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Question

# A charge −q is placed at the axis of a charged ring of radius r at a distance of 2√2r as shown in the figure. If the ring is fixed and carrying a charge +Q, the kinetic energy of charge −q when it is released and reaches the centre of the ring will be

A
qQ2πϵ0r
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B
qQ6πϵ0r
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C
qQ4πϵ0r
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D
qQ12πϵ0r
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Solution

## The correct option is B qQ6πϵ0r From work-energy theorem, Gain in KE = Loss in PE KE of −q at O=q(VO−VP) =q⎡⎢ ⎢⎣Q4πϵ0r−Q4πϵ0√r2+(2√2r)2⎤⎥ ⎥⎦ =qQ4πϵ0[1r−1√r2+8r2] =qQ4πϵ0[1r−13r]=qQ6πϵ0r

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