In the rectangle shown below the two corners have charges $q_1=-5\mu C$  and $q_2=+2.0\mu C$  The work done in moving a charge $+3.0\mu C$ From $B$ to $A$  is (take $1/4\pi {\epsilon }_0={10}^{10}N-m^2/C^2$)

Work done$V_A={10}^{10}\left[\frac{\left(-5\times {10}^{-6}\right)}{15\times {10}^{-2}}+\frac{2\times {10}^{-6}}{5\times {10}^{-2}}\right]=\frac{1}{15}\times {10}^{6}volt$

                       $$\begin{gathered} and{V}_B={10}^{10}\left[\frac{\left(2\times {10}^{-6}\right)}{15\times {10}^{-2}}-\frac{5\times {10}^{-6}}{5\times {10}^{-2}}\right]=-\frac{13}{15}\times {10}^{6}volt \\ \therefore W_{BA}=\hspace{0.17em}Q\left(V_A-\hspace{0.17em}{V}_B\right)=3\times {10}^{-6}\left[\frac{1}{15}\times {10}^6-\left(-\frac{13}{15}\times {10}^6\right)\right]=2.8J \end{gathered}$$