An electron is taken from point A to point B along the path AB in a uniform electric field of intensity  $E=10V{m}^{-1}.$ Side AB = 5 m, and side BC= 3 m. Then, the amount of work done on the electron by us is 

 $W_{AB}=W_{AC}+W_{CB}$

$W_{CB}$ should be zero, because in moving from C to B, we always move perpendicular to filed. Hence, force applied by field and displacement will be at ${90}^0.$

$W_{AC}=-e(V_C-V_A)$

$V_C-V_A=-E\times AC=-10\times 4=-40$

$\therefore W_{AC}=-e\times (-40)=40e$

So $W_{AB}=40eV$