Two thin wires rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding.…

Two thin wires rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding. The charges on the two rings are $+q$ and $-q$. The potential difference between the centres of the two rings is
  1. $\mathrm{QR} / 4 \pi \varepsilon_0 \mathrm{~d}^2$
  2. $\frac{Q}{2 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$
  3. zero
  4. $\frac{Q}{4 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$

Solution

$v_1=\frac{k q}{R}-\frac{k q}{\sqrt{R^2+d^2}}$ $v_2=\frac{-k q}{R}+\frac{k q}{\sqrt{R^2+d^2}}$

Asked in: JEE Main 2005

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