PhysicsCBSE Class 12 · Board exam 2024, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2024: question 31(a)

Set in the CBSE Class 12 board exam of 2024, paper 55/1/1, section E, question 31(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(a) · 5 marks · Long answerElectric Charges and Fields

(i) Derive an expression for potential energy of an electric dipole $p$ in an external uniform electric field $E$ . When is the potential energy of the dipole (1) maximum, and (2) minimum?
(ii) An electric dipole consists of point charges $- 1.0$ pC and $+ 1.0$ pC located at (0, 0) and (3 mm, 4 mm) respectively in x – y plane. An electric field $E = (\frac{1000 V}{m}) \hat{i}$ is switched on in the region. Find the torque $τ$ acting on the dipole.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Potential energy equals the work done by an external torque in rotating the dipole slowly.

Hint 2

Integrate $pE sin θ$ and take $U = 0$ when the dipole is perpendicular to the field.

Hint 3

For (ii), $p$ runs from −q to +q; then compute $p \times E$ with unit-vector cross products.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) A dipole in a uniform field $E$ feels a torque of magnitude $τ = pE sin θ$ . To rotate it slowly from $θ_{0}$ to $θ_{1}$ , an external torque equal and opposite to this must act. The work it does is $W = \int_{θ_{0}}^{θ_{1}} τ_{e x t} d θ = \int_{θ_{0}}^{θ_{1}} pE sin θ d θ$ .
½ mark
$W = pE (cos θ_{0} - cos θ_{1})$ . This work is stored as potential energy: $U (θ_{1}) - U (θ_{0}) = pE (cos θ_{0} - cos θ_{1})$ .
½ mark
Take the zero of potential energy at $θ_{0} = π /2$ (dipole perpendicular to the field) and write $θ_{1} = θ$ : $U (θ) = pE (cos \frac{π}{2} - cos θ)$ .
½ mark
$U (θ) = - pE cos θ = - p \cdot E$
½ mark
(1) U is maximum ( $+ pE$ ) when $θ = 18 0^{\circ}$ , i.e. $p$ is antiparallel to $E$ .
½ mark
(2) U is minimum ( $- pE$ ) when $θ = 0^{\circ}$ , i.e. $p$ is along $E$ (stable equilibrium).
½ mark
(ii) The dipole moment points from −q to +q: $p = q d = 1.0 \times 1 0^{- 12} (3 \times 1 0^{- 3} \hat{i} + 4 \times 1 0^{- 3} \hat{j})$ .
½ mark
$p = (3 \hat{i} + 4 \hat{j}) \times 1 0^{- 15}$ C m. The torque is $τ = p \times E$ .
½ mark
$τ = (3 \hat{i} + 4 \hat{j}) \times 1 0^{- 15} \times 1 0^{3} \hat{i} = 4 \times 1 0^{- 12} (\hat{j} \times \hat{i})$ , using $\hat{i} \times \hat{i} = 0$ .
½ mark
Since $\hat{j} \times \hat{i} = - \hat{k}$ : $τ = - 4 \times 1 0^{- 12} \hat{k}$ N m, i.e. magnitude $4 \times 1 0^{- 12}$ N m along the negative z-direction.
½ mark

Answer: (i) $U = - p \cdot E$ ; maximum at θ = 180°, minimum at θ = 0°. (ii) $τ = - 4 \times 1 0^{- 12} \hat{k}$ N m

Where the marks are

Writes work done as $\int pE sin θ d θ$ (½ mark)
Obtains $W = pE (cos θ_{0} - cos θ_{1})$ (½ mark)
Chooses the reference $θ_{0} = π /2$ (½ mark)
Obtains $U = - pE cos θ = - p \cdot E$ (½ mark)
Maximum when antiparallel (½ mark)
Minimum when parallel (½ mark)
Writes $p$ correctly (½ mark)
Uses $τ = p \times E$ (½ mark)
Magnitude $4 \times 1 0^{- 12}$ N m (½ mark)
Direction along −z (½ mark)

Revise the chapter

Chapter: Electric Charges and Fields (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Electric Charges and Fields, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2024: question 31(a)

Hints

Step-by-step solution

Revise the chapter

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