the test charge is so small that it does not distort the field of the source.

The electric field vector \vec E at a point is defined by the electric force \vec F acting on a positive test charge q placed at that point, divided by the magnitude of the test charge. It is a single vector quantity that carries both the strength and the direction of the field at that point. The strength of the field decides how much force a charge feels there. The direction in which a unit positive test charge moves, or tends to move, is the direction of the electric field.

Electric field intensity: The force per unit positive test charge at any point is called the electric field intensity at that point.

If \vec F is the force experienced by a test charge q placed in the field, the electric intensity at that point is

\vec E=\dfrac{\vec F}{q}

(10.6)

Rearranged, the force on a charge q in a field E is F=qE (this is the form the book uses on p. 193). Now use Coulomb's law for the force between the source charge Q and the test charge q.

Start with the definition of the field.
E=\dfrac{F}{q}
Write the force on the test charge q due to the source charge Q at distance r with Coulomb's law (in vector form, \hat r is the unit vector pointing away from Q).
\vec F=k\,\dfrac{Qq}{r^{2}}\,\hat r
Substitute and cancel the test charge q. The field depends only on the source charge and the distance, not on the test charge.
\vec E=k\,\dfrac{Q}{r^{2}}\,\hat r\qquad(10.7)

\vec E=k\,\dfrac{Q}{r^{2}}\,\hat r

(10.7)

The SI unit of E is newton per coulomb (\mathrm{N\,C^{-1}}) or volt per metre (\mathrm{V\,m^{-1}}). The strength of the field is proportional to the magnitude of the source charge, and it decreases as we move away from the source charge. E is always directed from positive towards negative charges.

Field of a point charge against distance

E=kQ/r^{2} for Q=9\,\mu\mathrm{C} (the charge of Example 10.2), with E in units of 10^{3}\,\mathrm{N\,C^{-1}}. At r=3\,\mathrm{m} the field is 9\times10^{3}\,\mathrm{N\,C^{-1}}.

Example 10.2: Field at the surface of a sphere

Determine the electric field at the surface of a sphere of radius 3.0\,\mathrm{m} if a point charge of 9\,\mu\mathrm{C} is placed at the centre.

Given

r = 3.0\,\mathrm{m} radius of the sphere
q = 9\,\mu\mathrm{C}=9\times10^{-6}\,\mathrm{C} point charge at the centre

To find

E at the surface

The surface of the sphere is at distance r=3.0\,\mathrm{m} from the point charge, so use the field of a point charge.
E=k\,\dfrac{q}{r^{2}}
Put in the values.
E=9\times10^{9}\times\dfrac{9\times10^{-6}}{(3)^{2}}=9\times10^{9-6}
Evaluate. For a positive charge the field points radially outward.
E=9\times10^{3}\,\mathrm{N\,C^{-1}}

E=9\times10^{3}\,\mathrm{N\,C^{-1}}

Assignment 10.2: Field outside a charged sphere

A conducting sphere carries a charge of 200\,\mu\mathrm{C}. Find the electric field intensity at a distance 60\,\mathrm{cm} from the centre of the sphere.

Given

Q = 200\,\mu\mathrm{C}=2.00\times10^{-4}\,\mathrm{C} charge on the sphere
r = 60\,\mathrm{cm}=0.60\,\mathrm{m} distance from the centre

To find

The point is outside the isolated conducting sphere, so the sphere acts as a point charge at its centre.
E=k\,\dfrac{Q}{r^{2}}
Put in the values.
E=9\times10^{9}\times\dfrac{2.00\times10^{-4}}{(0.60)^{2}}=\dfrac{1.8\times10^{6}}{0.36}
Evaluate. The field points away from the sphere if the charge is positive.
E=5.0\times10^{6}\,\mathrm{N\,C^{-1}}

E=5.0\times10^{6}\,\mathrm{N\,C^{-1}}

Worked example: field at a point on the perpendicular bisector of two equal charges

Two equal charges of +4.0\,\mu\mathrm{C} are 0.60\,\mathrm{m} apart. Find the electric field at the point P which is 0.40\,\mathrm{m} above the midpoint of the line joining them.

Given

Q = 4.0\times10^{-6}\,\mathrm{C}\ (\text{each})
r = \sqrt{0.30^{2}+0.40^{2}}=0.50\,\mathrm{m}
k = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

To find

the resultant field E at P

Fields at P from two equal positive charges. E_1 and E_2 have equal size and point away from their charges. Their horizontal parts cancel and the vertical parts add to the resultant E (accent).

Each charge is the same distance from P, found by Pythagoras' theorem: r=0.50\,\mathrm{m}. Both fields therefore have the same size.
E_1=E_2=k\dfrac{Q}{r^{2}}=\dfrac{(9\times10^{9})(4.0\times10^{-6})}{(0.50)^{2}}=1.44\times10^{5}\,\mathrm{N\,C^{-1}}
Positive charges produce fields directed away from them. Resolve each field into a horizontal and a vertical component. The line to P makes an angle with the vertical whose cosine is 0.40/0.50=0.8 and whose sine is 0.30/0.50=0.6.
E_{1y}=E_{2y}=E_1\times0.8=1.152\times10^{5}\,\mathrm{N\,C^{-1}}
The horizontal components E_1\times0.6 are equal and opposite, so they cancel (by symmetry).
E_x=E_1(0.6)-E_2(0.6)=0
Only the vertical components remain, and they add.
E=E_{1y}+E_{2y}=2(1.152\times10^{5})=2.30\times10^{5}\,\mathrm{N\,C^{-1}}

E=2.30\times10^{5}\,\mathrm{N\,C^{-1}}\ \text{directed away from the charges, along the bisector}

10.3Electric field linesOften examined

A way of visualising electric field patterns is to draw lines called electric field lines. They were first introduced by Faraday.

Electric lines of force: Imaginary lines drawn to represent the direction and intensity of the electric field in the vicinity of a charged body.

Electric fields are three-dimensional, so a drawing on paper is only a two-dimensional picture. Figure 10.5 shows the field lines of a single positive point charge; it shows only the lines that lie in the plane containing the point charge.

The field lines are actually directed radially outward from a positive charge in all directions. Instead of the flat “wheel” of lines in the figure, you should picture an entire spherical distribution of lines.

Figure 10.5: Electric field lines of an isolated positive charge in 2-dimension.

The field lines due to a single negative point charge are directed toward the charge (Fig. 10.6).

Figure 10.6: Electric field lines of an isolated negative charge in 2-dimension.

In either case the lines are along the radial direction and extend all the way to infinity. The lines become closer as they approach the charge, which shows that the strength of the field increases as we move towards the source charge.

Two opposite charges: the electric dipole

Figure 10.7 shows the symmetric field lines for two point charges of equal magnitude but opposite sign. This charge configuration is called an electric dipole. The field lines start from the positive charge and end at the negative charge. The high density of lines between the charges shows a strong field in this region. The field there is stronger because the resultant intensity is the sum of the intensities due to the positive and the negative charge, which point the same way between them. The number of lines that begin at the positive charge must equal the number that terminate at the negative charge.

Electric dipole: A configuration of two point charges of equal magnitude but opposite sign.

Figure 10.7: Two-dimensional drawing of electric field lines of two opposite charges placed near each other.

Two like charges

Figure 10.8 shows the field lines of two positive charges. At some points between the two charges the resultant intensity is zero, which creates a field-free region. Two negative charges produce a very similar field, except that the directions are reversed.

Figure 10.8: Two-dimensional drawing of electric field lines of two similar charges placed near each other. The dot midway between the charges marks the point where E=0.

