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Two thin rings of radius R are placed at a distance a between them such that their axes coincide. Potential difference between the centre of the ring will be , if charges on rings are $+q $ and $-q$

$(A)\;\frac{2q}{\pi \in _0} \bigg[ \frac{1}{R} -\frac{1}{\sqrt {R^2+a^2}}\bigg]\\ (B)\;\frac{2q}{2\pi \in _0} \bigg[ \frac{1}{R} -\frac{1}{\sqrt {R^2+a^2}}\bigg] \\ (C)\; \frac{q}{2\pi \in _0} \bigg[ \frac{1}{R} -\frac{1}{\sqrt {R^2+a^2}}\bigg] \\ (D)\;none $

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1 Answer

$V_A= V_1 +V_2$
$V_A= \large\frac{q}{4 \pi \in_0} \bigg(\frac{1}{R} -\frac{1}{\sqrt {R^2+a^2}}\bigg)$
$V_B=\large\frac{-1}{4 \pi \in _0} \frac{q}{R} +\frac{1}{4 \pi \in_0} \frac{q}{r}$
$V_A-V_B= \large\frac{2q}{4 \pi \in_0} \bigg[ \frac{1}{R} -\frac{1}{r}\bigg]$
$V_A-V_B=\large\frac{q}{2\pi \in _0} \bigg[ \frac{1}{R} -\frac{1}{\sqrt {R^2+a^2}}\bigg]$
Hence B is the correct answer.

 

answered Jan 15, 2014 by meena.p
edited Aug 1, 2014 by thagee.vedartham
 

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