Two 20 
C charges are placed at a distance of 0.5 m apart. How much work is done when a 5 C charge is moved 0.1 m along the line joining the two charges from the point midway between these two charges?
The work done against the electric fields is equal to the change in potential energy of the 5 C charge. Since potential energy is a scalar quantity, we can simply compute the change in potential energy within each individual field (generated by each charge) and sum up to obtain the answer.
Change in P.E. within field of left charge =  "\frac{20\mu\times 5\mu}{4 \pi \epsilon_{0}}\left(\frac{1}{0.15}-\frac{1}{0.25}\right)")
J
This change in P.E. is positive since work has to be done against the field of the 20 {tex}\mu{\tex}C one. (Work done against repulsion)
Change in P.E. within field right charge = "\frac{20\mu\times 5\mu}{4 \pi \epsilon_{0}}\left(\frac{1}{0.35}-\frac{1}{0.25}\right)")
J (Negative, since motion is along direction of repulsion.)
Therefore, the work done is 1.4 J.
