Electrostatics deals with the study of forces, fields and potentials arising from static charges. Force and el

Question

Figure — Figure — CBSE 2023 55/4/1 Q34 — FigureFigure — CBSE 2023 55/4/1 Q34

Q. Electrostatics deals with the study of forces, fields and potentials arising from static charges. Force and electric field due to a point charge is basically determined by Coulomb's law. For symmetric charge configurations Gauss's law, which is also based on Coulomb's law, helps us to find the electric field. A charge / a system of charges like a dipole experiences a force / torque in an electric field. Work is required to be done to provide a specific orientation to a dipole with respect to an electric field. Answer the following questions based on the above :

(a) Consider a uniformly charged thin conducting shell of radius

R

. Plot a graph showing the variation of

|\vec{E}|

with distance

r

from the centre, for points

0 \le r \le 3R

(b) The figure shows the variation of potential

V

with

\dfrac{1}{r}

for two point charges

Q_1

and

Q_2

, where

V

is the potential at a distance

r

due to a point charge. Find

\dfrac{Q_1}{Q_2}

6 \times 10^{-7}\ \text{C-m}

is kept in a uniform electric field of

10^4\ \text{N/C}

such that the dipole moment and the electric field are parallel. Calculate the potential energy of the dipole.

(OR)

\vec{p}

is initially kept in a uniform electric field

\vec{E}

such that

\vec{p}

is perpendicular to

\vec{E}

. Find the amount of work done in rotating the dipole to a position at which

\vec{p}

becomes antiparallel to

\vec{E}

CBSECBSE Class XII Board 2023Subjective· 4mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Electric Potential

Electric Potential

The Idea

When you lift a book onto a shelf you do work against gravity, and the book gains gravitational potential energy that depends on where it sits. Electric charges behave the same way in an electric field. Electric potential is the electrical analogue of "height" in a gravitational field: it tells you the potential energy a unit charge would have at a given point.

Formally, the electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, slowly (without giving it kinetic energy), against the electric field.

V = \frac{W}{q_0}

Here $W$ is the work done to move a small test charge $q_0$ from infinity to the point. Because both $W$ and $q_0$ are scalars, electric potential is a scalar quantity — it has magnitude but no direction. Its SI unit is the volt:

1\ \text{V} = 1\ \text{J/C}

Potential Due to a Point Charge

For a single point charge $q$ , the potential at a distance $r$ from it is

V = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}

Notice it falls off as $1/r$ , whereas the electric field of a point charge falls off as $1/r^2$ . The potential is taken as zero at infinity, our chosen reference. A positive charge makes the potential around it positive; a negative charge makes it negative.

Potential Difference

Usually we care about the potential difference between two points $A$ and $B$ :

V_A - V_B = \frac{W_{B\to A}}{q_0}

the work per unit charge needed to move a charge from $B$ to $A$ . This is the quantity a voltmeter reads and the "voltage" that drives current in a circuit.

Relation Between Field and Potential

Field and potential are two views of the same thing. The electric field is the negative rate of change of potential with distance:

E = -\frac{dV}{dr}

The minus sign says the field points from high potential toward low potential — a positive charge, left free, rolls "downhill" in potential. Where the potential changes steeply, the field is strong.

Superposition

Because potential is a scalar, the potential due to several charges is just the algebraic sum of their individual potentials — no vectors, no angles:

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots\right)

This makes potential far easier to compute than the field, which needs vector addition. Once you have $V$ everywhere, you can get the field by differentiating.

Equipotential Surfaces

An equipotential surface is a surface on which the potential has the same value everywhere.

No work is done in moving a charge along an equipotential surface (since $\Delta V = 0$ ).
The electric field is always perpendicular to an equipotential surface. …

Part (b)Concept understanding — Dipole Alignment Energy

Dipole Alignment Energy — From Intuition to Formula

Imagine you have a tiny bar magnet — a compass needle. You know it always turns to point north. But what if you try to hold it pointing east? You feel a torque, a twisting force that wants to rotate it back. If you let go, it snaps to align with the field.

