(a) An electric dipole of dipole moment vecp is placed in a uniform electric field vecE at an angle θ with it.

Question

\vec{p}

(i) maximum, and

(ii) half of maximum.

(b) Two point charges

q_1 = +1\ \mu

C and

q_2 = +4\ \mu

C are placed 2 m apart in air. At what distance from

q_1

along the line joining the two charges, will the net electric field be zero ?

(OR)

(a) Derive an expression for the energy stored in a parallel plate capacitor of capacitance C when charged up to voltage V. How is this energy stored in the capacitor ?

(b) A capacitor of capacitance 1

\mu

F is charged by connecting a battery of negligible internal resistance and emf 10 V across it. Calculate the amount of charge supplied by the battery in charging the capacitor fully.

CBSECBSE Class XII Board 2020Subjective· 5mImportance★★★★★

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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 — Capacitor Energy Storage

Capacitor Energy Storage

The Intuition

Charging a capacitor is like piling sand onto a growing heap. The first grain of charge lands on an empty plate easily. But every later bit of positive charge must be pushed onto a plate that is already positive, and it resists. So more and more work is needed as the plate fills up. All of that work does not disappear — it is stored in the capacitor as electrostatic potential energy, ready to be released later.

Building the Formula

Suppose at some instant during charging the capacitor already holds charge $q$ , so the voltage across it at that moment is $v = q/C$ . Moving one more small charge $dq$ onto the plate costs work:

dW = v\,dq = \frac{q}{C}\,dq

Adding up (integrating) all these small contributions as the charge builds from $0$ to a final value $Q$ gives the total work done:

W = \int_0^{Q} \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{Q^2}{2} = \frac{Q^2}{2C}

This work is exactly the energy $U$ stored in the charged capacitor.

Three Equivalent Forms

Using $Q = CV$ , the same stored energy can be written three ways — pick whichever matches the quantities you know:

U = \frac{Q^2}{2C} = \frac{1}{2}QV = \frac{1}{2}CV^2

Use $\dfrac{Q^2}{2C}$ when the charge is fixed (capacitor disconnected from the source).
Use $\dfrac{1}{2}CV^2$ when the voltage is fixed (capacitor stays connected to a battery).

Watch out

The factor of $\tfrac{1}{2}$ is essential. A common error is writing $U = QV$ . That would only be true if the full voltage $V$ acted while all the charge moved — but the voltage climbs steadily from $0$ to $V$ as the plates fill, so the effective average voltage is $V/2$ , giving $U = \tfrac{1}{2}QV$ .

Where the Energy Lives — Energy Density

The energy is stored in the electric field occupying the space between the plates, not on the plates themselves. For a parallel-plate capacitor this leads to a general result: energy stored per unit volume of field is

u = \frac{1}{2}\,\varepsilon_0 E^2

where $E$ is the field strength. Wherever an electric field exists, energy is stored there, with density proportional to $E^2$ .

A Quick Example

A $10\ \mu\text{F}$ capacitor is charged to $100\ \text{V}$ . The stored energy is:

U = \frac{1}{2}CV^2 = \frac{1}{2}\times(10\times10^{-6})\times(100)^2 = 0.05\ \text{J}

That 0.05 J can be released almost instantly — which is exactly how a camera flash works: charge slowly, discharge fast. …

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Part (a): $\vec\tau=\vec p\times\vec E$ , $\tau=pE\sin\theta$ (max at $90^\circ$ , half-max at $30^\circ$ / $150^\circ$ ); the net field of the two like charges is zero at $x=2/3$ m from $q_1$ .

Part (b): a charged capacitor stores $U=\tfrac12CV^2=Q^2/2C$ in the electric field between its plates; a $1\,\mu$ F capacitor at $10$ V draws $Q=CV=10^{-5}$ C.

Part (a) — Torque on a dipole and the zero-field point

Torque. A dipole $\vec p=q(2\vec a)$ in a uniform field $\vec E$ (angle $\theta$ between them) feels forces $+q\vec E$ and $-q\vec E$ : equal, opposite, non-collinear — a couple. The perpendicular distance between the force lines is $2a\sin\theta$ , so

\tau=(qE)(2a\sin\theta)=pE\sin\theta,\qquad\vec\tau=\vec p\times\vec E.

(i) Maximum torque $\tau_{max}=pE$ when $\sin\theta=1$ , i.e. $\theta=90^\circ$ (dipole $\perp$ field).
(ii) Half of maximum $\tau=\tfrac12pE$ when $\sin\theta=\tfrac12$ , i.e. $\theta=30^\circ$ or $150^\circ$ .

Zero-field point. $q_1=+1\,\mu$ C at $x=0$ , $q_2=+4\,\mu$ C at $x=2$ m. Between two like charges the fields oppose; let the null be at distance $x$ from $q_1$ :

$\frac{kq_1}{x^2}=\frac{kq_2}{(2-x)^2}\Rightarrow\frac{1}{x^2}=\frac{4}{(2-x)^2}.$ …

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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 19 on this concept.