Two oppositely charged parallel plates

Figure 10.9 shows the electric lines of force between two oppositely charged parallel plates separated by a small distance. One plate has a uniform positive charge distribution and the other a uniform negative charge distribution. The field lines start from the positive plate and end at the negative plate. The field between the plates is uniform in strength. If the plates have finite length, the field lines at the ends of the plates curve outward. Such a curved field at the ends of the plates is called a fringing field; it shows that the field is not uniform at the ends.

Fringing field: The curved electric field at the ends of two finite oppositely charged parallel plates, which shows that the field is not uniform at the ends.

Figure 10.9: Fringing field at the ends of oppositely charged plates.

Properties of electric field lines

From the above discussion we conclude that:

The field lines never intersect each other.
To represent field lines starting or ending at infinity, a single charge must be used.
Electric lines of force originate from a positive charge and terminate at a negative charge.
The tangent drawn to any point on a field line indicates the direction of the electric intensity at that point. The arrows indicate the direction of the electric field at various points.
The intensity of the electric field is proportional to the number of electric field lines per unit area.
Electric lines of force are close together where the field is strong and far apart in the weaker regions.
Electric lines of force have the tendency to contract in length. This explains the attraction between oppositely charged bodies.

10.4Potential gradientOften examined

The potential gradient represents the rate of change of electric potential with displacement. In other words, it represents the slope along which the potential is changing.

The potential difference between two points is the difference between the electric potential at point A and the electric potential at point B, as shown in Fig. 10.10 (a). If V_A and V_B are the potentials measured at these two points, then V_A-V_B is the potential difference. If \Delta V is the change in potential between two points separated by a distance \Delta r, then \Delta V/\Delta r is the potential gradient.

Potential gradient: The rate of change of electric potential \Delta V with respect to displacement \Delta r.

Derivation Relation between electric field and potential gradient

Aim. Derive E=-\dfrac{\Delta V}{\Delta r} for the uniform field between two charged plates.

Assumptions.

The field E is uniform (Fig. 10.10 a).
A test charge q moves a small distance \Delta r from point A to point B.
\Delta V=V_B-V_A is the change in potential from A to B.

Figure 10.10 (a): Potential difference between two points A and B in the uniform field between two oppositely charged plates.

The work done on the test charge is force times displacement.
W=F\,\Delta r\qquad(10.8)
In an electric field E the force on a charge q is F=qE. Substitute it in Eq. (10.8).
W=qE\,\Delta r\qquad(10.9)
The work done on the test charge q in moving through a potential difference \Delta V is equal to the decrease of its electric potential energy, which is q\,\Delta V (potential energy per unit charge is potential).
W=-q\,\Delta V\qquad(10.10)
Compare Eqs. (10.9) and (10.10): both give the same work.
qE\,\Delta r=-q\,\Delta V
Cancel the charge q from both sides.
E\,\Delta r=-\Delta V
Divide by \Delta r.
E=-\dfrac{\Delta V}{\Delta r}\qquad(10.11)

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

The strength of the electric field is equal to the potential gradient \Delta V/\Delta r. The negative sign means that the electric potential increases in the direction opposite to the electric field vector, that is, potential falls along the field. The unit of E follows as volt per metre.

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

(10.11)

Potential falls along the field

In a uniform field the potential changes steadily with distance. The slope of the line is \Delta V/\Delta r=-E.

Equation (10.11) also shows that the unit of the electric field E is volt per metre. It is the same as newton per coulomb, which is what Assignment 10.3 asks you to show (below).

Example 10.3: Distance from a point charge

The electric field at a point due to a point charge is 24\,\mathrm{N\,C^{-1}} and the electric potential at that point is 6\,\mathrm{J\,C^{-1}}. Calculate the distance of the point from the charge.

Given

E = 24\,\mathrm{N\,C^{-1}} electric field at the point
V = 6\,\mathrm{J\,C^{-1}} electric potential at the point

To find

Use the relation between potential and field strength for the point.
V=Er
Solve for r.
r=\dfrac{V}{E}
Put in the values.
r=\dfrac{6}{24}=0.25\,\mathrm{m}

r=0.25\,\mathrm{m}

Assignment 10.3: Units of the electric field

Show that \dfrac{\text{volt}}{\text{metre}}=\dfrac{\text{newton}}{\text{coulomb}}.

A volt is a joule per coulomb (energy per unit charge).
1\,\mathrm{V}=1\,\mathrm{J\,C^{-1}}
A joule is a newton metre (work is force times distance).
1\,\mathrm{J}=1\,\mathrm{N\,m}
Combine the two and divide by the metre.
\dfrac{\mathrm{V}}{\mathrm{m}}=\dfrac{\mathrm{J}}{\mathrm{C\,m}}=\dfrac{\mathrm{N\,m}}{\mathrm{C\,m}}=\dfrac{\mathrm{N}}{\mathrm{C}}

\dfrac{\text{volt}}{\text{metre}}=\dfrac{\text{newton}}{\text{coulomb}}

Worked example: field between plates from the potential graph

Two parallel plates are 4.0\,\mathrm{cm} apart. The potential falls uniformly from 120\,\mathrm{V} at the positive plate to 0\,\mathrm{V} at the negative plate, as the graph shows. (a) Find the electric field between the plates. (b) Find the force on a charge of 5.0\,\mathrm{nC} placed between them. (c) State the direction of the field.

Potential between the plates

The potential falls steadily from 120\,\mathrm{V} at the positive plate to 0 at the negative plate. The slope is \Delta V/\Delta r=-30\,\mathrm{V\,cm^{-1}}=-3000\,\mathrm{V\,m^{-1}}, so E=3000\,\mathrm{V\,m^{-1}}.

(a) Read the potential difference and the separation from the graph. Change centimetres to metres, because E is in \mathrm{V\,m^{-1}}.
\Delta V=0-120=-120\,\mathrm{V},\qquad \Delta r=4.0\,\mathrm{cm}=0.040\,\mathrm{m}
Use E=-\Delta V/\Delta r. The minus sign cancels the negative slope, so E is positive.
E=-\dfrac{\Delta V}{\Delta r}=-\dfrac{-120}{0.040}=3.0\times10^{3}\,\mathrm{V\,m^{-1}}=3.0\times10^{3}\,\mathrm{N\,C^{-1}}
(b) The force on a charge in a field is F=qE.
F=qE=(5.0\times10^{-9})(3.0\times10^{3})=1.5\times10^{-5}\,\mathrm{N}
(c) The field points in the direction in which the potential falls, that is from the positive plate (high V) towards the negative plate. A positive charge is pushed the same way.

\begin{aligned}E&=3.0\times10^{3}\,\mathrm{V\,m^{-1}}\\ F&=1.5\times10^{-5}\,\mathrm{N}\\ &\text{from the positive to the negative plate}\end{aligned}

Potential gradient in electrical engineering and technology

10.4.1Motion of a charged particle in an electric field

When a charge +q is placed in an electric field E, as shown in Fig. 10.10 (b), it experiences a force F=qE in the direction of the field.

Figure 10.10 (b): A charged particle +q in an electric field E between two plates a distance d apart with a potential difference V.

If the charge is allowed to move freely in the electric field, it moves from the positive plate to the negative plate and gains kinetic energy.
If the charge is moved against the electric field, an external force (F=+qE) is required. This increases the potential energy of the charge. When the charge is released from its position, it moves back and gains an equivalent amount of kinetic energy.

