A block of mass m containing a net negative charge -q is placed on a frictionless horizontal table and is connected to a wall through an unstretched spring of spring constant k. If the horizontal electric field E parallel to the spring is switched on, then the maximum compression of the spring is

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qE k

$2 qE / k$

$\sqrt{qE / k}$

Zero

answer is B.

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Detailed Solution

Force on the block : F=qE toward left.Let spring be compressed maximum by x. Then
$Fx = \frac{1}{2} {kx}^{2} or qEx = \frac{1}{2} {kx}^{2} or x = \frac{2 qE}{k}$

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