PhysicsCBSE Class 12 · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2026-27: question 31(B)

From the official CBSE sample question paper for 2026-27, 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 answerElectrostatic Potential and Capacitance

I. Show that the electric field at any point is equal to the negative of the potential gradient at that point.
II. A small metallic sphere is gradually charged in open air. It is observed that electrical breakdown of air occurs when the electric field at its surface reaches $2.5 \times 1 0^{6}$ V m $^{- 1}$ . If the sphere is allowed to attain a maximum potential of $5 \times 1 0^{5}$ V without causing any discharge, determine the minimum radius the insulated sphere must have.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Relate the work done by the field on a test charge to the change in potential.

Hint 2

For a sphere, compare the formulas for V and E at the surface — their ratio is simple.

Hint 3

The field must not exceed the breakdown value at the given potential; solve for r.

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. Let a small test charge $+ q$ be in field $E$ . Force on it: $F = q E$ .
½ mark
Work done by the field in a small displacement $d r$ : $d W = F \cdot d r = q E \cdot d r$ .
½ mark
Work done by the field equals decrease in potential energy: $d W = - q d V$ , so $d V = - \frac{d W}{q}$ .
½ mark
Substituting: $d V = - \frac{q E \cdot d r}{q} = - E \cdot d r$ .
½ mark
Taking $d r$ along $E$ : $d V = - E d r$ .
½ mark
$\Rightarrow E = - \frac{d V}{d r}$ : the electric field equals the negative of the potential gradient, and points in the direction of steepest decrease of potential.
½ mark
II. For a charged sphere of radius r: $V = \frac{k Q}{r}$ and surface field $E = \frac{k Q}{r ^{2}}$ , so $E = \frac{V}{r}$ .
½ mark
No breakdown requires $E \leq 2.5 \times 1 0^{6}$ V m $^{- 1}$ when $V = 5 \times 1 0^{5}$ V, so $r \geq \frac{V}{E _{ma x}}$ .
½ mark
$r_{min} = \frac{5 \times 1 0 ^{5}}{2.5 \times 1 0 ^{6}}$
½ mark
$r_{min} = 0.2$ m (20 cm).
½ mark

Answer: I. E = −dV/dr; II. minimum radius = 0.2 m

Where the marks are

F = qE on test charge (½ mark)
dW = qE·dr (½ mark)
dV = −dW/q (½ mark)
Substitution dV = −E·dr (½ mark)
Along field: dV = −E dr (½ mark)
E = −dV/dr with statement (½ mark)
Relation E = V/r for sphere (½ mark)
r = V/E_max condition (½ mark)
Substitution (½ mark)
r = 0.2 m (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 paperSample paper 2026-27: every question, by section.

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CBSE Class 12 Physics sample paper 2026-27: question 31(B)

Hints

Step-by-step solution

Revise the chapter

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