Electric potential is the work done in bringing unit positive charge .
Here work(w) is given by
[tex]\begin{gathered} W=\text{ FS}\cos\theta;\begin{cases}F={force=\text{ 3880N}} \\ s={displacement=\text{ 0.357m}}\end{cases} \\ \therefore W=\text{ 3880}\times0.357\begin{cases}\theta={0\text{ \lparen for same direction\rparen}} \\ \cos\theta={1}\end{cases} \\ \therefore W=1385.16\text{ J} \end{gathered}[/tex]Again
Total change in electric potential energy (V) is given by
[tex]\begin{gathered} V=\text{ W}\times Q\begin{cases}V={electric\text{ potential}} \\ Q={charge=-5.62\times10^{-4^{\placeholder{⬚}}}}\end{cases} \\ V=\text{ 1385.16}\times-5.62\times10^{-4}\text{ = -0.778J} \end{gathered}[/tex]Final answer is -0.778J
