First slide

Pontial due to

Question

Two identical thin rings, each of radius R metres, are coaxially placed at a distance of R metres from each other. If $Q_{1}$ coulombs and $Q_{2}$ coulombs are the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of first ring to that of the second is

Moderate

A

Zero
B

$q (Q_{1} - Q_{2}) (\sqrt{2} - 1) / \sqrt{2} \times 4 π ε_{0} R$
C

$q \sqrt{2} (Q_{1} + Q_{2}) / 4 π ε_{0} R$
D

$q (Q_{1} + Q_{2}) (\sqrt{2} + 1) / \sqrt{2} \times 4 π \in_{0} R$

Solution

$V_{1} = \frac{k Q_{1}}{R} + \frac{k Q_{2}}{\sqrt{2} R}$
$\begin{array}{l} V_{2} = \frac{k Q_{1}}{\sqrt{2} R} + \frac{k Q_{2}}{R} \\ W = q (V_{2} - V_{1}) = kq (Q_{2} - Q_{1}) (1 - \frac{1}{\sqrt{2}}) \\ = \frac{1}{4 {πε}_{0}} \frac{(\sqrt{2} - 1) q (Q_{2} - Q_{1})}{\sqrt{2}} . \end{array}$

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