(i) An electric dipole (dipole moment p = pi), consisting of charges -q and q, separated by distance 2a, is placed along the x-axis, with its centre at the origin. Show that the potential V, due to this dipole, at a point x, (x >> a) is equal to (1)/(4πε_0)·px². (ii) Two isolated metallic spheres S_1 and S_2 of radii 1 cm and 3 cm respectively are charged such that both have the same charge density ((2)/(π)×10^-9 )C/m². They are placed far away from each other and connected by a thin wire. Calculate the new charge on sphere S_1.

(i) V = (1)/(4πε_0)( p· i)/(x²) for x ≫ a; (ii) new charge on S₁ = 2 × 10⁻¹² C

PhysicsCBSE Class 12 · Board exam 2024, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2024: question 31(b)

Set in the CBSE Class 12 board exam of 2024, paper 55/1/1, 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) An electric dipole (dipole moment $p = p \hat{i}$ ), consisting of charges $- q$ and $q$ , separated by distance 2a, is placed along the x-axis, with its centre at the origin. Show that the potential V, due to this dipole, at a point x, (x >> a) is equal to $\frac{1}{4 π ε _{0}} \cdot \frac{p \cdot i ^}{x ^{2}}$ .
(ii) Two isolated metallic spheres $S_{1}$ and $S_{2}$ of radii 1 cm and 3 cm respectively are charged such that both have the same charge density $(\frac{2}{π} \times 1 0^{- 9}) C/ m^{2}$ . They are placed far away from each other and connected by a thin wire. Calculate the new charge on sphere $S_{1}$ .

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/1/1.

Hints

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

Hint 1

Add the potentials of +q and −q at P, using their distances $x - a$ and $x + a$ .

Hint 2

Combine the fractions, identify $2 q a = p$ , then drop $a^{2}$ for x ≫ a.

Hint 3

For (ii), find each charge from $σ \cdot 4 π r^{2}$ ; once connected, total charge is conserved and the potentials $k Q / r$ become equal.

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) Figure (described): $- q$ at $x = - a$ and $+ q$ at $x = + a$ on the x-axis, centre O at the origin, and point P at distance x on the axis beyond $+ q$ . Then $p = 2 q a \hat{i}$ .
½ mark
Potential of a point charge is $\frac{1}{4 π ε _{0}} \frac{q}{r}$ . By superposition, $V = V_{+ q} + V_{- q} = \frac{1}{4 π ε _{0}} [\frac{q}{x - a} - \frac{q}{x + a}]$ .
½ mark
$V = \frac{q}{4 π ε _{0}} \frac{( x + a ) - ( x - a )}{x ^{2} - a ^{2}} = \frac{1}{4 π ε _{0}} \frac{2 q a}{x ^{2} - a ^{2}} = \frac{1}{4 π ε _{0}} \frac{p}{x ^{2} - a ^{2}}$
½ mark
Since $p$ is along $\hat{i}$ , $p = p \cdot \hat{i}$ , so $V = \frac{1}{4 π ε _{0}} \frac{p \cdot i ^}{x ^{2} - a ^{2}}$ .
½ mark
For $x ≫ a$ , $a^{2}$ can be neglected compared with $x^{2}$ : $V = \frac{1}{4 π ε _{0}} \cdot \frac{p \cdot i ^}{x ^{2}}$ . Hence proved.
½ mark
(ii) Initial charges are $Q = σ \cdot 4 π r^{2}$ . $Q_{1} = \frac{2}{π} \times 1 0^{- 9} \times 4 π (1 \times 1 0^{- 2})^{2} = 8 \times 1 0^{- 13}$ C.
½ mark
$Q_{2} = \frac{2}{π} \times 1 0^{- 9} \times 4 π (3 \times 1 0^{- 2})^{2} = 72 \times 1 0^{- 13}$ C
½ mark
When joined by the wire, charge is conserved: $Q_{1}^{'} + Q_{2}^{'} = Q_{1} + Q_{2} = 80 \times 1 0^{- 13}$ C.
½ mark
The spheres reach a common potential: $\frac{Q _{1}^{'}}{4 π ε _{0} r _{1}} = \frac{Q _{2}^{'}}{4 π ε _{0} r _{2}}$ , so $\frac{Q _{2}^{'}}{Q _{1}^{'}} = \frac{r _{2}}{r _{1}} = 3$ and $Q_{2}^{'} = 3 Q_{1}^{'}$ .
½ mark
$4 Q_{1}^{'} = 80 \times 1 0^{- 13}$ , so $Q_{1}^{'} = 2 \times 1 0^{- 12}$ C. This agrees with $C_{1} V$ , where $C_{1} = \frac{1}{9} \times 1 0^{- 11}$ F and $V = 18$ V.
½ mark

Answer: (i) $V = \frac{1}{4 π ε _{0}} \frac{p \cdot i ^}{x ^{2}}$ for x ≫ a; (ii) new charge on S₁ = 2 × 10⁻¹² C

Where the marks are

Diagram with charges and point P (½ mark)
Superposition expression for V (½ mark)
Simplifies to $p / [4 π ε_{0} (x^{2} - a^{2})]$ (½ mark)
Writes $p = p \cdot \hat{i}$ (½ mark)
Approximation for x ≫ a (½ mark)
Q₁ = 8 × 10⁻¹³ C (½ mark)
Q₂ = 72 × 10⁻¹³ C (½ mark)
Charge conservation (½ mark)
Equal potentials, so $Q_{2}^{'} / Q_{1}^{'} = r_{2} / r_{1}$ (½ mark)
Q₁' = 2 × 10⁻¹² C (½ mark)

Revise the chapter

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

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CBSE Class 12 Physics 2024: question 31(b)

Hints

Step-by-step solution

Revise the chapter

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