What the force F=qE does to the particle

Following Eq. (10.9) and Fig. 10.10 (b)

A positive charge feels a force along E. A negative charge feels a force opposite to E (so a negative particle moves from the negative plate towards the positive plate).
\vec F=q\vec E
In a uniform field E is the same everywhere between the plates, so the force is constant. By Newton's second law the particle (mass m) has a constant acceleration, and a particle released from rest moves in a straight line along the field.
a=\dfrac{F}{m}=\dfrac{qE}{m}
The work done by the force over the plate separation d is the kinetic energy gained. With E=V/d from Eq. (10.11) (magnitudes), this is the charge times the potential difference.
W=Fd=qEd=qV

Worked example: a charged particle accelerated between plates

A particle of charge +2.0\,\mu\mathrm{C} and mass 4.0\,\mathrm{mg} is released from rest at the positive plate of a pair of parallel plates 5.0\,\mathrm{cm} apart, with a potential difference of 400\,\mathrm{V} between them. Find (a) the field, (b) the force and acceleration, (c) the kinetic energy and speed on reaching the negative plate. Is the weight of the particle important?

Given

q = 2.0\times10^{-6}\,\mathrm{C}
m = 4.0\,\mathrm{mg}=4.0\times10^{-6}\,\mathrm{kg}
d = 0.050\,\mathrm{m}
V = 400\,\mathrm{V}

To find

(a) E
(b) F and a
(c) kinetic energy and speed

(a) For a uniform field, the size of E is the potential difference over the separation.
E=\dfrac{V}{d}=\dfrac{400}{0.050}=8.0\times10^{3}\,\mathrm{V\,m^{-1}}
(b) The force is constant, so the acceleration is constant and along the field.
\begin{aligned}F&=qE=(2.0\times10^{-6})(8.0\times10^{3})=1.6\times10^{-2}\,\mathrm{N}\\ a&=\dfrac{F}{m}=\dfrac{1.6\times10^{-2}}{4.0\times10^{-6}}=4.0\times10^{3}\,\mathrm{m\,s^{-2}}\end{aligned}
(c) The work done by the field over the plate separation is the kinetic energy gained, W=qEd=qV.
K=qV=(2.0\times10^{-6})(400)=8.0\times10^{-4}\,\mathrm{J}
Put K=\tfrac12mv^{2} and solve for the speed.
v=\sqrt{\dfrac{2K}{m}}=\sqrt{\dfrac{2(8.0\times10^{-4})}{4.0\times10^{-6}}}=\sqrt{400}=20\,\mathrm{m\,s^{-1}}
Check with kinematics: v^{2}=2ad=2(4.0\times10^{3})(0.050)=400, so v=20\,\mathrm{m\,s^{-1}}, which agrees.
Weight: mg=(4.0\times10^{-6})(9.8)=3.9\times10^{-5}\,\mathrm{N}, about 400 times smaller than the electric force 1.6\times10^{-2}\,\mathrm{N}. So the weight can be ignored.

E=8.0\times10^{3}\,\mathrm{V\,m^{-1}},\ F=1.6\times10^{-2}\,\mathrm{N},\ a=4.0\times10^{3}\,\mathrm{m\,s^{-2}},\ K=8.0\times10^{-4}\,\mathrm{J},\ v=20\,\mathrm{m\,s^{-1}}

10.5Ferrofluid

Ferrofluid: A liquid which becomes highly magnetised in the presence of a magnetic field. Ferrofluids are colloidal suspensions of magnetic nanoparticles in a liquid.

A ferrofluid is composed of very small, nanometre-sized particles (diameter usually 10\,\mathrm{nm} or less) of magnetite, hematite or some other compound containing iron, and a liquid (usually oil). These particles are suspended in the liquid. The nanoparticles are usually iron oxide (\ce{Fe3O4}) and the liquids are water or an organic solvent like kerosene.

Figure 10.12: Microscopic view of a ferrofluid. Magnetic particles carry a coating of surfactant and float in the carrier liquid.

The parts of a ferrofluid as named in Fig. 10.12.
Part of the ferrofluid	What it is
Magnetic particle	nanometre-sized particle (diameter usually 10\,\mathrm{nm} or less) of magnetite, hematite or another iron compound; usually iron oxide \ce{Fe3O4}
Surfactant	the layer around each magnetic particle, labelled in Fig. 10.12
Carrier liquid	the liquid that carries the particles: usually oil, water or an organic solvent such as kerosene

Behaviour with and without a magnetic field

No magnetic field: the ferrofluid acts like an ordinary liquid, and the magnetite particles move freely in the fluid.
In a magnetic field: the particles are temporarily magnetised and react to the poles of a magnet. Figure 10.11 is a magnified photograph of a ferrofluid influenced by a magnetic field; it shows the liquid rising into spikes.
Field removed: the induced magnetic field disappears.

Applications

Ferrofluids are fascinating materials with a range of applications in different fields such as electronics, optics and medical physics. These materials are used in:

magnetic resonance imaging (MRI)
bio-sensing
medical imaging
medicinal therapy
dynamic loudspeakers
magneto-optic sensors
heat transfer and dissipation

In the near future, ferrofluids may be used to carry medications to specific locations in the body.

Faraday cage

Faraday cage: A cage made of a conducting material. A Faraday cage distributes charge or radiation around the cage's exterior and cancels out electric charges or radiation within the cage. In short, it is a hollow conductor in which the charge remains on the external surface of the cage.

The fields within a conductor cancel out with any external fields, so the electric field within a conductor is zero. Faraday cages act as big hollow conductors. We can put things into a Faraday cage to shield them from electrical fields.

How a Faraday cage works

Suppose a cage is placed in an external electric field. The free charges in the conducting cage are pushed by the field and move along the metal until they pile up on the outside surface, as drawn in the figure. These charges induce an internal electric field that points opposite to the external field. Inside the cage the two fields cancel, so the net field is zero and the inside is shielded from the influence of the external field.

Faraday cage: an external field makes the free charges of the hollow conductor collect on its outer surface. They set up an internal field that cancels the external one, so E=0 inside.

Formula sheet

Coulomb's law

F=k\dfrac{q_1q_2}{r^{2}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^{2}}

q_1,q_2: the charges; r: distance between them

Unit: \mathrm{N}

Coulomb's constant

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

\varepsilon_0=8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}: permittivity of free space

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

Coulomb's law, vector form

\vec F_{12}=k\dfrac{q_1q_2}{r^{2}}\hat r_{12},\qquad \vec F_{21}=-\vec F_{12}

\vec F_{12}: force on q_1 by q_2; \hat r_{12}: unit vector in the direction of \vec F_{12}

Unit: \mathrm{N}

Superposition of forces

\vec F=\vec F_{12}+\vec F_{13}+\vec F_{14}

forces on q_1 due to q_2, q_3, q_4, added as vectors

Unit: \mathrm{N}

Electric field intensity

\vec E=\dfrac{\vec F}{q}

F: force on a test charge q

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

Force on a charge in a field

F=qE

q: charge, E: electric field intensity

Unit: \mathrm{N}

Field of a point charge

E=k\dfrac{Q}{r^{2}}

Q: source charge; r: distance from it

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

Potential gradient

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

\Delta V: change in potential over a small displacement \Delta r in a uniform field

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

Key terms

Electric charge: A property of matter that causes a force when it is near another charge. It comes in two forms, positive and negative.
Electrostatics: The study of electric charges at rest and the forces between them.
Coulomb's law: The force of attraction or repulsion between two charged bodies is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them: F=kq_1q_2/r^{2}.
Coulomb's constant: The constant k=\dfrac{1}{4\pi\varepsilon_0} in Coulomb's law; 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} in a vacuum.
Permittivity of free space: The constant \varepsilon_0=8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} used to write k=1/4\pi\varepsilon_0.
Point charge: An imaginary charge located at a single point in space. An isolated charged sphere acts as a point charge at its centre for points outside it.
Superposition principle of force: When several separate charges act on a charge, each exerts a force; the forces are computed separately and added as vectors.
Electric field: The region around a charge in which a test charge can feel an electric force.
Electric field intensity: The force per unit positive test charge at a point, E=F/q. Unit: \mathrm{N\,C^{-1}} or \mathrm{V\,m^{-1}}.
Electric lines of force: Imaginary lines that represent the direction and intensity of the electric field around a charged body.
Electric dipole: Two point charges of equal magnitude and opposite sign placed near each other.
Fringing field: The curved electric field at the ends of two finite oppositely charged parallel plates.
Potential gradient: The rate of change of electric potential \Delta V with respect to displacement \Delta r; E=-\Delta V/\Delta r.
Ferrofluid: A colloidal suspension of magnetic nanoparticles (usually iron oxide, \ce{Fe3O4}) in a liquid (water or an organic solvent like kerosene) that becomes highly magnetised in a magnetic field.
Faraday cage: A cage made of a conducting material (a hollow conductor) that shields its inside from external electric fields, because the field within a conductor is zero.

