PhysicsCBSE Class 12 · Board exam 2025, set 55/7/1 · 5 marks · Long answer

CBSE Class 12 Physics 2025: question 31(a)

Set in the CBSE Class 12 board exam of 2025, paper 55/7/1, 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 answerElectrostatic Potential and Capacitance

(i) A parallel plate capacitor with plate area A and plate separation d has a capacitance $C_{0}$ . A slab of dielectric constant K having area A and thickness $(\frac{d}{4})$ is inserted in the capacitor, parallel to the plates. Find the new value of its capacitance.
(ii) You are provided with a large number of 1 $μ$ F identical capacitors and a power supply of 1200 V. The dielectric medium used in each capacitor can withstand up to 200 V only. Find the minimum number of capacitors and their arrangement, required to build a capacitor system of equivalent capacitance of 2 $μ$ F for use with this supply.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/7/1.

Hints

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

Hint 1

Inside the slab the field is reduced by a factor K; add the potential drops across the air gap and the slab.

Hint 2

For (ii), first decide how many capacitors in series share 1200 V safely, then how many such rows in parallel give 2 μF.

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) Without the slab, $C_{0} = \frac{ε _{0} A}{d}$ . With the slab, $E_{0} = \frac{σ}{ε _{0}}$ in the air gap (thickness $d - t$ ) and $E = \frac{E _{0}}{K}$ inside the dielectric (thickness $t$ ).
½ mark
$V = E_{0} (d - t) + \frac{E _{0}}{K} t = \frac{Q}{ε _{0} A} (d - t + \frac{t}{K})$ , so $C = \frac{Q}{V} = \frac{ε _{0} A}{d - t + \frac{t}{K}}$ .
½ mark
With $t = \frac{d}{4}$ : $C = \frac{ε _{0} A}{\frac{3 d}{4} + \frac{d}{4 K}} = \frac{ε _{0} A}{d} \cdot \frac{4 K}{3 K + 1}$ .
1 mark
$C = \frac{4 K}{3 K + 1} C_{0}$
1 mark
(ii) Each capacitor can take at most 200 V, so each series row must have $n_{s} = \frac{1200}{200} = 6$ capacitors. Capacitance of one row $= \frac{1}{6} μ$ F.
1 mark
Let m such rows be connected in parallel: $C_{e q} = \frac{m}{6} = 2 \Rightarrow m = 12$ .
½ mark
Minimum number $= 6 \times 12 = 72$ capacitors: 12 parallel rows, each with 6 capacitors in series.
½ mark

Answer: (i) C = 4K C₀/(3K+1); (ii) 72 capacitors: 12 rows in parallel, each with 6 capacitors in series.

Where the marks are

Fields/potential difference across the air gap and the slab (½ + ½ mark)
Substitution with t = d/4 (1 mark)
C = 4KC₀/(3K+1) (1 mark)
6 capacitors in series per row, row capacitance 1/6 μF (1 mark)
12 rows in parallel (½ mark)
Total 72 capacitors (½ mark)

Revise the chapter

Chapter: Electrostatic Potential and Capacitance (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Electrostatic Potential and Capacitance, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 55/7/1: 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(a)

Hints

Step-by-step solution

Revise the chapter

More questions from Electrostatic Potential and Capacitance