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JEE Main 2005 (AIEEE) · Electric Potential & Equipotentials

Electric Potential & Equipotentials question from JEE Main 2005 (AIEEE)

Q125JEE Main 2005 · AIEEEMedium

Electric Potential & Equipotentials

Two thin wire 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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Answer: (a)

$V_1 = \dfrac{kq}{R} - \dfrac{kq}{\sqrt{R^2 + d^2}}$ , $V_2 = -\dfrac{kq}{R} + \dfrac{kq}{\sqrt{R^2 + d^2}}$
$V_1 - V_2 = 2kq\left[\dfrac1R - \dfrac{1}{\sqrt{R^2 + d^2}}\right]$
$= \dfrac{q}{2\pi\epsilon_0}\left[\dfrac1R - \dfrac{1}{\sqrt{R^2 + d^2}}\right]$
So the answer is (a).

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