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A ring of radius R carries charge $+q$ uniformly distributed on it. If a charge $- 90$ is released on ring's axis at a distance $\sqrt 3 R$ from its centre , the kinetic energy of $-90$ when it reaches the centre of ring will be :

$(A)\;\frac{1}{4 \pi \in _0} \frac{q\;90}{\sqrt 3 R} \\ (B)\;\frac{1}{4 \pi \in _0} \frac{q\;90}{3R} \\ (C)\; \frac{1}{4 \pi \in _0} \frac{q\;90}{ R} \\ (D)\;\frac{1}{4 \pi \in _0} \frac{q\;90}{2 R} $

1 Answer

Potential due to ring at distance $\sqrt 3 R$
$\qquad= V_1=\large\frac{1}{4 \pi \in _0} \frac{q}{\sqrt {R^2+3R^2}}$
$\qquad= \large\frac{1}{4 \pi \in _0} \frac{q}{2R}$
At centre potential $V_2 =\large\frac{1}{4 \pi \in _0} \frac{q}{R}$
$K.E= q_0 (V_2-V_1)=\large\frac{1}{4 \pi \in _0} \frac{q 90}{2R}$
Hence D is the correct answer.

 

answered Jan 13, 2014 by meena.p
edited Aug 21, 2014 by thagee.vedartham
 

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