That "snap" releases energy. The energy that was stored in the misaligned configuration is called dipole alignment energy (or potential energy of a dipole in an external field).

The Core Intuition

A dipole (like a compass needle or a polar molecule) has two opposite "poles" — a north and a south, or a positive and a negative charge. When placed in an external field:

Aligned (parallel to the field): the dipole is in its lowest energy state — like a ball at the bottom of a valley.
Anti-aligned (opposite to the field): the dipole is in its highest energy state — like a ball balanced at the top of a hill.
Perpendicular: the energy is somewhere in between.

The energy depends on how much the dipole is twisted away from the field direction. The more you force it to point against the field, the more energy you store — like winding a spring.

The Precise Statement

For an electric dipole with dipole moment $\vec{p}$ placed in a uniform external electric field $\vec{E}$ , the potential energy of alignment is:

U = -\vec{p} \cdot \vec{E} = -pE\cos\theta

where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$ .

For a magnetic dipole (like a current loop or a compass needle) with magnetic moment $\vec{\mu}$ in a magnetic field $\vec{B}$ :

U = -\vec{\mu} \cdot \vec{B} = -\mu B\cos\theta

Why the Negative Sign?

This is the part that confuses most students. Let's break it down.

When $\theta = 0^\circ$ (aligned), $\cos\theta = 1$ , so $U = -pE$ . This is the minimum energy — the most stable configuration.

When $\theta = 180^\circ$ (anti-aligned), $\cos\theta = -1$ , so $U = +pE$ . This is the maximum energy — the least stable.

The negative sign is a convention that makes the aligned state the lowest energy. Think of it this way: the field does positive work to rotate the dipole from anti-aligned to aligned, so the dipole loses potential energy. The formula captures that loss as a negative value relative to the zero-energy reference (which is usually taken at $\theta = 90^\circ$ , where $U = 0$ ).

Watch out

A common mistake: thinking $U = \vec{p} \cdot \vec{E}$ (without the minus sign). That would make the aligned state highest energy — physically wrong. The dipole wants to align, so aligned must be lowest energy.

What It Physically Means

The alignment energy tells you:

How much work an external agent must do to rotate the dipole from aligned to some angle $\theta$ .
How stable the dipole is in a given orientation — the deeper the energy well (larger $p$ or $E$ ), the harder to knock it out of alignment.
The torque on the dipole: $\tau = -\frac{dU}{d\theta} = -pE\sin\theta$ , which matches the familiar $\vec{\tau} = \vec{p} \times \vec{E}$ .

A Quick Example

A water molecule has a permanent electric dipole moment $p = 6.2 \times 10^{-30} \text{ C·m}$ . In an electric field of $10^6 \text{ N/C}$ (a strong laboratory field): …

Start your 14-day free trial to unlock the full solution →

(a) A charged conducting shell has $E=0$ inside and $E\propto1/r^2$ outside (discontinuous at $r=R$ ); (b) from the $V$ vs $1/r$ slopes $Q_1/Q_2=3$ ; (c) a dipole parallel to the field has $U=-pE=-6\times10^{-3}\ \text{J}$ ; and the alternative (b) part gives work $W=pE$ to rotate a dipole from perpendicular to antiparallel.

Part (a)

(a) Field of a uniformly charged thin conducting shell.

By Gauss's law, a spherical Gaussian surface inside the shell encloses no charge, so $E=0$ for $0\le r<R$ . Just outside the surface the field is $\dfrac{\sigma}{\varepsilon_0}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$ , and for $r>R$ the whole charge acts as if concentrated at the centre:

E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}.

Graph: a flat line at $E=0$ from $r=0$ to $r=R$ ; a jump up to $kQ/R^2$ at $r=R$ ; then a $1/r^2$ curve for $r>R$ , reaching $kQ/(9R^2)$ at $r=3R$ . The field is discontinuous at the conductor's surface.

