Two positive charges of magnitude ‘q’ are placed, at the end of a side (side1) of a square of side ‘2a’.Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is

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$\frac{1}{4 π ε_{0}} \frac{2 q Q}{a} (1 - \frac{2}{\sqrt{5}})$

$\frac{1}{4 π ε_{0}} \frac{2 q Q}{a} (1 - \frac{1}{\sqrt{5}})$

$\frac{1}{4 π ε_{0}} \frac{2 q Q}{a} (1 + \frac{1}{\sqrt{5}})$

answer is D.

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Detailed Solution

Potential at point A,
$V_{A} = \frac{2 k q}{a} - \frac{2 k q}{a \sqrt{5}}$
$∴$ (potential due to each $q = \frac{k a}{a}$ and potential due to each $- q = \frac{- k q}{a \sqrt{5}} (∵ \sqrt{{(2 a)}^{2} + a^{2}} = a \sqrt{5})$
Potential at point .B,
$V_{B} = 0$
( $∵$ point B is equidistant from all the four charges)
$∴$ Using work energy theorem,
${(W_{A B})}_{e l e c t r i c} = Q (V_{A} - V_{B})$
$= \frac{2 k q Q}{a} [1 - \frac{1}{\sqrt{5}}]$
$= (\frac{1}{4 π ε_{0}}) \frac{2 Q q}{a} [1 - \frac{1}{\sqrt{5}}]$

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