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

CBSE Class 12 Physics 2025: question 31(a)

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

(i) Two point charges 5 μC and –1 μC are placed at points (–3 cm, 0, 0) and (3 cm, 0, 0) respectively. An external electric field $E = \frac{A}{r ^{2}} \overset{r}{^}$ where $A = 3 \times 1 0^{5}$ Vm is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.
(ii) A system of two conductors is placed in air and they have net charge of +80μC and –80μC which causes a potential difference of 16 V between them.
(1) Find the capacitance of the system.
(2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors ?
(3) If the charges on two conductors are changed to +160 μC and –160 μC, will the capacitance of the system change ? Give reason for your answer.

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

Find the potential of the external field from $V = - \int E \cdot d r$ . Both charges are 3 cm from the origin.

Hint 2

The mutual interaction energy does not change, so only $q V$ terms are added. In (ii), the charge is fixed when the dielectric is inserted.

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) The potential of the external field: $V (r) = - \int_{\infty}^{r} E \cdot d r = \frac{A}{r} = \frac{3 \times 1 0 ^{5}}{r}$ .
½ mark
Without the field, $U_{i} = \frac{k q _{1} q _{2}}{r _{12}}$ . With the field, $U_{f} = \frac{k q _{1} q _{2}}{r _{12}} + q_{1} V (r_{1}) + q_{2} V (r_{2})$ .
½ mark
So $Δ U = q_{1} V (r_{1}) + q_{2} V (r_{2})$ , with $r_{1} = r_{2} = 3$ cm $= 3 \times 1 0^{- 2}$ m, giving $V = \frac{3 \times 1 0 ^{5}}{3 \times 1 0 ^{- 2}} = 1 0^{7}$ V at both charges.
½ mark
$Δ U = (5 \times 1 0^{- 6} - 1 \times 1 0^{- 6}) \times 1 0^{7} = 40$ J.
½ mark
(ii)(1) $C = \frac{Q}{V} = \frac{80 μ C}{16 V} = 5 μ$ F.
1 mark
(ii)(2) With the dielectric, $C^{'} = K C = 3 \times 5 = 15 μ$ F. The charge stays 80 μC, so $V^{'} = \frac{Q}{C ^{'}} = \frac{80}{15} \approx 5.33$ V.
1 mark
(ii)(3) No. Capacitance depends only on the size, shape and separation of the conductors and on the medium. Doubling the charge doubles V (to 32 V), so $C = Q / V$ stays 5 μF.
1 mark

Answer: (i) ΔU = 40 J; (ii)(1) 5 μF; (2) ≈ 5.33 V; (3) No, capacitance depends on geometry and medium, not charge

Where the marks are

Obtains $V = A / r$ from the given field (½ mark)
Writes $U_{i}$ and $U_{f}$ , giving $Δ U = q_{1} V_{1} + q_{2} V_{2}$ (1 mark)
Substitutes to get $Δ U = 40$ J (½ mark)
$C = 5 μ$ F (1 mark)
$C^{'} = 15 μ$ F and $V^{'} = 5.33$ V (1 mark)
Answers 'No' with the geometry reason (½ + 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 55/1/1: every question, by section.

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CBSE Class 12 Physics 2025: question 31(a)

Hints

Step-by-step solution

Revise the chapter

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