Multiple choice questionsp. 195

Which of the following is equivalent to 1 volt?

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

Why

1\,\mathrm{V}=1\,\mathrm{J\,C^{-1}}, because the potential difference is the work done per unit charge, V=W/q. Joule per second (B) is the watt, not the volt. Newton per second (A) and coulomb per joule (D) are not units of potential.

Two positive charges are placed on a screen. Which statement describes the electric field produced by the charges?

AIt is constant everywhere.
BIt is zero near each charge.
CIt is zero halfway between the charges.Answer
DIt is strongest halfway between the charges.

Why

For two equal positive charges, the fields at the midpoint are equal in size and point in opposite directions (each points away from its own charge), so they cancel and the resultant field is zero. This is the field-free region between two similar charges in Figure 10.8.

Option B is wrong because the field is strongest near each charge, where E=kQ/r^{2} is large. Option D is the opposite of the truth, and option A is wrong because the field changes from point to point.

Note. The question does not say the two charges are equal. The answer assumes equal magnitudes, which is the case the book describes.

Four charges are placed at the corners of a square ABCD as shown. The force on a positive charge placed at the centre of the square is

Four charges at the corners of a square ABCD, with a point at its centre.

A0
Balong diagonal AC
Calong diagonal BD
Dperpendicular to side ABAnswer

Why

Put the centre at the origin, with the sides parallel to the axes. Each corner is at distance d=a/\sqrt{2} from the centre, so each charge produces a field of magnitude kQ/d^{2} there (doubled for \pm2q). +q at B and +2q at D point away from those corners, while -q at C and -2q at A point towards them.

\begin{aligned}E_x&=\frac{kq}{\sqrt{2}\,d^{2}}\,(1+1-2-2)=-\frac{2kq}{\sqrt{2}\,d^{2}}\\ E_y&=\frac{kq}{\sqrt{2}\,d^{2}}\,(-1+1+2-2)=0\end{aligned}

The net field is non-zero and points along the -x direction, which is horizontal, parallel to sides AD and BC. It is therefore perpendicular to side AB, and the force on the positive test charge is in the same direction. Option A is wrong because the field is not zero, and options B and C are wrong because the net field is not along either diagonal.

The bob of a pendulum with mass m, length l and a charge q is in the rest position in a uniform horizontal electric field of E. The tension in the string of the pendulum is:

Amg
BqE
C[(mg)^2+(qE)^2]^{1/2}Answer
D[(mg)^2+(qE)^2]^{1/4}

Why

At rest the bob is in equilibrium under three forces: its weight mg downwards, the horizontal electric force qE, and the tension T along the string. The tension must balance the resultant of the other two, so its magnitude is the magnitude of that resultant.

T=\sqrt{(mg)^{2}+(qE)^{2}}=\left[(mg)^{2}+(qE)^{2}\right]^{1/2}

Options A and B use only one of the two forces. Option D takes the fourth root instead of the square root; it has the wrong unit (\mathrm{N}^{1/2} instead of \mathrm{N}).

52025 Annual I, Set B, MCQ 112025 Annual I, Set D, MCQ 12

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. Because q is negative, the force is opposite to \vec E, that is from North to South. The particle therefore accelerates towards the South.

If two points are at the same potential, are there any electric field lines connecting them?

Ayes
BNoAnswer
Cmay or may not
Dinsufficient information

Why

Field lines run in the direction in which the potential falls. From E=-\Delta V/\Delta r, a non-zero field along a line means the potential changes between any two points on that line, so two points at the same potential cannot lie on one field line. Option C is tempting, but the answer does not depend on the arrangement of the charges.

Coulomb law is usually applied in the area of:

AMagnetism
BElectrostaticsAnswer
CElectromagnetism
DGravitation

Why

Coulomb's law gives the force between stationary point charges, which is the subject of electrostatics (the study of electric charges at rest). Magnetism and electromagnetism concern magnetic fields and moving charges, and gravitation is governed by Newton's law, which has the same mathematical form with masses in place of charges.

Let F be the force between two equal point charges at some distance. If the distance between them is doubled and individual charges are also doubled, what will be the force acting between the charges?

AFAnswer
B2F
C4F
DF/2

Why

Coulomb's law gives F=kq_1q_2/r^{2}. With each charge doubled and the distance doubled:

F'=\frac{k(2q)(2q)}{(2r)^{2}}=\frac{4kq^{2}}{4r^{2}}=F

The four-fold increase in the product of charges is cancelled by the four-fold increase in r^{2}, so the force is unchanged. Options B and C would need only one of the changes to be made, and D would need the charges to be halved.

Short questionsp. 195–196

10.13 marks

When an air passenger touches the knob of the toilet door during a flight, he may get an electric shock. Why is it so?

Solution

During the flight the passenger's body is charged by friction with the seat, clothing and air, in the same way that a comb is charged by rubbing it through the hair. The body acquires a net electric charge and keeps it while he sits in the cabin.

The metal knob is a conductor. When the passenger touches it, the charge on his body flows through his hand to the knob in a sudden discharge, which he feels as a shock.

10.22025 Annual I, Set B, Q.2 (xii)3 marks

How will the radius of a flexible ring change if it is given a positive charge?

Solution

The positive charge spreads over the ring, so every part of the ring carries positive charge. Like charges repel each other: for two charges with the same sign the product q_1q_2 is positive and Coulomb's law gives a repulsive force. The parts of the ring therefore push one another apart.

The ring expands outwards, so its radius increases, and the charges move farther apart, which reduces the repulsion between them.

10.33 marks

Write some applications of electrostatics in real life.

Solution

The unit gives the following applications of electrostatics:

Induction with a charged comb. A comb charged by rubbing attracts small pieces of paper, which is due to induction. The charge on the comb induces the opposite kind of charge on the paper, and the paper is attracted to the comb.
Faraday cage. A cage made of a conducting material shields what is inside it from external electric fields and charges, so things placed in it are protected.
Cable insulation in power systems. The dielectric insulation of a cable must be kept higher than the potential gradient of the conductor, otherwise the cable is not safe. A high potential gradient in power systems can also harm a person who touches the cable.

10.43 marks

What are the limitations of Coulomb's law?

Solution

Coulomb's law F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^{2}} holds only under the following conditions:

It applies to point charges, or to a uniform spherical charge distribution, for which the charge acts as if it were at the centre. For any other arrangement the force must be found by adding the forces from the separate parts of the charge (the superposition principle).
The value k=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} applies only when there is vacuum between the charges. In another medium the value of k is different.
It describes the force between charges that are at rest, which is the subject of electrostatics.

10.53 marks

If two points are at the same potential, are there any electric field lines connecting them?

Solution

No. Electric field lines point in the direction in which the potential decreases. Along a field line the field is not zero, so E=-\Delta V/\Delta r gives a change in potential between any two points on that line.

E=-\frac{\Delta V}{\Delta r}\neq0 \;\Rightarrow\; \Delta V\neq0 \text{ between any two points on the same line}

Two points at the same potential therefore lie on the same equipotential, and no line of force joins them.

10.63 marks

What is the difference between electric field and electric field strength?

