(a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacit

Question

Q. (a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.

(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.

(OR)

(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

(b) Two identical point charges, q each, are kept 2 m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.

CBSECBSE Class XII Board 2019Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Energy Stored in a Capacitor

Charging a capacitor to charge Q and voltage V stores energy $U=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV$ in its field -- exactly HALF the total energy $QV$ that the charging battery actually supplies, since the battery delivers every increment of charge at its own full, constant voltage V while the capacitor's own voltage rises only gradually from 0 to V during charging. The other half is unavoidably dissipated as heat in the connecting wires and the battery's own internal resistance. …

Part (b)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: $\tau_{\text{max}} = pE$ . …

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(a) The battery does work against the rising capacitor voltage, storing $U=\frac12CV^2$ ; sharing the charge with an identical uncharged capacitor halves the stored energy, ratio $\frac12$ .

(b) The equatorial field of a dipole is $E=\frac{p}{4\pi\varepsilon_0(r^2+a^2)^{3/2}}$ (anti-parallel to $\vec p$ ), and the balancing charge $Q$ sits at the midpoint with magnitude $q/4$ and opposite sign.

Part (a)

Charging a capacitor and energy stored

Connecting a capacitor to a battery makes the battery pump electrons from the plate joined to its positive terminal to the plate joined to its negative terminal — a current flows in the wires only, since the dielectric between the plates is an insulator. Charge accumulates until the capacitor voltage rises to the battery emf, giving $Q=CV$ with an electric field (and stored energy) in the gap.

To find the energy, note that when the instantaneous charge is $q$ the voltage is $v=q/C$ , so transferring a further $dq$ costs

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

Integrating from $0$ to $Q$ ,

U=\int_0^Q\frac{q}{C}\,dq=\frac{Q^2}{2C}=\frac12CV^2=\frac12QV.

(b) Charge redistribution between two identical capacitors

The charged capacitor holds $Q=CV$ and $U_i=\frac12CV^2$ . After disconnecting the battery and connecting an identical uncharged capacitor, the two are in parallel: $C_{tot}=2C$ . Charge is conserved, so the common voltage is

V'=\frac{Q}{C_{tot}}=\frac{CV}{2C}=\frac{V}{2}.

The final energy is

$U_f=\frac12C_{tot}V'^2=\frac12(2C)\Big(\frac{V}{2}\Big)^2=\frac{CV^2}{4},\qquad \frac{U_f}{U_i}=\frac{CV^2/4}{CV^2/2}=\frac12.$ …

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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 25 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 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 D1 markMCQ
$Q. A capacitor of 100 μF is charged to 100 volt. The energy stored in it will be (A) 0.5 joule (B) 5 joule (C) 50 joule (D) 100 joule$
›Reveal solutionSolution

Energy stored in a capacitor U = ½CV² = 0.5 J.

The energy stored in a charged capacitor is

$U = \tfrac{1}{2}CV^{2}$

Substitute C = 100 μF = 100×10⁻⁶ F and V = 100 V: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The energy stored in a capacitor is in the form of$
(a) Kinetic energy
(b) Potential energy
(c) Heat energy
(d) Magnetic energy
›Reveal solutionSolution

Charging a capacitor does work against the electric field building up between its plates; this work is stored as electrostatic potential energy.

When a capacitor of capacitance $C$ is charged to potential difference $V$ (charge $Q=CV$ ), external work is done to move charge against the growing field between the plates. This work does not appear as kinetic, heat, or magnetic e …

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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 A11 markMCQ
$6\times10^{-6}$
(a) 0.12 J
(b) 0.24 J
(c) 0.6 J
(d) 12 J
›Reveal solutionSolution

(a) 0.12 J …

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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 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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Q.(a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.

Part (a)

Charging a capacitor and energy stored

(b) Charge redistribution between two identical capacitors

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