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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
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The correct answer is:
$\frac{Q}{2 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$
$\frac{Q}{2 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$
$v_1=\frac{k q}{R}-\frac{k q}{\sqrt{R^2+d^2}}$
$v_2=\frac{-k q}{R}+\frac{k q}{\sqrt{R^2+d^2}}$
$v_2=\frac{-k q}{R}+\frac{k q}{\sqrt{R^2+d^2}}$
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