Solution

The electric field is the region around a charge in which a test charge feels an electric force. It describes a region of space. The electric field strength (electric field intensity) is the quantity that measures the field at a particular point, namely the force per unit positive test charge placed there:

\vec E=\frac{\vec F}{q}

E is a vector. Its magnitude is the field strength, in \mathrm{N\,C^{-1}} or \mathrm{V\,m^{-1}}, and its direction is the direction of the force on a positive charge. The field is the region; the field strength is the quantity that describes it at each point.

10.73 marks

Draw electric field lines for (a) an isolated positive charge (b) oppositely charged parallel plates.

Solution

(a) Isolated positive charge. The lines are straight and radial, pointing outwards in all directions and evenly spread. They start on the charge and extend to infinity, becoming farther apart as the distance from the charge increases, because the field weakens.

(b) Oppositely charged parallel plates. The lines start on the positive plate, meet the plates at right angles and end on the negative plate. Between the plates they are parallel and equally spaced, which shows that the field is uniform there. At the ends of the plates they curve outwards, which is the fringing field.

Figure 10.5 and Figure 10.9: field lines of an isolated positive charge (a) and of oppositely charged parallel plates (b).

10.83 marks

How can we say that 'a charged sphere also behaves as a point charge for a point outside the spherical conductor'?

Solution

The field at any point outside a spherically symmetric charge distribution is the same as the field that would be produced by the same charge placed at its centre. A charged sphere carries its charge symmetrically, so for a point outside it the field is that of a point charge Q at the centre:

E=\frac{kQ}{r^{2}}

Here r is measured from the centre of the sphere. This holds only while the sphere is isolated. If another charged object is brought near it, the charge on the sphere is redistributed, is no longer spherically symmetric, and the sphere cannot be treated as a point charge.

10.93 marks

Describe the effect of a uniform electric field on the motion of charged particles.

Solution

A uniform field exerts a constant force \vec F=q\vec E on a charge. The force is in the direction of \vec E for a positive charge and opposite to \vec E for a negative charge. Since the force is constant, the acceleration is constant:

a=\frac{F}{m}=\frac{qE}{m}

A positive charge released between the plates moves from the positive plate towards the negative plate and gains kinetic energy. A negative charge moves the other way. If a charge is moved against the field, an external force is needed and its potential energy increases; when it is released it moves back and regains the same amount of kinetic energy.

Comprehensive questionsp. 196

10.1

Define and explain Coulomb's law.

Solution

Statement

Coulomb's law: the force of attraction or repulsion between two charged bodies is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges.

Mathematical form

Derivation Coulomb's law as an equation

Aim. Write the force F between two point charges q_1 and q_2 separated by a distance r in vacuum.

Assumptions.

Both charges are point charges at rest in vacuum.
The force acts along the line joining them.

Figure 10.2: (a) like charges repel each other; (b) unlike charges attract each other.

The force is directly proportional to the product of the charges, so double one charge and the force doubles.
F\propto q_1q_2
The force is inversely proportional to the square of the distance.
F\propto\dfrac{1}{r^{2}}
Combine the two proportionalities with Coulomb's constant k.
F=k\,\dfrac{q_1q_2}{r^{2}}
Write k in terms of the permittivity of free space \varepsilon_0=8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}.
k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

ResultF=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{q_1q_2}{r^{2}}

Like charges (product positive) repel and unlike charges (product negative) attract. Doubling the distance reduces the force to one quarter.

Vector form

Force is a vector. The force on q_1 due to q_2 is \vec F_{12}=k\,\dfrac{q_1q_2}{r^{2}}\,\hat r_{12}, where \hat r_{12} is the unit vector in the direction of \vec F_{12}. By Newton's third law the force on q_2 is equal and opposite, \vec F_{21}=-\vec F_{12}.

Conditions

The law applies to point charges and to uniform spherical charge distributions.
The value of k applies in vacuum; it changes in another medium.

10.22025 Annual I, Set D, Q.3 OR

What is electric field strength? Explain with the help of examples.

Solution

Definition

The electric field is the region around a charge in which another charge feels an electric force. The electric field strength (electric field intensity) \vec E at a point is the force per unit positive test charge placed at that point:

\vec E=\frac{\vec F}{q}

The test charge q must be very small so that it does not disturb the field of the source charge.

Field of a point charge

Derivation Field strength of a point charge

Aim. Find the field strength at distance r from a point charge Q in vacuum.

Assumptions.

Q is a point charge in vacuum.
The test charge is small and positive.

Figure 10.4: A small positive test charge q placed near a source charge Q experiences the electric field \vec E established by Q.

The force on the test charge q is given by Coulomb's law.
F=k\,\dfrac{Qq}{r^{2}}
Divide by the test charge, as in the definition of \vec E. The charge q cancels.
E=\dfrac{F}{q}=k\,\dfrac{Q}{r^{2}}
The field depends only on the source charge Q and the distance r, not on the test charge.
E=\dfrac{kQ}{r^{2}}

ResultE=\dfrac{kQ}{r^{2}}

The field strength falls off as 1/r^{2}. It points away from Q if Q is positive and towards Q if Q is negative.

Units and direction

The SI unit is \mathrm{N\,C^{-1}} or, equivalently, \mathrm{V\,m^{-1}}. The field points radially outwards from a positive source, the direction in which a positive test charge is pushed.

Examples

Field at the surface of a charged sphere (book Example 10.2)

Write the field of a point charge.
E=k\,\dfrac{q}{r^{2}}
A sphere of radius 3.0\,\mathrm{m} carries a point charge of 9\,\mu\mathrm{C} at its centre. The field at its surface is at distance r=3.0\,\mathrm{m} from the charge.
E=9\times10^{9}\times\dfrac{9\times10^{-6}}{(3.0)^{2}}
Evaluate.
E=9\times10^{3}\,\mathrm{N\,C^{-1}}

Field of a point charge at a given distance

A 2\,\mu\mathrm{C} point charge is placed in vacuum. Find the field at 20\,\mathrm{cm} from it. Convert the distance to metres: r=0.20\,\mathrm{m}.
E=9\times10^{9}\times\dfrac{2\times10^{-6}}{(0.20)^{2}}
Evaluate.
E=4.5\times10^{5}\,\mathrm{N\,C^{-1}}, \text{ directed away from the charge}

In both examples the field is found from the source charge and the distance alone. For several charges the field at a point is found by adding the field vectors, as with forces.

10.32026 Annual I, Set B, Q.2 (xi) OR2026 Annual I, Set D, Q.62025 Annual I, Set D, Q.3 OR

What is meant by the term potential gradient? Show that E=\Delta V/\Delta r.

Solution

Potential gradient

The potential gradient is the rate of change of electric potential \Delta V with respect to displacement \Delta r, that is \dfrac{\Delta V}{\Delta r}. It is the slope of the potential along the direction in which the potential changes.

Relation between field and potential gradient

Derivation Field strength as the potential gradient

Aim. Show that the field strength equals the potential gradient. Written as in the book, E=-\dfrac{\Delta V}{\Delta r}, and the magnitude of the gradient is E=\dfrac{\Delta V}{\Delta r}.

Assumptions.

A uniform field E exists between two parallel plates.
The test charge moves a small distance \Delta r from A to B along the field.

Figure 10.10 (a): Potential difference between two points A and B.

Work done on a test charge q moved through a displacement \Delta r is force times displacement.
W=F\,\Delta r
The force on a charge in a uniform field is F=qE.
W=qE\,\Delta r
The same work also equals the decrease in electric potential energy, because moving q through a potential difference \Delta V changes its potential energy by q\,\Delta V.
W=-q\,\Delta V
Equate the two expressions for W and cancel q. The minus sign shows that the work is done against the field.
qE\,\Delta r=-q\,\Delta V\;\Rightarrow\;E\,\Delta r=-\Delta V
Divide by \Delta r.
E=-\dfrac{\Delta V}{\Delta r}

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

The field strength is the potential gradient. Ignoring the sign, which shows that the potential falls in the direction of the field, E=\dfrac{\Delta V}{\Delta r}. The unit is volt per metre.

