First slide

Pontial due to

Question

Two identical thin rings, each of radius R metre are co-axially placed at distance R metre apart. If Q₁and Q₂ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is :

Moderate

A

Zero
B

$q (Q_{1} - Q_{2}) (\sqrt{2} - 1) / \sqrt{(2)} 4 {πε}_{0} R$
C

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

$q (Q_{1} + Q_{2}) (\sqrt{2} + 1) / \sqrt{(2)} 4 {πε}_{0} R$

Solution

Potential energy of point charge placed on the axial line of the ring is
$\begin{array}{rcl} U & = & \frac{q}{4 {πε}_{0}} \frac{Q}{\sqrt{R^{2} + x^{2}}} \\ U_{i} & = & \frac{q}{4 {πε}_{0}} [\frac{Q_{1}}{R} + \frac{Q_{2}}{\sqrt{2} R}] \\ U_{f} & = & \frac{q}{4 {πε}_{0}} [\frac{Q_{2}}{R} + \frac{Q_{1}}{\sqrt{2} R}]; \\ W & = & U_{f} - U_{i} \\ = & \frac{q}{4 {πε}_{0}} [\frac{Q_{2}}{R} + \frac{Q_{1}}{\sqrt{2} R}] - \frac{q}{4 {πε}_{0}} [\frac{Q_{1}}{R} + \frac{Q_{2}}{\sqrt{2} R}] \\ = & \frac{q}{4 \sqrt{2} {πε}_{0} R} [Q_{2} - Q_{1}] [\sqrt{2} - 1] \\ Q_{1} - Q_{2} or Q_{2} - Q_{1} depends on starting and beginning of q \end{array}$

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