Question 195 of 311

A two point charges $4q$ and $–q$ are fixed on the $x$ -axis at $x=\dfrac{-d}{2}$ and $x=\dfrac{d}{2}$, respectively. If the 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:

AIncrease by $\dfrac{3{q}^{2}}{4\pi {\in }_{0}d}$BIncrease by $\dfrac{2{q}^{2}}{3\pi {\in }_{0}d}$Cdecrease by $\dfrac{{q}^{2}}{4\pi {\in }_{0}d}$Ddecrease by $\dfrac{4{q}^{2}}{3\pi {\in }_{0}d}$

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Answer: (D) decrease by $\dfrac{4{q}^{2}}{3\pi {\in }_{0}d}$

Potential at origin $O$, the potential due to two charges is given by
${V}_{0}=\dfrac{4q}{4\pi {\epsilon }_{0}\left(\dfrac{d}{2}\right)}+\dfrac{\left(-q\right)}{4\pi {\epsilon }_{0}\left(\dfrac{d}{2}\right)}=\dfrac{6q}{4\pi {\epsilon }_{0}d}$
Potential at $P$ at a distance, $x=d$ is given by
${V}_{P}=\dfrac{4q}{4\pi {\epsilon }_{0}\left(\dfrac{3d}{2}\right)}+\dfrac{\left(-q\right)}{4\pi {\epsilon }_{0}\left(\dfrac{d}{2}\right)}=\dfrac{q}{6\pi {\epsilon }_{0}d}$
Change in potential energy due the charge motion of charge $q$ is given by
$∆U=q\Delta V=q\left({V}_{f}-{V}_{i}\right)$
$\Rightarrow ∆U=q\left({V}_{P}-{V}_{O}\right)$
$∆U=q\left(\dfrac{q}{6\pi {\epsilon }_{0}d}-\dfrac{3q}{2\pi {\epsilon }_{0}d}\right)=-\dfrac{4{q}^{2}}{3\pi {\epsilon }_{0}d}$