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

CBSE Class 12 Physics 2023: question 31(b)

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

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

(i) Derive an expression for the electric field at a point on the equatorial plane of an electric dipole consisting of charges q and $-$ q separated by a distance 2a.
(ii) The distance of a far off point on the equatorial plane of an electric dipole is halved. How will the electric field be affected for the dipole ?
(iii) Two identical electric dipoles are placed along the diagonals of a square ABCD of side $2$ m as shown in the figure. Obtain the magnitude and direction of the net electric field at the centre (O) of the square.

Figure in the paper: Square ABCD with diagonals meeting at centre O. Charges: +q at A (top left), $-$ q at B (top right), $-$ q at C (bottom right), +q at D (bottom left). Dipoles lie along the diagonals AC and BD. See it in the official paper.

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

On the equatorial line both charges are equidistant; resolve their fields into components.

Hint 2

At far points $(r^{2} + a^{2})^{3/2} \approx r^{3}$ .

Hint 3

For the square, find O's distance from each corner (half the diagonal) and combine fields pair by pair.

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) Dipole: $- q$ at A and $+ q$ at B, AB = 2a, $p$ from $- q$ to $+ q$ . Point P on the perpendicular bisector at distance r from the centre O; each charge is at distance $r^{2} + a^{2}$ from P. Magnitudes: $E_{+ q} = E_{- q} = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2} + a ^{2}}$
½ mark
$E_{+ q}$ is directed away from +q and $E_{- q}$ towards −q. Their components normal to the dipole axis cancel; components parallel to the axis add, both pointing opposite to $p$ . With $cos θ = \frac{a}{r ^{2} + a ^{2}}$ : $E = 2 E_{+ q} cos θ = \frac{1}{4 π ε _{0}} \frac{2 q a}{( r ^{2} + a ^{2} ) ^{3/2}}$
1 mark
In vector form, with $p = q (2 a) \overset{p}{^}$ : $E = - \frac{p}{4 π ε _{0} ( r ^{2} + a ^{2} ) ^{3/2}}$
½ mark
(ii) For a far point ( $r ≫ a$ ): $E = - \frac{p}{4 π ε _{0} r ^{3}}$ , so $E \propto 1/ r^{3}$ .
½ mark
When r is halved: $E^{'} = \frac{p}{4 π ε _{0} ( r /2 ) ^{3}} = \frac{8 p}{4 π ε _{0} r ^{3}} = 8 E$ . The field becomes 8 times (direction unchanged).
½ mark
(iii) Diagonal = $2 \times 2 = 2$ m, so each charge is 1 m from O. Pair A(+q), C(−q): field of A at O points away from A (towards C), field of C points towards C; each $k q / 1^{2}$ , so $E_{1} = 2 k q$ along OC.
½ mark
Pair B(−q), D(+q): field of B points towards B, field of D points away from D (towards B); so $E_{2} = 2 k q$ along OB.
½ mark
OB ⟂ OC, so $E_{n e t} = E_{1}^{2} + E_{2}^{2} = 22 k q = \frac{2 2 q}{4 π ε _{0}} = \frac{q}{2 π ε _{0}}$ N/C
½ mark
Direction: along the bisector of OB and OC, i.e. parallel to DC (from D towards C, horizontally to the right).
½ mark

Answer: (i) $E = - \frac{p}{4 π ε _{0} ( r ^{2} + a ^{2} ) ^{3/2}}$ ; (ii) becomes 8 times; (iii) $E = 22 k q = \frac{q}{2 π ε _{0}}$ , along DC

Where the marks are

Expressions for fields due to +q and −q (½ mark)
Normal components cancel, axial components add; magnitude $2 q a / [4 π ε_{0} (r^{2} + a^{2})^{3/2}]$ (1 mark)
Vector form opposite to $p$ (½ mark; ½ deducted if not vector)
$E \propto 1/ r^{3}$ at far point (½ mark)
Field becomes 8 times (½ mark)
Field due to A and C: $2 k q$ along OC (½ mark)
Field due to B and D: $2 k q$ along OB (½ mark)
Net magnitude $22 k q$ (½ mark)
Direction along DC (½ mark)

Revise the chapter

Chapter: Electric Charges and Fields (notes, formulas, where marks are lost and every past question)
Topic: Electric Dipole

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(b)

Hints

Step-by-step solution

Revise the chapter

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