Question

Two identical thin rings each of radius R, are coaxially placed a distance R apart. If Q₁ and Q₂ 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

• A. Zero
• B. $\frac{q(Q_{1} - Q_{2})(\sqrt{2} - 1)}{4\pi\varepsilon_{0}R\sqrt{2}}$
• C. $\frac{q(Q_{1} + Q_{2})\sqrt{2}}{4\pi\varepsilon_{0}R}$
• D. $\frac{q\left( \frac{Q_{1}}{Q_{2}} \right)(\sqrt{2} - 1)}{4\pi\varepsilon_{0}R\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{q(Q_{1} - Q_{2})(\sqrt{2} - 1)}{4\pi\varepsilon_{0}R\sqrt{2}}$

Answer 2

Potential at the centre of first ring

$V_{A} = \frac{Q_{1}}{4\pi\varepsilon_{0}R} + \frac{Q_{2}}{4\pi\varepsilon_{0}\sqrt{R^{2} + R^{2}}}$

Potential at the centre of second ring

$V_{B} = \frac{Q_{2}}{4\pi\varepsilon_{0}R} + \frac{Q_{1}}{4\pi\varepsilon_{0}\sqrt{R^{2} + R^{2}}}$Potential difference between the two centres $V_{A} - V_{B} = \frac{(\sqrt{2} - 1)(Q_{1} - Q_{2})}{4\pi\varepsilon_{0}R\sqrt{2}}$

∴Workdone$W = \frac{q(\sqrt{2} - 1)(Q_{1} - Q_{2})}{4\pi\varepsilon_{0}R\sqrt{2}}$