(b) Ratio $Q_1/Q_2$ from the $V$ – $1/r$ graph.

For a point charge $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}=kQ\left(\dfrac1r\right)$ , so a plot of $V$ against $1/r$ is a straight line through the origin with slope $kQ$ . The line for $Q_1$ makes $60^\circ$ and that for $Q_2$ makes $30^\circ$ with the $1/r$ axis:

m_1=\tan60^\circ=\sqrt3,\qquad m_2=\tan30^\circ=\frac{1}{\sqrt3}.

Since $Q\propto m$ ,

\frac{Q_1}{Q_2}=\frac{m_1}{m_2}=\frac{\sqrt3}{1/\sqrt3}=3.

(c) Potential energy of a dipole parallel to the field.

U=-\vec p\cdot\vec E=-pE\cos\theta,\qquad \theta=0^\circ.

$U=-(6\times10^{-7}\ \text{C·m})(10^4\ \text{N/C})(1)=-6\times10^{-3}\ \text{J}.$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 48 on this concept.

CBSE 2026Set 55/1/11 markMCQ
$r_1$
›Reveal solutionSolution
When two widely separated conducting spheres are connected by a wire, they reach the same electric potential. Since surface field $E = \frac{kQ}{r^2}$ and potential $V = \frac{kQ}{r}$ , combining these gives $E \propto 1/r$ . Therefore the ratio of surface fields is $E_A/E_B = r_2/r_1$ , which corresponds to option (B).

The key insight here is about what happens when conductors are connected by a wire. Charge flows until both spheres are at the same electric potential — that's the fundamental condition for electrostatic equilibrium in a conductor. Once you grasp that, the rest is just algebra.

Let's think about why potential equality is the right starting point. A conducting wire means the two spheres form a single conductor. In electrostatics, the entire surface of a conductor is an equipotential. So spheres A and B must have the same potential $V$ .

Now, for an isolated conducting sphere of radius $r$ carrying charge $Q$ , the potential at its surface (taking infinity as zero) is:
$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$
And the electric field just outside its surface is:
$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$
Notice the relationship: $E = V/r$ . That's a neat shortcut we'll use.

Tip

For any isolated conducting sphere, $E = V/r$ directly. This saves you from carrying the $Q$ through the algebra — just remember it comes from $V = kQ/r$ and $E = kQ/r^2$ .

Let's work through it step by step.

Set potentials equal. Since the wire connects them, $V_A = V_B$ . Using $V = kQ/r$ (where $k = 1/4\pi\epsilon_0$ ):

$k\frac{Q_A}{r_1} = k\frac{Q_B}{r_2}$
Cancel $k$ and rearrange:
$\frac{Q_A}{Q_B} = \frac{r_1}{r_2}$

Write the surface field ratio. For each sphere, $E = kQ/r^2$ . So: $\frac{E_A}{E_B} = \frac{k Q_A / r_1^2}{k Q_B / r_2^2} = \frac{Q_A}{Q_B} \cdot \frac{r_2^2}{r_1^2}$ …
Start your 14-day free trial to see the full solution →
CBSE 2026Set 55/3/11 markMCQ
$m$
›Reveal solutionSolution
Work done by a constant electric field equals force times displacement; since the particle starts from rest, all that work converts to kinetic energy, giving $K = q E_0 x$ .

The heart of this problem is the work-energy theorem: the work done by all forces on a particle equals its change in kinetic energy. When a charged particle moves through an electric field, the field exerts a force that does work, and if the particle starts from rest, every joule of work becomes kinetic energy.

A uniform electric field $\vec{E} = E_0\,\hat{i}$ exerts a force $\vec{F} = q\vec{E}$ on a charge $q$ . This force is constant in magnitude and direction, so the work done is simply force times displacement along the direction of the force.

