(a) (i) Derive an expression for the potential energy of an electric dipole vecp in an external uniform electr

Question

Q. (a)

(i) Derive an expression for the potential energy of an electric dipole

\vec{p}

in an external uniform electric field

\vec{E}

. When is the potential energy of the dipole

(1) maximum, and

(2) minimum ?

(ii) An electric dipole consists of point charges

-1.0\ \text{pC}

and

+1.0\ \text{pC}

located at

(0, 0)

and

(3\ \text{mm}, 4\ \text{mm})

respectively in the

x

–

y

plane. An electric field

\vec{E} = \left(1000\ \dfrac{\text{V}}{\text{m}}\right)\hat{i}

is switched on in the region. Find the torque

\vec{\tau}

acting on the dipole.

(OR)

(b)

(i) An electric dipole (dipole moment

\vec{p} = p\hat{i}

), consisting of charges

-q

and

q

, separated by distance

2a

, is placed along the

x

-axis with its centre at the origin. Show that the potential

V

, due to this dipole, at a point

x

(x \gg a)

is equal to

\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{\vec{p}\cdot\hat{i}}{x^2}

(ii) Two isolated metallic spheres

S_1

and

S_2

of radii

1\ \text{cm}

and

3\ \text{cm}

respectively are charged such that both have the same charge density

\left(\dfrac{2}{\pi}\times10^{-9}\right)\ \text{C/m}^2

. They are placed far away from each other and connected by a thin wire. Calculate the new charge on sphere

S_1

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Part (a)Concept understanding — Electric Dipole in a Field

Electric Dipole in a Uniform Field — First Look

Imagine you have a tiny bar magnet. If you place it in a uniform magnetic field, it doesn't get pulled anywhere — but it does twist to align with the field. An electric dipole behaves in exactly the same way when placed in a uniform electric field.

An electric dipole is simply a pair of equal and opposite charges, $+q$ and $-q$ , separated by a small distance $d$ . Think of it as a tiny "stretched" charge. The dipole has a dipole moment $\vec{p}$ , a vector that points from the negative charge to the positive charge, with magnitude $p = qd$ .

Now place this dipole in a uniform electric field $\vec{E}$ . Uniform means the field has the same strength and direction everywhere in that region.

What happens? Two forces, one twist

The positive charge feels a force $F_+ = +qE$ in the direction of the field. The negative charge feels a force $F_- = -qE$ opposite to the field. These two forces are equal in magnitude but opposite in direction — so they cancel out as far as net force is concerned. The dipole as a whole does not accelerate linearly.

But the forces are not along the same line. They are separated by the distance $d$ , so they form a couple — a pair of equal, opposite, parallel forces that produce a torque. This torque tries to rotate the dipole so that its dipole moment $\vec{p}$ aligns with the field $\vec{E}$ .

Note

No net force means the centre of mass of the dipole stays put. Only rotation happens.

The torque formula

Let the dipole make an angle $\theta$ with the field direction (so $\theta = 0$ when $\vec{p}$ and $\vec{E}$ point the same way). The lever arm for each force about the centre is $(d/2)\sin\theta$ . The torque from each force is $F \times \text{lever arm} = qE \cdot (d/2)\sin\theta$ . Since both forces contribute in the same rotational sense, the total torque is:

\tau = 2 \cdot qE \cdot \frac{d}{2}\sin\theta = qd\,E\sin\theta

But $qd = p$ , the dipole moment. So:

\tau = pE\sin\theta

The direction of the torque is such that it tries to reduce $\theta$ — to bring $\vec{p}$ into alignment with $\vec{E}$ . In vector form:

\vec{\tau} = \vec{p} \times \vec{E}

What does this mean physically?

When $\theta = 0$ (dipole aligned with field), $\sin 0 = 0$ , so torque is zero. This is the stable equilibrium position.
When $\theta = 90^\circ$ (dipole perpendicular to field), torque is maximum: $\tau_{\text{max}} = pE$ . …

Part (b)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. …

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Part (a): $U=-\vec p\cdot\vec E$ , minimum ( $-pE$ ) when aligned, maximum ( $+pE$ ) when anti-aligned; for the given dipole $\vec\tau=-4.0\times10^{-12}\,\hat k$ N·m.

