Question

Two identical thin rings, each of radius a meter, are coaxially placed at a distance $a$ meter 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 $q$ coulomb from the centre of one ring to that of the other is :

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

Answer 1

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

Work required will be equal to the change in potential energy, so just calculate final and initial energy, and the difference will give us the work required:
$W=U_{final}-U_{initial}$
$W=[\frac{K{Q}_{1}q}{\sqrt{R^2+0^2}}+\frac{K{Q}_{2}q}{\sqrt{R^2+R^2}}]-[\frac{K{Q}_{1}q}{\sqrt{R^2+R^2}}+\frac{K{Q}_{2}q}{\sqrt{R^2+0^2}}]$
Using the formula of potential energy due to a charged ring at a po-nt P on it's axis at a distance a = $\frac{KQ}{\sqrt{R^2+a^2}}$
$\Rightarrow W=\frac{Kq(Q_1-Q_2)(\sqrt{2}-1)}{\sqrt{2}R}$