Question

Two identical thin rings, each of radius $R$ metres are coaxially placed at distance $R$ metres apart. If $Q_1$ coulumb and $Q_2$ coulumb 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

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

Answer 1

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


Potential due to an elemental charge $dQ$ on a ring of radius $r$ at a distance $x$ on its axis is given by

${\int }_0^{V}dV={\int }_0^Q\frac{kdQ}{\sqrt{r^2+x^2}}$
$\Rightarrow V=\frac{kQ}{\sqrt{r^2+x^2}}$



Work done in moving charge $q$ from the centre of one ring to the other $=$ Change in potential energy of the charge

i.e $W=\Delta U=U_f-U_i=U_A-U_B$

$U_A=\frac{k{Q}_{1}q}{R}+\frac{k{Q}_{2}q}{\sqrt{2}R}$
$=\frac{kq}{R}(Q_1+\frac{Q_2}{\sqrt{2}})$

$U_B=\frac{k{Q}_{2}q}{R}+\frac{k{Q}_{1}q}{\sqrt{2}R}$
$=\frac{kq}{R}(Q_2+\frac{Q_1}{\sqrt{2}})$

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