PhysicsCBSE Class 12 · Board exam 2025, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2025: question 31(b)

Set in the CBSE Class 12 board exam of 2025, paper 55/1/1, section E, question 31(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

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

(i) Consider three metal spherical shells A, B and C, each of radius R. Each shell is having a concentric metal ball of radius R/10. The spherical shells A, B and C are given charges +6q, –4q, and 14q respectively. Their inner metal balls are also given charges –2q, +8q and –10q respectively. Compare the magnitude of the electric fields due to shells A, B and C at a distance 3R from their centres.
(ii) A charge –6 μC is placed at the centre B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge +5 μC is moved from point 'C' to point 'A' along the circumference. Calculate the work done on the charge.

Figure in the paper: Along a horizontal line: point D (charge +6q) on the left, then point C, then point B (charge –6q) at the centre of a semicircle, and point A at the right end. The semicircle of radius 5 cm is drawn above the line from C to A with B at the centre. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/1/1.

Hints

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

Hint 1

Outside a spherically symmetric charge distribution, only the total charge matters. Add the shell and ball charges.

Hint 2

Work in moving a charge depends only on the end potentials. Find the distances of C and A from D and B, then use $W = q (V_{A} - V_{C})$ .

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) At $r = 3 R$ (outside each shell) the field is that of the total charge enclosed, as if placed at the centre: $E = \frac{k Q _{t o t a l}}{r ^{2}}$ .
½ mark
Total charges: A: $6 q - 2 q = 4 q$ ; B: $- 4 q + 8 q = 4 q$ ; C: $14 q - 10 q = 4 q$ .
½ mark
$E = \frac{k ( 4 q )}{( 3 R ) ^{2}} = \frac{4 k q}{9 R ^{2}}$ for each, so $E_{A} : E_{B} : E_{C} = 1 : 1 : 1$ (all equal).
1 mark
(ii) From the figure, $D C = 5$ cm, $B C = B A = 5$ cm and $D A = 15$ cm, with +6 μC at D and −6 μC at B. $V_{C} = \frac{k ( 6 \times 1 0 ^{- 6} )}{0.05} - \frac{k ( 6 \times 1 0 ^{- 6} )}{0.05} = 0$ .
1 mark
$V_{A} = k (6 \times 1 0^{- 6}) (\frac{1}{0.15} - \frac{1}{0.05}) = 9 \times 1 0^{9} \times 6 \times 1 0^{- 6} \times (- \frac{40}{3})$ .
½ mark
$V_{A} = - 7.2 \times 1 0^{5}$ V.
½ mark
Work done (by external agent) on the +5 μC charge: $W = q (V_{A} - V_{C}) = 5 \times 1 0^{- 6} \times (- 7.2 \times 1 0^{5} - 0) = - 3.6$ J. The work done by the electric field is +3.6 J. The work is path independent, so moving along the semicircle makes no difference.
1 mark

Answer: (i) E_A = E_B = E_C = 4kq/(9R²), ratio 1:1:1; (ii) W = −3.6 J

Where the marks are

Uses total enclosed charge (4q for each) and $E = k Q / r^{2}$ (1 mark)
Concludes the fields are equal, 1 : 1 : 1 (1 mark)
Finds $V_{C} = 0$ (1 mark)
Finds $V_{A} = - 7.2 \times 1 0^{5}$ V (1 mark)
Gets $W = q (V_{A} - V_{C}) = - 3.6$ J (1 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 paper2025 board paper, set 55/1/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Physics 2025: question 31(b)

Hints

Step-by-step solution

Revise the chapter

More questions from Electric Charges and Fields