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A two point charges 4q and $- q$ are fixed on the x-axis at $x = - \frac{d}{2}$ and $x = \frac{d}{2}$ , respectively. If a third point charge ‘q’ is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will :

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increase by $\frac{2 q^{2}}{3 π \in_{0} d}$

increase by $\frac{3 q^{2}}{4 π \in_{0} d}$

decrease by $\frac{q^{2}}{4 π \in_{0} d}$

decrease by $\frac{4 q^{2}}{3 π \in_{0} d}$

answer is C.

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Detailed Solution

The change is only due to 4q
Charge q distance from $" - q "$ is not changing.
$S o Δ U = \frac{(4 q) (q)}{4 π \in_{0}} (\frac{1}{(3 d / 2)} - \frac{1}{(d / 2)}) = - \frac{4 q^{2}}{3 π \in_{0} d}$

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