10.4

Describe the motion of a charged particle in a uniform electric field.

Solution

Force on the charge

In a uniform field \vec E between two parallel plates, a charge q experiences the force \vec F=q\vec E. The field has the same strength and direction everywhere between the plates, so the force is the same at every point in that region.

Acceleration

Newton's second law relates the force to the acceleration of a particle of mass m.
F=ma
Equate the electric force to ma and solve for a.
qE=ma\;\Rightarrow\;a=\dfrac{qE}{m}
q, E and m are all constant, so the acceleration is constant. A positive charge accelerates in the direction of \vec E, and a negative charge accelerates opposite to \vec E.

Positive charge released freely

A positive charge released in the field moves from the positive plate towards the negative plate, in the direction of the field. It speeds up and gains kinetic energy. The field does work on it, and this work equals the decrease in potential energy:

\text{gain in kinetic energy}=q\,\Delta V

Figure 10.10 (b): Charged particle in an electric field.

Charge moved against the field

If the charge is moved against the field, an external force equal to qE must be applied, and the potential energy of the charge increases. When it is released, the electric force carries it back and it regains kinetic energy equal to the potential energy it had gained.

Negative charge

A negative charge is pushed opposite to \vec E. Released between the plates, it moves from the negative plate towards the positive plate.

10.5

What are ferrofluids? Explain.

Solution

Definition and structure

A ferrofluid is a colloidal suspension of magnetic nanoparticles in a liquid. The particles are usually iron oxide (\ce{Fe3O4}), with diameters of about 10\,\mathrm{nm} or less, and the liquid is usually water or an organic solvent such as kerosene. A surfactant coats each particle so that the particles stay spread out in the liquid (Figure 10.12).

Figure 10.12: Microscopic view of a ferrofluid.

Behaviour

Without a magnetic field a ferrofluid acts like an ordinary liquid, because the particles move freely about in the carrier liquid. In a magnetic field the particles become temporarily magnetised and respond to the poles of the magnet, so the liquid is pulled into spikes (Figure 10.11). Removing the external field makes the induced magnetism disappear.

Uses

magnetic resonance imaging (MRI) and medical imaging
bio-sensing and medicinal therapy
dynamic loudspeakers
magneto-optic sensors
heat transfer and dissipation

In the near future ferrofluids may be used to carry medicines to specific locations in the body.

10.62025 Annual I, Set B, Q.2 (x)

Explain how a Faraday cage works?

Solution

What a Faraday cage is

A Faraday cage is a cage made of a conducting material. It is a hollow conductor, and any charge on it remains on its external surface.

How it shields the inside

An external electric field acts on the free charges of the conductor. Electrons move against the field and collect on one side of the cage, leaving the other side with an excess of positive charge.
These induced charges set up their own field inside the cage, opposite to the external field. Charge keeps moving until the two fields cancel and the field inside the conductor is zero.
E_{\text{inside}}=E_{\text{external}}+E_{\text{induced}}=0
All the induced charge lies on the outside of the cage, and there is no field inside it. Anything inside the cage is therefore shielded from the external field.

Charge sits on the outside of the cage, and the field inside it is zero.

Summary

The cage distributes charge over its exterior, so the field inside the conducting material and the cage is zero. Sensitive equipment and people can therefore be shielded from external electric fields by placing them inside a conducting enclosure.

10.7

Describe the effect of a uniform electric field on the motion of charged particle.

Solution

A uniform field has the same magnitude and direction at every point. A charged particle of charge q placed in it experiences a constant force \vec F=q\vec E.

A positive charge is pushed in the direction of \vec E and a negative charge opposite to \vec E.
\vec F=q\vec E
Because the force is constant, the acceleration is also constant.
a=\dfrac{F}{m}=\dfrac{qE}{m}

A positive charge released between the plates moves from the positive plate towards the negative plate and gains kinetic energy, equal to the decrease in its potential energy. A negative charge moves the other way, from the negative plate towards the positive plate.

If the particle is moved against the field, an external force equal to qE is needed. This increases its potential energy, which is released as kinetic energy when the particle is let go.

Numerical problemsp. 196

10.1

Charges of magnitude 100\,\mu\mathrm{C} each are located in vacuum at the corners A, B and C of an equilateral triangle measuring 4\,\mathrm{m} on each side. If the charge at A and C is positive and the charge at B is negative, what is the magnitude and direction of the total force on the charge at C?

Solution

Given

q_A=q_C = +100\,\mu\mathrm{C}
q_B = -100\,\mu\mathrm{C}
a = 4\,\mathrm{m}
k = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

To find

F_{\text{net}} on the charge at C, in magnitude and direction

Each pair of charges is 4\,\mathrm{m} apart with charges of 100\,\mu\mathrm{C}, so each pair exerts a force of the same magnitude F by Coulomb's law.
F=k\,\dfrac{q^{2}}{a^{2}}=9\times10^{9}\times\dfrac{(100\times10^{-6})^{2}}{4^{2}}
Evaluate.
F=\dfrac{9\times10^{9}\times10^{-8}}{16}=5.625\,\mathrm{N}
On C the charge at A is like C, so it repels C along AC. The charge at B is unlike C, so it attracts C along CB. Each force makes 60^\circ with side AB (see the figure).
Resolve each force along AB and perpendicular to AB. The perpendicular components are equal and opposite and cancel. The components along AB add.
F_x=F\cos60^\circ+F\cos60^\circ=2F\cos60^\circ
Substitute and evaluate. The perpendicular components cancel, so there is no perpendicular component.
F_x=2\times5.625\times\tfrac12=5.625\,\mathrm{N},\qquad F_y=F\sin60^\circ-F\sin60^\circ=0

Forces on C: F_CA (repulsion from A) and F_CB (attraction towards B) add to the resultant, which is parallel to AB.

F_{\text{net}}=5.625\,\mathrm{N}\ \text{parallel to AB (from A towards B)}

Book answer. 5.625\,\mathrm{N}, force is parallel to AB

Note. Agrees with the book. Both pair forces have magnitude 5.625\,\mathrm{N}, and their perpendicular components cancel, which leaves a resultant of 5.625\,\mathrm{N} along AB.

10.2

What is the magnitude of a point charge that would create an electric field of 1.00\,\mathrm{N\,C^{-1}} at a point 1.00\,\mathrm{m} away?

Solution

Given

E = 1.00\,\mathrm{N\,C^{-1}}
r = 1.00\,\mathrm{m}
k = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

To find

Solve E=\dfrac{kQ}{r^{2}} for Q.
Q=\dfrac{Er^{2}}{k}
Substitute.
Q=\dfrac{1.00\times(1.00)^{2}}{9\times10^{9}}
Evaluate.
Q=1.11\times10^{-10}\,\mathrm{C}

Q\approx1.11\times10^{-10}\,\mathrm{C}

Book answer. 1.11\times10^{-10}\,\mathrm{C}

10.3

Consider a point charge +q placed at the origin and another point charge -2q placed at a distance of 9\,\mathrm{m} from the charge +q. Determine the point between the two charges at which the electric potential is zero.

