PhysicsCBSE Class 12 · Board exam 2026, set 55/1/1 · 1 mark · MCQ

CBSE Class 12 Physics 2026: question 2

Set in the CBSE Class 12 board exam of 2026, paper 55/1/1, section A, question 2, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q2 · 1 mark · MCQElectrostatic Potential and Capacitance

A conducting wire connects two charged metallic spheres A and B of radii $r_{1}$ and $r_{2}$ respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields, $(E_{A} / E_{B})$ at the surfaces of spheres A and B will be

(A) $\frac{r _{1}}{r _{2}}$
(B) $\frac{r _{2}}{r _{1}}$
(C) $\frac{r _{1}^{2}}{r _{2}^{2}}$
(D) $\frac{r _{2}^{2}}{r _{1}^{2}}$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/1/1.

Hints

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

Hint 1

What is common to two conductors joined by a wire?

Hint 2

Write E at the surface in terms of V and 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.

The connected spheres are at the same potential: $\frac{k q _{1}}{r _{1}} = \frac{k q _{2}}{r _{2}}$ .
$E = \frac{k q}{r ^{2}} = \frac{V}{r}$ , so $\frac{E _{A}}{E _{B}} = \frac{r _{2}}{r _{1}}$ .
1 mark

Answer: $r_{2} / r_{1}$ , option (B)

Where the marks are

Equal potentials give $E \propto 1/ r$ , so the ratio is $r_{2} / r_{1}$ (1 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 paper2026 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2026: question 2

Hints

Step-by-step solution

Revise the chapter

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