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 SEM31 markMCQ
$Q. The capacitances of two conductors X and Y are 3C and C respectively. The conductor X is given a charge Q which the conductor shares with Y. The ratio of total shared energy to its total initial energy is$
(a) 9 : 16
(b) √3 : 2
(c) 3 : 4
(d) 4 : 3
›Reveal solutionSolution

Charge Q on X (capacitance 3C) redistributes over X and Y (total 4C in parallel). Energy = Q²/2C_total, so ratio = (1/4C)/(1/3C) = 3/4 — some energy is lost. Option (c).

Step 1 — initial energy (all charge Q on X, capacitance 3C):

U_i = Q²/(2·3C) = Q²/6C.

Step 2 — after connection, X and Y share charge at a common potential; total capacitance = 3C + C = 4C, and total charge is conserved at Q.

U_f = Q²/(2·4C) = Q²/8C.

…

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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 ANNUAL1 mark
$Q. Define the electric dipole moment.$
›Reveal solutionSolution

An electric dipole moment measures the strength and orientation of a pair of equal and opposite charges separated by a small distance.

An electric dipole consists of two equal and opposite point charges $+q$ and $-q$ separated by a small distance $2a$ . Its electric dipole moment is defined as the vector:

$\vec{p} = q(2\vec{a})$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. What orientation of an electric dipole in uniform electric field is said to be in stable equilibrium ?$
(a) θ = 0
(b) θ = 90
(c) θ = 120
(d) θ = 180
›Reveal solutionSolution
A dipole is in stable equilibrium when its dipole moment is parallel to the field, i.e. θ = 0°.

The torque on a dipole in a uniform field is $\tau = pE\sin\theta$ , and its potential energy is $U = -pE\cos\theta$ .

Both θ = 0° and θ = 180° give zero torque (equilibrium positions), but they are not equally stable:

At θ = 0°, $U = -pE$ (minimum energy) → any small angular displacement creates a restoring torque that brings p back to alignment with E. This is stable equilibrium. …
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CBSE 2024Set A11 markMCQ
$Q. The electric dipole placed in uniform electric field is unstable, if the angle between electric field and dipole moment is$
(a) $0^\circ$
(b) $60^\circ$
(c) $90^\circ$
(d) $180^\circ$
›Reveal solutionSolution

(d) $180^\circ$ …

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CBSE 2024Set IMPROVEMENT1 markMCQ
$_2$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Water is a polar molecule because its bent geometry separates the centres of positive and negative charge, giving it a permanent dipole moment.

Assertion (A) is correct: the water molecule H $_2$ O possesses a permanent electric dipole moment even without any external field, because it is a polar molecule. Reason (R) is also correct: a permanent dipole moment arises in molecules whose positive and negative charge centres do not coincide — in H $_2$ O, the bent (non-linear) shape of the molecule means the centre of the negative charge (near the oxygen) does …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The capacity of a condenser is 2\times10⁻⁶ F and its potential is 200 V. The energy released on discharging it fully will be$
(a) 0.02 J
(b) 0.04 J
(c) 0.08 J
(d) 0.16 J.
›Reveal solutionSolution

Energy stored in a charged capacitor is $E = (1/2)CV^2$ ; substituting the given values gives 0.04 J.

When a capacitor of capacitance $C$ is charged to potential difference $V$ , it stores energy

$E = (1/2)CV^2$

This energy is released (as heat/spark) when the capacitor is fully discharged.

Given: $C = 2 \times 10^{-6}$ F, $V = 200$ V.

…

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CBSE 2023Set F1 markMCQ
$Q. What is the angle between the electric dipole moment P and the electric field strength E when the dipole is in a stable equilibrium? (A) pi/4 (B) pi (C) pi/2 (D) 0$
›Reveal solutionSolution
Stable equilibrium ⇒ P parallel to E ⇒ angle = 0.

The potential energy of a dipole in a uniform field is $U = -pE\cos\theta$ , and the torque is $\tau = pE\sin\theta$ .

At $\theta = 0$ : $\tau = 0$ and $U = -pE$ (minimum energy) → stable equilibrium. …
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CBSE 2022Set M1 mark
$Q. Define 'electric dipole moment'.$
›Reveal solutionSolution

Dipole moment = charge × separation, directed −q → +q.

An electric dipole is a pair of equal and opposite charges $+q$ and $-q$ separated by a small distance $2a$ .

The electric dipole moment is defined as the product of the magnitude of either charge and the distance of separation between them:

$\vec{p} = q \,(2\vec{a})$ …
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Q.(a) An electric dipole of dipole moment p⃗\vec{p}p​ is placed in a uniform electric field E⃗\vec{E}E at an angle θ\thetaθ with it. Derive the expression for torque (τ\tauτ) acting on it. Find the orientation of the dipole relative to the electric field for which torque on it is

Part (a) — Torque on a dipole and the zero-field point

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Q.(a) An electric dipole of dipole moment $\vec{p}$ is placed in a uniform electric field $\vec{E}$ at an angle $\theta$ with it. Derive the expression for torque ( $\tau$ ) acting on it. Find the orientation of the dipole relative to the electric field for which torque on it is

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