Solution

Given

q_1 = +q\text{ at the origin}
q_2 = -2q\text{ at } 9\,\mathrm{m}
V = 0

To find

the distance x of the point from +q

Take the point at distance x from +q, so it is 9-x from -2q. Electric potential is a scalar, so the potentials of the two charges add directly.
V=\dfrac{kq}{x}+\dfrac{k(-2q)}{9-x}
Set the potential to zero.
\dfrac{kq}{x}=\dfrac{2kq}{9-x}
Cancel kq and cross-multiply.
9-x=2x
Solve for x.
x=3\,\mathrm{m}
Check: the potentials at x=3\,\mathrm{m} are \dfrac{kq}{3} and -\dfrac{2kq}{6}, which cancel.

x=3\,\mathrm{m}\ \text{from }+q

Book answer. 3\,\mathrm{m} from +q

10.4

An object with a net charge of 24\,\mu\mathrm{C} is placed in a uniform electric field of 610\,\mathrm{N\,C^{-1}} directed vertically. What is the mass of this object if it floats in the field?

Solution

Given

q = 24\,\mu\mathrm{C}
E = 610\,\mathrm{N\,C^{-1}}
g = 9.8\,\mathrm{m\,s^{-2}}

To find

Floating means the upward electric force balances the weight.
qE=mg
Make m the subject.
m=\dfrac{qE}{g}
Substitute.
m=\dfrac{24\times10^{-6}\times610}{9.8}
Evaluate.
m=1.49\times10^{-3}\,\mathrm{kg}

m\approx1.5\times10^{-3}\,\mathrm{kg}=1.5\,\mathrm{g}

Book answer. 1.5\,\mathrm{g}

Note. Agrees with the book. With g=10\,\mathrm{m\,s^{-2}} the mass would be 1.46\,\mathrm{g}, which also rounds to 1.5\,\mathrm{g}.

10.5

What will be the electric field at a distance of 30\,\mathrm{cm} from a 3\,\mu\mathrm{C} point charge?

Solution

Given

q = 3\,\mu\mathrm{C}
r = 30\,\mathrm{cm}=0.30\,\mathrm{m}

To find

Use the field of a point charge.
E=\dfrac{kq}{r^{2}}
Substitute with r in metres.
E=\dfrac{9\times10^{9}\times3\times10^{-6}}{(0.30)^{2}}
Evaluate.
E=\dfrac{2.7\times10^{4}}{0.09}=3.0\times10^{5}\,\mathrm{N\,C^{-1}}

E=3.0\times10^{5}\,\mathrm{N\,C^{-1}}\ \text{directed away from the charge}

Book answer. 3\times10^{5}\,\mathrm{N\,C^{-1}}

10.62026 Annual I, Set D, Q.2 (xiii) OR

The electric field at a point due to a point charge is 26\,\mathrm{N\,C^{-1}} and the electric potential at that point is 13\,\mathrm{J\,C^{-1}}. Calculate the distance of the point from the charge.

Solution

Given

E = 26\,\mathrm{N\,C^{-1}}
V = 13\,\mathrm{J\,C^{-1}}

To find

For a point charge V=\dfrac{kQ}{r} and E=\dfrac{kQ}{r^{2}}. Dividing V by E cancels kQ and leaves r, so V=Er.
\dfrac{V}{E}=\dfrac{kQ/r}{kQ/r^{2}}=r
Make r the subject and substitute.
r=\dfrac{V}{E}=\dfrac{13}{26}
Evaluate.
r=0.5\,\mathrm{m}

r=0.5\,\mathrm{m}

Book answer. 0.5\,\mathrm{m}

Note. Agrees with the book. The relation V=Er is the one used in the book's Example 10.3, and holds for the field of a point charge with potential zero at infinity. The working of that example on page 192 has a stray minus sign (-r=6/24), a misprint.

10.7

The electric field at a point 0.5\,\mathrm{m} from a charge is 2.5\,\mathrm{N\,C^{-1}}. Find the value of electric potential at that point.

Solution

Given

E = 2.5\,\mathrm{N\,C^{-1}}
r = 0.5\,\mathrm{m}

To find

For a point charge V=\dfrac{kQ}{r} and E=\dfrac{kQ}{r^{2}}, so V=Er.
V=Er
Substitute.
V=2.5\times0.5
Evaluate. Note 1\,\mathrm{J\,C^{-1}}=1\,\mathrm{V}.
V=1.25\,\mathrm{V}

V=1.25\,\mathrm{V}

Book answer. 1.25\,\mathrm{V}

Note. Agrees with the book.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 111 mark

An electron is placed in a uniform electric field directed from south to north. In which direction will electron move after it is released?

AEast
BSouthAnswer
CNorth
DWest

Why

The force on a charge is \vec F=q\vec E. The electron is negative, so the force is opposite to \vec E. The field points from south to north, so the force, and the acceleration, point towards the south. The answer is B.

Option C (north) is the direction in which a positive charge would move. East and west are perpendicular to the field and the force has no component along them.

2026 Annual I, Set B, MCQ 121 mark

Electric field strength is:

AWork done
BEnergy
CForce per unit chargeAnswer
DCharge per force

Why

The electric field strength at a point is the force per unit positive test charge placed there, E=\dfrac{F}{q} (unit \mathrm{N\,C^{-1}}), so C is correct.

Option A (work done) is wrong: work per unit charge is the potential difference, V=W/q, not the field strength. Option D inverts the definition (it would be q/F).

2025 Annual I, Set B, Q.2 (x)3 marks

What is Faraday's cage? Briefly explain its working.

Solution

Faraday cage (1 mark). A Faraday cage is a cage made of a conducting material. It is a hollow conductor in which any charge stays on the outside surface, and the electric field inside it is zero.

Working (2 marks).

An external electric field pushes on the free charges of the conductor. Electrons move against the field and collect on one side of the cage; the other side is left with positive charge.
These induced charges set up an internal field opposite to the external one. Charge keeps moving until the two cancel.
E_{\text{inside}}=E_{\text{external}}+E_{\text{induced}}=0
All the induced charge sits on the outside, so the inside is shielded. People and sensitive equipment placed inside are protected from the external field.

Charge sits on the outside of the cage, and the field inside it is zero.

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

What is meant by potential gradient? Derive its formula.

Solution

Potential gradient (1 mark). The potential gradient is the rate of change of electric potential \Delta V with distance \Delta r, that is \dfrac{\Delta V}{\Delta r}. It is the slope of the potential along the direction in which the potential changes.

Formula (2 marks). Take a uniform field E between two parallel plates and move a test charge q a small distance \Delta r from A to B along the field.

The work done on the charge is force times displacement, and the force is F=qE.
W=F\,\Delta r=qE\,\Delta r
The same work equals the decrease of potential energy, because q\,\Delta V is the change of potential energy.
W=-q\,\Delta V
Equate the two expressions, cancel q and divide by \Delta r.
qE\,\Delta r=-q\,\Delta V\;\Rightarrow\;E=-\dfrac{\Delta V}{\Delta r}

So the field strength equals the potential gradient. The minus sign shows that the potential falls in the direction of the field; ignoring it, E=\Delta V/\Delta r, with the unit volt per metre.

2025 Annual I, Set B, Q.2 (xii)3 marks

How will the radius of a flexible ring change if it is given negative charge?

Solution

The negative charge spreads over the ring, so every part of the ring carries negative charge. Like charges repel: for two charges of the same sign q_1q_2 is positive and Coulomb's law gives a repulsive force.

The parts of the ring therefore push one another apart, and the radius of the ring increases. As the charges move farther apart the repulsion between them falls. The ring behaves the same way as for a positive charge, since only the sign of both charges is changed.

2025 Annual I, Set D, MCQ 121 mark

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

ANorth
BSouthAnswer
CEast
DWest

Why

The force on a charge is \vec F=q\vec E. For a negative charge q<0, so the force is opposite to the field. The field points from south to north, so the particle accelerates towards the south. The answer is B.

Option A (north) would be correct for a positive charge. East and west are at right angles to the field and the force has no component in those directions.

