Four objects W, X, Y and Z, each with charge +q are held fixed at four points of a square of side d as shown in the figure. Objects X and Z are on the midpoints of the sides of the square. The electrostatic force exerted by object W on object X is F. Then the magnitude of the force exerted by object W on Z is

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F/3

F/2

F/5

F/7

answer is B.

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Detailed Solution

$F = \frac{1}{4 {πε}_{0}} \cdot \frac{q \cdot q}{{(\frac{d}{2})}^{2}} \Rightarrow \frac{1}{4 {πε}_{0}} \frac{q^{2}}{d^{2}} = \frac{F}{4}$
Separation between W and Z, $d^{'} = \sqrt{d^{2} + {(\frac{d}{2})}^{2}}$
$d^{'} = \sqrt{\frac{5}{4} d^{2}}$
Force by W on Z
$\begin{array}{l} F^{'} = \frac{1}{4 {πε}_{0}} \cdot \frac{q \cdot q}{d^{' 2}} = \frac{1}{4 {πε}_{0}} \frac{q^{2}}{{(\sqrt{\frac{5}{4} d^{2}})}^{2}} from (2) \\ = \frac{1}{4 {πε}_{0}} \cdot \frac{q^{2}}{\frac{5}{4} d^{2}} = \frac{4}{5} (\frac{1}{4 {πε}_{0}} \frac{q^{2}}{d^{2}}) from (1) \\ = \frac{F}{5} \end{array}$