Step-by-step reasoning

Identify the force on the particle. The electric force on a charge $q$ in field $\vec{E}$ is

$\vec{F} = q\vec{E} = q E_0\,\hat{i}$
The magnitude is $F = q E_0$ , directed along the positive $x$ -axis.

Calculate the work done by this force. The particle moves a distance $x$ along the $x$ -axis, in the same direction as the force. Work done by a constant force is

$W = F \cdot d = q E_0 \cdot x = q E_0 x$

Apply the work-energy theorem. The particle starts from rest, so initial kinetic energy $K_i = 0$ . The work-energy theorem states

$W = \Delta K = K_f - K_i$
Therefore,
$K_f = W = q E_0 x$
The kinetic energy after travelling distance $x$ is simply the work done by the electric field. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. SI unit of electric potential is:$
(a) Ohm
(b) Volt
(c) Coulomb
(d) Ampere
›Reveal solutionSolution

Electric potential is defined as work done per unit charge, so its SI unit is the Volt.

Electric potential at a point is $V = \dfrac{W}{q}$ , i.e. the work done in bringing a unit positive charge from infinity to that point. Since work is measured in joules (J) and charge in coulombs (C), the unit o …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. [Case study, continued] The charges q1 and q2 produce a potential, which at any point P will be$
(a) V1,2 = 1/(4πε₀) × (q1/r1P² + q2/r2P²)
(b) V1,2 = 1/(4πε₀) × (q1/r1P + q2/r2P)
(c) V1,2 = 1/(4πε₀) × (q1/r1P - q2/r2P)
(d) V1,2 = 1/(4πε₀) × (2q1/r1P + 3q2/r2P)
›Reveal solutionSolution

The potential due to a group of point charges at any point is just the plain algebraic (scalar) sum of the potentials each charge produces there — the superposition principle for potential.

Electric potential obeys the superposition principle: the total potential at any point due to several charges equals the SCALAR sum (not vector sum, since potential is a scalar) of the potentials due to each individual charge, each given by $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ (distance to the first power).

For two charges $q_1$ (at distance $r_{1P}$ from P) and $q_2$ (at distance $r_{2P}$ from P):

$V_{1,2} = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_{1P}} + \frac{q_2}{r_{2P}}\right)$

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The standard potential of earth is :$
(a) Zero
(b) Infinite
(c) One
(d) None of the above
›Reveal solutionSolution

The Earth is the chosen reference for potential, so its standard potential is taken as zero.

Electric potential is always measured relative to some reference. Because the Earth is a very large conductor whose potential is practically unaffected by adding or removing charge, it is universally chosen as the reference (zero) level of potential.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Match the Column A with Column B and write the correct pair. Column A: Potential gradient. Column B:$
(i) μ₀nI,
(ii) Volt × second,
(iii) μ₀nI/2,
(iv) Volt × meter,
(v) Volt × meter⁻¹,
(vi) Volt.
›Reveal solutionSolution

Potential gradient dV/dx has unit V/m = Volt × metre⁻¹, option (v).

The potential gradient is the rate of change of electric potential with distance, dV/dx. Its SI unit is volt per metre (V/m), i.e. Volt × metre⁻¹. In magnitude it equals th …

Start your 14-day free trial to see the full solution →
CBSE 2025Set 55/5/11 markMCQ
$\vec{E}=\dfrac{\alpha}{r^{2}}\,\hat{r}$
›Reveal solutionSolution
For a radial electric field $\vec{E} = \frac{\alpha}{r^2}\,\hat{r}$ , integrate $-\vec{E} \cdot d\vec{l}$ along a radial path from infinity to find the potential; the magnitude is $\frac{\alpha}{r}$ .

The connection between electric field and potential is one of the most fundamental relationships in electrostatics. The electric field points in the direction of steepest decrease of potential, and its magnitude tells us how rapidly the potential drops. Mathematically, $\vec{E} = -\nabla V$ , or in one dimension, $E = -\frac{dV}{dr}$ for a radial field.

