PhysicsCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2025-26: question 31(II)

From the official CBSE sample question paper for 2025-26, section E, question 31(II), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(II) · 5 marks · Long answerElectrostatic Potential and Capacitance

(A) If a charge of 1µC is placed at the origin and another charge of 3 µC placed at the point (20m,0m,0m) in an external uniform electric field of $40 V/m \hat{i}$ with the electric potential at origin to be zero. Find the electrical potential energy of system.
(B) If one charge particle is moved from A to C To B and another charge particle of equal magnitude is moved from A to D to B, In uniform external magnetic field. Than for which charge particle more work will be needed? (use fig for reference)
(C) Electrostatic potential is constant throughout the volume of conductor has the same value on its surface why?

Figure in the paper: Uniform field E drawn as horizontal parallel arrows pointing right. Points A (bottom left) and D (bottom right) on the lower line, B at upper right; a curved path from A through C to B, and a straight path A to D then D to B. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

Potential falls along the direction of the field: $V = - E \cdot r$ when $V (0) = 0$ .

Hint 2

Include the mutual energy and each charge's energy in the external field.

Hint 3

Work done by a conservative field depends only on initial and final points.

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.

(A) For a uniform field $E = 40 \hat{i}$ V/m with $V (0) = 0$ : $V (x) = - E x = - 40 x$ . So $V (r_{1}) = 0$ and $V (r_{2}) = - 40 \times 20 = - 800$ V.
1 mark
$U = \frac{k q _{1} q _{2}}{r _{12}} + q_{1} V (r_{1}) + q_{2} V (r_{2}) = \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 6} \times 3 \times 1 0 ^{- 6}}{20} + 0 + 3 \times 1 0^{- 6} (- 800)$
1 mark
$U = 1.35 \times 1 0^{- 3} - 2.4 \times 1 0^{- 3} = - 1.05 \times 1 0^{- 3}$ J
1 mark
(B) Equal work is needed for both charges. The electrostatic field is conservative, so work done depends only on the end points A and B, not on the path (A–C–B or A–D–B). (Even if the field were magnetic, it would do no work on either charge.)
1 mark
(C) Inside a conductor in electrostatic equilibrium $E = 0$ , so $d V = - E \cdot d l = 0$ : no work is needed to move a test charge inside or on the surface. Hence potential is constant throughout the volume and equals its value at the surface.
1 mark

Answer: (A) U = −1.05×10⁻³ J; (B) same work for both paths; (C) E = 0 inside, so V is constant and equal to the surface value

Where the marks are

Potentials at the two charge positions using $V = - E x$ (1 mark)
Correct expression for potential energy in external field (1 mark)
Correct value $- 1.05 \times 1 0^{- 3}$ J (1 mark)
Work same as field is conservative (1 mark)
E = 0 inside conductor ⇒ potential constant (1 mark)

Revise the chapter

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

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 2025-26: every question, by section.

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CBSE Class 12 Physics sample paper 2025-26: question 31(II)

Hints

Step-by-step solution

Revise the chapter

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