Question

Two identical thin rings, each of radius R metres, are coaxially placed R metres apart. If $Q_1$ coulombs and $Q_2$ coulombs are the charges uniformly spread on the two rings respectively, then the work done in moving a charge q from the centre of one ring to that of the other is :-

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

Answer 1

Answer: (A), (B)

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

Potential of a ring at some axial point , $V=\frac{1}{4\pi {ϵ}_0}\frac{Q}{\sqrt{x^2+R^2}}$

Potential of ring of charge $Q_1$ at centre $=$ potential due to $Q_1+$ potential due to $Q_2$ ,i.e,$V_1=\frac{1}{4\pi {ϵ}_0}\frac{Q_1}{\sqrt{0+R^2}}+\frac{1}{4\pi {ϵ}_0}\frac{Q_2}{\sqrt{R^2+R^2}}=\frac{1}{4\pi {ϵ}_0}\frac{Q_1}{R}+\frac{1}{4\pi {ϵ}_0}\frac{Q_2}{\sqrt{2}R}$

Potential of ring of charge $Q_2$ at centre $=$ potential due to $Q_2+$ potential due to $Q_1$,i.e.,$V_2=\frac{1}{4\pi {ϵ}_0}\frac{Q_2}{\sqrt{0+R^2}}+\frac{1}{4\pi {ϵ}_0}\frac{Q_1}{\sqrt{R^2+R^2}}=\frac{1}{4\pi {ϵ}_0}\frac{Q_2}{R}+\frac{1}{4\pi {ϵ}_0}\frac{Q_1}{\sqrt{2}R}$

The work done,$W=q(V_1-V_2)=q(\frac{1}{4\pi {ϵ}_0}\frac{Q_1}{R}+\frac{1}{4\pi {ϵ}_0}\frac{Q_2}{\sqrt{2}R}-\frac{1}{4\pi {ϵ}_0}\frac{Q_2}{R}-\frac{1}{4\pi {ϵ}_0}\frac{Q_1}{\sqrt{2}R})$

$W=\frac{q}{4\pi {ϵ}_{0}R}(Q_1-Q_2)+\frac{q}{4\pi {ϵ}_{0}R}\frac{(Q_2-Q_1)}{\sqrt{2}}$

$W=\frac{q}{4\pi {ϵ}_{0}R}(Q_1-Q_2)(1-\frac{1}{\sqrt{2}})=\frac{q(Q_1-Q_2)}{4\sqrt{2}\pi {ϵ}_{0}R}(\sqrt{2}-1)$