Question

Two identical thin rings each of radius R meters are coaxially placed at a distance R meters apart. If $Q_1$ coulomb and $Q_2$ 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 other is

• A. Zero
• B. $\frac{q(Q_2-Q_1)(\sqrt{2}-1)}{\sqrt{2}.4\pi {ϵ}_{0}R}$
• C. $\frac{q\sqrt{2}(Q_1+Q_2)}{4\pi {ϵ}_{0}R}$
• D. $\frac{q(Q_1+Q_2)(\sqrt{2}+1)}{\sqrt{2}.4\pi {ϵ}_{0}R}$

Answer 1

The correct option is B $\frac{q(Q_2-Q_1)(\sqrt{2}-1)}{\sqrt{2}.4\pi {ϵ}_{0}R}$


$W=q(V_{{\sigma }_2}-V_{{\sigma }_1})$
where $V_{{\sigma }_1}=\frac{Q_1}{4\pi ϵR}+\frac{Q_2}{4\pi {ϵ}_{0}R\sqrt{2}}$
and $V_{{\sigma }_2}=\frac{Q_2}{4\pi ϵR}+\frac{Q_1}{4\pi {ϵ}_{0}R\sqrt{2}}$
$\Rightarrow V_{{\sigma }_2}-V_{{\sigma }_1}=\frac{(Q_2-Q_1)}{4\pi {ϵ}_{0}R}[1-\frac{1}{\sqrt{2}}]$
So, $W=\frac{q.(Q_2-Q_1)}{4\pi {ϵ}_{0}R}\frac{(\sqrt{2}-1)}{\sqrt{2}}$