To find the potential at a point, we integrate the electric field along a path. The potential difference between two points is:
$V(r) - V(r_0) = -\int_{r_0}^{r} \vec{E} \cdot d\vec{l}$
We conventionally choose $r_0 = \infty$ as our reference point where $V(\infty) = 0$ , so:
$V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{l}$
Now let's work through this problem step by step.

Set up the line integral for a radial field.

Since both $\vec{E}$ and the path element $d\vec{l}$ point radially (we choose a radial path for simplicity), we have:

$\vec{E} \cdot d\vec{l} = E_r \, dr = \frac{\alpha}{r^2} \, dr$

Evaluate the integral from infinity to $r$ .

$V(r) = -\int_{\infty}^{r} \frac{\alpha}{r^2} \, dr$
Reversing the limits to make the calculation cleaner:
$V(r) = \int_{r}^{\infty} \frac{\alpha}{r^2} \, dr$

Perform the integration.

$V(r) = \alpha \int_{r}^{\infty} \frac{1}{r^2} \, dr = \alpha \left[-\frac{1}{r}\right]_{r}^{\infty}$ $V(r) = \alpha \left(0 - \left(-\frac{1}{r}\right)\right) = \frac{\alpha}{r}$

Interpret the result. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set IMPROVEMENT1 markMCQ
$Q. Electron volt is the unit of which of the following?$
(a) Electric Potential
(b) Energy
(c) Electric Field
(d) Electric Current
›Reveal solutionSolution

Electron volt (eV) is a unit of energy, not of potential, field or current.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set D1 markMCQ
$Q. The speed of an electron accelerated from rest under a potential difference V is (A) proportional to V (B) proportional to \sqrtV (C) proportional to 1/V (D) proportional to V^2$
›Reveal solutionSolution

The kinetic energy gained equals eV; since ½mv² = eV, the speed v ∝ √V.

An electron accelerated from rest through a potential difference V gains kinetic energy equal to the work done by the field:

$eV = \frac{1}{2}mv^2$

Solving for the speed:

$v = \sqrt{\frac{2eV}{m}}$

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Two charges 1C and -1C are placed 1 m apart. The potential at the centre of the line joining the two charges will be$
(a) 2 V
(b) -2 V
(c) zero
(d) 0.5 V
›Reveal solutionSolution

At the midpoint of the line joining equal-and-opposite charges, the two charges are equidistant, so their potential contributions are equal in magnitude but opposite in sign and cancel exactly.

Electric potential due to a point charge q at distance r is V = kq/r (k = 1/4-pi-epsilon0).

The two charges are q1 = +1 C and q2 = -1 C, separated by 1 m, so each is r = 0.5 m from the midpoint.

V = k(+1)/0.5 + k(-1)/0.5 = 2k - 2k = 0

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. When a photon is accelerated (from rest) through a potential difference of one volt, the kinetic energy gained by it is equal to$
(a) 1837eV
(b) 1 eV
(c) $\frac{1}{1837}$ eV
(d) none of the above.
›Reveal solutionSolution

Energy gained by a charge accelerated through a potential difference depends only on charge × voltage, not on mass — so it is 1 eV regardless of which particle is being accelerated.

Work-energy theorem for a charge q accelerated from rest through potential difference V:

$KE = qV$

For an elementary charge $q = e = 1.6\times10^{-19}\ \text{C}$ and $V = 1\ \text{V}$ :

$KE = (1.6\times10^{-19})(1) = 1.6\times10^{-19}\ \text{J} = 1\ \text{eV}$ (by definition of the electron-volt)

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Electric potential 'V' at a distance 'r' from a point charge is directly proportional to ______.$
(a) r
(b) r²
(c) 1/r
(d) 1/r²
›Reveal solutionSolution

The electric potential of a point charge falls off as the inverse of the distance from it.

The electric potential at a distance $r$ from a point charge $q$ is

$V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$

…

Start your 14-day free trial to see the full solution →

Part (a)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days