2026 Annual I, Set D, MCQ 131 mark

If 800J energy is used to move a charge of 4C between two points in an electric field. Potential difference is:

A5V
B100V
C200VAnswer
D5\times10^{-3}V

Why

The potential difference is the work done per unit charge, V=\dfrac{W}{q}=\dfrac{800\,\mathrm{J}}{4\,\mathrm{C}}=200\,\mathrm{V}, so C is correct.

Option D, 5\times10^{-3}\,\mathrm{V}, comes from dividing the wrong way round (q/W=4/800). Option B (100\,\mathrm{V}) would need a charge of 8\,\mathrm{C}.

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

The electric field at a point due to a point charge is 26\,\mathrm{N/C} and the electric potential at that point is 13\,\mathrm{J/C}. Calculate the distance of the point from the charge.

Solution

Given

E = 26\,\mathrm{N\,C^{-1}}
V = 13\,\mathrm{J\,C^{-1}}

To find

Formulas (1 mark). For a point charge the field and the potential are E=\dfrac{kQ}{r^{2}} and V=\dfrac{kQ}{r}.
**Divide V by E (1 mark).** kQ cancels and leaves r.
\dfrac{V}{E}=\dfrac{kQ/r}{kQ/r^{2}}=r
Substitute with unit (1 mark).
r=\dfrac{V}{E}=\dfrac{13}{26}=0.5\,\mathrm{m}

r=0.5\,\mathrm{m}

2025 Annual I, Set D, Q.3 OR7 marks

What is meant by electric field strength? Show that: E = -\dfrac{\Delta V}{\Delta r}

Solution

Electric field strength

The electric field is the region around a charge in which another charge feels an electric force. The electric field strength (intensity) \vec E at a point is the force per unit positive test charge placed there:

\vec E=\dfrac{\vec F}{q}

The test charge must be so small that it does not disturb the source field. The unit is \mathrm{N\,C^{-1}} or \mathrm{V\,m^{-1}}, and the direction is that of the force on a positive charge. For a point charge Q, Coulomb's law gives F=kQq/r^{2}, so E=kQ/r^{2}.

Show that E=-\Delta V/\Delta r

Derivation Field strength as the potential gradient

Aim. Show that the field strength equals the potential gradient. Written as in the book, E=-\dfrac{\Delta V}{\Delta r}, and the magnitude of the gradient is E=\dfrac{\Delta V}{\Delta r}.

Assumptions.

A uniform field E exists between two parallel plates.
The test charge moves a small distance \Delta r from A to B along the field.

Figure 10.10 (a): Potential difference between two points A and B.

Work done on a test charge q moved through a displacement \Delta r is force times displacement.
W=F\,\Delta r
The force on a charge in a uniform field is F=qE.
W=qE\,\Delta r
The same work also equals the decrease in electric potential energy, because moving q through a potential difference \Delta V changes its potential energy by q\,\Delta V.
W=-q\,\Delta V
Equate the two expressions for W and cancel q. The minus sign shows that the work is done against the field.
qE\,\Delta r=-q\,\Delta V\;\Rightarrow\;E\,\Delta r=-\Delta V
Divide by \Delta r.
E=-\dfrac{\Delta V}{\Delta r}

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

The field strength is the potential gradient. Ignoring the sign, which shows that the potential falls in the direction of the field, E=\dfrac{\Delta V}{\Delta r}. The unit is volt per metre.

The ratio \Delta V/\Delta r is the potential gradient. The field strength is equal to the potential gradient, and the minus sign shows that the potential falls in the direction of the field.

2026 Annual I, Set D, Q.63+3 = 6 marks

a. What is meant by term protentional gradient? Derive its formula. b. Show that \dfrac{\text{volt}}{\text{meter}} = \dfrac{\text{newton}}{\text{columb}}

Solution

(a) Potential gradient and its formula (3 marks)

The potential gradient is the rate of change of electric potential \Delta V with displacement \Delta r, that is \dfrac{\Delta V}{\Delta r}. It is the slope along which the potential changes.

Derivation Field strength as the potential gradient

Aim. Show that the field strength equals the potential gradient. Written as in the book, E=-\dfrac{\Delta V}{\Delta r}, and the magnitude of the gradient is E=\dfrac{\Delta V}{\Delta r}.

Assumptions.

A uniform field E exists between two parallel plates.
The test charge moves a small distance \Delta r from A to B along the field.

Figure 10.10 (a): Potential difference between two points A and B.

Work done on a test charge q moved through a displacement \Delta r is force times displacement.
W=F\,\Delta r
The force on a charge in a uniform field is F=qE.
W=qE\,\Delta r
The same work also equals the decrease in electric potential energy, because moving q through a potential difference \Delta V changes its potential energy by q\,\Delta V.
W=-q\,\Delta V
Equate the two expressions for W and cancel q. The minus sign shows that the work is done against the field.
qE\,\Delta r=-q\,\Delta V\;\Rightarrow\;E\,\Delta r=-\Delta V
Divide by \Delta r.
E=-\dfrac{\Delta V}{\Delta r}

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

The field strength is the potential gradient. Ignoring the sign, which shows that the potential falls in the direction of the field, E=\dfrac{\Delta V}{\Delta r}. The unit is volt per metre.

(b) Volt per metre equals newton per coulomb (3 marks)

By definition the volt is the joule per coulomb, because potential difference is work per unit charge, V=W/q. The work is force times distance, so the joule is the newton metre.

Write the volt in terms of newton and coulomb.
1\,\mathrm{V}=1\,\dfrac{\mathrm{J}}{\mathrm{C}}=1\,\dfrac{\mathrm{N\,m}}{\mathrm{C}}
Divide by the metre to get the unit of \Delta V/\Delta r. The metre cancels.
\dfrac{\mathrm{volt}}{\mathrm{metre}}=\dfrac{\mathrm{N\,m}}{\mathrm{C\,m}}=\dfrac{\mathrm{newton}}{\mathrm{coulomb}}

Both are units of the electric field strength, since E=-\Delta V/\Delta r and also E=F/q.

Exam focus

What is potential gradient? Derive E=-\dfrac{\Delta V}{\Delta r} for the field between two parallel plates. (long)
The most repeated question of the unit: 2026 Set B Q.2 (xi) OR, 2026 Set D Q.6 (a) (6 marks) and 2025 Set D Q.3 OR (7 marks); also comprehensive question 3 of the book.
Go to answer
Show that volt per metre equals newton per coulomb. (short)
2026 Set D Q.6 (b); follows from V=W/q and J=N\,m.
Go to answer
Numerical: find r from the field E and potential V of a point charge, using r=V/E. (numerical)
2026 Set D Q.2 (xiii) OR (the same numbers as numerical question 6 of the book).
Go to answer
Define electric field strength and give E=kQ/r^{2}. (long)
2026 Set B MCQ 12 (force per unit charge) and the first half of 2025 Set D Q.3 OR; comprehensive question 2.
Go to answer
Direction of the force on a positive or negative charge in a uniform field. (mcq)
2025 Set B MCQ 11 (electron) and 2025 Set D MCQ 12 (negative particle), both field south to north, answer south; MCQ 5 of the book.
Go to answer
What is a Faraday cage? Explain its working. (short)
2025 Set B Q.2 (x); comprehensive question 6.
Go to answer
How does the radius of a flexible ring change when it is charged? (short)
2025 Set B Q.2 (xii) (negative charge); short question 2 of the book asks it for positive charge.
Go to answer
Coulomb's law: statement, vector form, and numericals on the force and field of point charges. (long)
Not in the 2025 or 2026 new-syllabus papers, but it is the base of the unit; comprehensive question 1 and numericals 1 to 7 of the book.
Go to answer
Motion of a charged particle in a uniform electric field. (long)
Not in the 2025 or 2026 papers (only the direction MCQs were asked); comprehensive questions 4 and 7 and the unit SLOs.
Go to answer