Part (b): the axial potential of a short dipole is $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec p\cdot\hat i}{x^2}$ ; after connecting the two spheres, the new charge on $S_1$ is $2\times10^{-12}$ C.

Part (a) — Dipole potential energy and torque

(i) Potential energy. In a uniform field the dipole feels no net force but a torque $\tau=pE\sin\theta$ (from the couple of forces $\pm q\vec E$ ). Rotating it by $d\theta$ needs work $dW=\tau\,d\theta=pE\sin\theta\,d\theta$ . Choosing the reference $U=0$ at $\theta=90^\circ$ ,

U(\theta)=\int_{90^\circ}^{\theta}pE\sin\theta'\,d\theta'=pE[-\cos\theta']_{90^\circ}^{\theta}=-pE\cos\theta.

\boxed{U=-\vec p\cdot\vec E}

Minimum $U=-pE$ at $\theta=0^\circ$ (stable, aligned with $\vec E$ ).
Maximum $U=+pE$ at $\theta=180^\circ$ (unstable, anti-aligned).

(ii) Torque. Displacement from $-q$ to $+q$ : $\vec d=(3\hat i+4\hat j)\times10^{-3}$ m. With $q=1.0\times10^{-12}$ C,

\vec p=q\vec d=(3\hat i+4\hat j)\times10^{-15}\ \text{C·m}.

\vec\tau=\vec p\times\vec E=(3\hat i+4\hat j)\times10^{-15}\times(1000\,\hat i).

Since $\hat i\times\hat i=0$ and $\hat j\times\hat i=-\hat k$ :

$\vec\tau=4\times10^{-15}\times1000\,(-\hat k)=-4.0\times10^{-12}\,\hat k\ \text{N·m}.$ …

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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 60 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: $\frac{E_A}{E_B} = \frac{k Q_A / r_1^2}{k Q_B / r_2^2} = \frac{Q_A}{Q_B} \cdot \frac{r_2^2}{r_1^2}$ …
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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. …
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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 …

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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):

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

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The size of ideal dipole is :$
(a) Zero
(b) Infinite
(c) One
(d) None of the above
›Reveal solutionSolution

An ideal dipole is the point-dipole limit: separation → 0 with dipole moment finite, so its size is zero.

An electric dipole consists of two equal and opposite charges +q and −q separated by a small distance 2a, with dipole moment p = q(2a). An 'ideal' (or point) dipole is the mathematical limit in which the separation 2a is made vanishingly small (2a → 0) while simultaneously making q very large …

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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.

…

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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 …

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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. …
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CBSE 2025Set X11 mark
$Q. An electric dipole placed in a uniform electric field experiences a net ________.$
›Reveal solutionSolution

torque In a uniform field the forces on the two charges of the dipole are equal and opposite, so the net force is zero, but they act along different lines and constitute a couple that produces a …

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CBSE 2025Set ANNUAL1 mark
$+q$
(i) Write the unit of electric dipole moment.
›Reveal solutionSolution

Dipole moment $p=q\times2a$ has units of charge $\times$ distance.

The electric dipole moment is defined as $p = q\times(2a)$ , the product of the magnitude of either charge and the separation between the two charges. Since charge is measured in coulombs (C) and separation in metres (m), the SI unit …

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CBSE 2025Set ANNUAL1 mark
$+q$
(iii) What is polar molecule?
›Reveal solutionSolution

Polar molecules have a built-in charge asymmetry that gives them a permanent dipole moment.

A polar molecule is one in which the centre of the positive charge distribution and the centre of the negative charge distribution do not coincide, due to the asymmetric arrangement of atoms and the unequal sharing of electrons between them (unequal electronegativities). As a result, such a molecule possesses a permanent (built-in) electric dipole moment, which exists even in the absence of any external electric field. Common examples include water (H₂O), HCl, and NH₃. This is in contrast to a non-polar molecule (like $\text{O}_2$ or $\text{CO}_2$ ), where the positive and negative charge centres coincide, giving zero perma …

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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.

…

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Q.(a)

Part (a) — Dipole potential energy and torque

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