PhysicsCBSE Class 12 · Board exam 2023, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2023: question 31(a)

Set in the CBSE Class 12 board exam of 2023, paper 55/1/1, section D, question 31(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(a) · 5 marks · Long answerElectric Charges and Fields

(i) State Coulomb's law in electrostatics and write it in vector form, for two charges.
(ii) 'Gauss's law is based on the inverse-square dependence on distance contained in the Coulomb's law.' Explain.
(iii) Two charges A (charge q) and B (charge 2q) are located at points (0, 0) and (a, a) respectively. Let $\hat{i}$ and $\hat{j}$ be the unit vectors along x-axis and y-axis respectively. Find the force exerted by A on B, in terms of $\hat{i}$ and $\hat{j}$ .

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 55/1/1.

Hints

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

Hint 1

For (ii), compute the flux of a point charge through a sphere and watch what happens to r.

Hint 2

For (iii), the force on B points along the vector from A to B.

Hint 3

Use $F = \frac{k q _{1} q _{2}}{r ^{3}} r$ with $r = 2 a$ .

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) Coulomb's law: the electrostatic force between two stationary point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them; it acts along the line joining them. $F = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}}$
1 mark
Vector form: force on $q_{1}$ due to $q_{2}$ : $F_{12} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}^{2}} \overset{r}{^}_{12} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}^{3}} r_{12}$ , where $r_{12}$ is the vector from $q_{2}$ to $q_{1}$ .
1 mark
(ii) Flux through a sphere of radius r around a charge q is $Φ = E \times 4 π r^{2} = \frac{q}{4 π ε _{0} r ^{2}} \times 4 π r^{2} = \frac{q}{ε _{0}}$ . The $1/ r^{2}$ of Coulomb's law exactly cancels the $r^{2}$ of the surface area, making flux independent of r and of the shape of the surface. If the force law were not inverse-square, this cancellation would fail and Gauss's law would not hold.
1 mark
(iii) Position vector of B relative to A: $r_{A B} = a \hat{i} + a \hat{j}$ , $∣ r_{A B} ∣ = a^{2} + a^{2} = 2 a$ ; $F = \frac{1}{4 π ε _{0}} \frac{q ( 2 q )}{r ^{3}} r_{A B}$
½ mark
$F = \frac{1}{4 π ε _{0}} \frac{2 q ^{2} ( a i ^ + a j ^ )}{( 2 a ) ^{3}} = \frac{1}{4 π ε _{0}} \frac{2 q ^{2} a ( i ^ + j ^ )}{2 2 a ^{3}}$
½ mark
$F = \frac{q ^{2}}{4 2 π ε _{0} a ^{2}} (\hat{i} + \hat{j})$ (magnitude $\frac{q ^{2}}{4 π ε _{0} a ^{2}}$ , directed away from A along the line AB).
1 mark

Answer: $F_{B A} = \frac{q ^{2}}{4 2 π ε _{0} a ^{2}} (\hat{i} + \hat{j})$

Where the marks are

Statement of Coulomb's law (1 mark)
Vector form with meaning of $r_{12}$ (1 mark)
Explanation: $1/ r^{2}$ cancels $r^{2}$ of area so flux is independent of r (1 mark)
Writes $r_{A B} = a \hat{i} + a \hat{j}$ , $r = 2 a$ and formula (½ mark)
Substitution (½ mark)
Final vector $\frac{q ^{2}}{4 2 π ε _{0} a ^{2}} (\hat{i} + \hat{j})$ (1 mark; only magnitude earns 1 mark)

Revise the chapter

Chapter: Electric Charges and Fields (notes, formulas, where marks are lost and every past question)
Topic: Coulomb's Law and Superposition

Practise this chapterBoard-style questions from Electric Charges and Fields, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 55/1/1: 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 2023: question 31(a)

Hints

Step-by-step solution

Revise the chapter

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