Physics - Electrostatics Question with Solution | TestHub

Potential at origin $O$, the potential due to two charges is given by

$V_{\mathit{0}}\mathit{=}\frac{\mathit{4}q}{\mathit{4}\pi {\epsilon }_{\mathit{0}}\left(\frac{\mathit{d}}{\mathit{2}}\right)}\mathit{+}\frac{\left(\mathit{-}q\right)}{\mathit{4}\pi {\epsilon }_{\mathit{0}}\left(\frac{\mathit{d}}{\mathit{2}}\right)}\mathit{=}\frac{\mathit{6}q}{\mathit{4}\pi {\epsilon }_{\mathit{0}}d}$

Potential at $P$ at a distance, $x=d$ is given by

$V_P\mathit{=}\frac{\mathit{4}q}{\mathit{4}\pi {\epsilon }_{\mathit{0}}\left(\frac{\mathit{3}\mathit{d}}{\mathit{2}}\right)}\mathit{+}\frac{\left(\mathit{-}q\right)}{\mathit{4}\pi {\epsilon }_{\mathit{0}}\left(\frac{\mathit{d}}{\mathit{2}}\right)}\mathit{=}\frac{q}{\mathit{6}\pi {\epsilon }_{\mathit{0}}d}$

Change in potential energy due the charge motion of charge $q$ is given by

 $\mathit{∆}U\mathit{=}q\Delta V\mathit{=}q\left(V_f\mathit{-}{V}_i\right)$

$\mathit{\Rightarrow }\mathit{∆}U\mathit{=}q\left(V_P\mathit{-}{V}_O\right)$

$∆U=q\left(\frac{q}{6\pi {\epsilon }_{\mathit{0}}d}\mathit{-}\frac{3q}{2\pi {\epsilon }_{\mathit{0}}d}\right)\mathit{=}\mathit{-}\frac{4q^{\mathit{2}}}{3\pi {\epsilon }_{\mathit{0}}d}$