Question

Four charges each of magnitude +Q are placed at the corners of a square of side 'a'.What is the work done in carrying an electron from the point of intersection of diagonals to the center of a side of the square?
Ans: -4ekQ[1+(1/?5)-?2] /a

Answer 1

The work done is equal to change in potential energy. Now the potential energy at the centre of the suqare (two diagonals intersect at the centre of the square) is
$-4\times \frac{keQ}{r}=-\frac{4keQ}{{\displaystyle \frac{a\sqrt{2}}{2}}}=-\frac{4keQ\sqrt{2}}{a}$
Potential energy at the centre of the side of the square is
$-2\times \frac{keQ}{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-2\times \frac{keQ}{{\displaystyle \raisebox{1ex}{$a\sqrt{5}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-\frac{4keQ}{a}\left(1+\frac{1}{\sqrt{5}}\right)$
So the work done is
$W=-\frac{4keQ}{a}\left(1+\frac{1}{\sqrt{5}}\right)-\left(-\frac{4keQ\sqrt{2}}{a}\right)=-\frac{4keQ}{a}\left(1+\frac{1}{\sqrt{5}}-\sqrt{2}\right)$