Solve this:

15. Four charges are placed at the corners of a square of side a.


Work done by external agent in carrying an electron from point O to O' is 
$$\begin{gathered} \left(1\right) \frac{-4\mathrm{ekQ}}{\mathrm{a}}\left(\sqrt{5}-2\sqrt{2}\right) \left(2\right) \frac{4\mathrm{ekQ}}{\mathrm{a}}\left(\sqrt{5}-2\sqrt{2}\right) \\ \left(3\right) \frac{4\mathrm{ekQ}}{\mathrm{a}}\left(\sqrt{5}+2\sqrt{2}\right) \left(4\right) \frac{-4\mathrm{ekQ}}{\mathrm{a}}\left(1+\frac{1}{\sqrt{5}}-\sqrt{2}\right) \end{gathered}$$

 

Dear Student,
           As we know work done in moving charge from one point to another point = change in potential energy.
                Change in potential energy = final P.E. - initial P.E.
                  Now, initial P.E.  at 'O' due to all four charges = $-\frac{4KQe}{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}}$          [ where, K =$\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}$ ]
          Now final P.E. at O'  due to two distant charges and two near charges collectively =
                                                                                                                 $-\left[\frac{2KQe}{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{2KQe}{{\displaystyle \sqrt{5}\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]$ respectively.

               Change in P.E.  =    $-\left[\frac{2KQe}{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{2KQe}{{\displaystyle \sqrt{5}\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]$$-\left[-\frac{4KQe}{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}}\right]$
                                         = $-\left[\frac{4KQe}{{\displaystyle a}}+\frac{4KQe}{{\displaystyle \sqrt{5}a}}- \frac{4\sqrt{2}KQe}{{\displaystyle a}}\right]$
                                       = $-\frac{4KQe}{a}\left[1+\frac{1}{\sqrt{5}}-\sqrt{2}\right]$
Hence the correct option is (4).


Regards