Question

17. 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

Answer 1

Given that:

Side of a square, $=a$

Assuming that all the charges positive $Q$ as shown in figure,

$W_{o{o}^{\prime }}=q\Delta V$

Potential difference is, $\Delta V=V_{o^{\prime }}-V_o$

$V_{o^{\prime }}=\frac{kQ}{\frac{a}{2}}+\frac{kQ}{\frac{a}{2}}+\frac{kQ}{C{O}^{\prime }}+\frac{kQ}{D{O}^{\prime }}$

$C{O}^{\prime }=D{O}^{\prime }=\frac{a\sqrt{5}}{2}$

$V_{o^{\prime }}=\frac{4kQ}{a}(1+\frac{1}{\sqrt{5}})$

And potentia duat O point will be, $V_o=\frac{4kQ}{\frac{\sqrt{2}a}{2}}=\frac{8kQ}{\sqrt{2}a}=\frac{4\sqrt{2}kQ}{a}$

So, $\Delta V=V_{o^{\prime }}-V_o$ =$\frac{4kQ}{a}(1+\frac{1}{\sqrt{5}})-$ $\frac{4\sqrt{2}kQ}{a}$

And work done will be, $W=q\Delta V$

Here $q=e$ ,so work done to move an electron,

$W=e\times \Delta V=\frac{4kQe}{a}\left[\frac{\sqrt{5}+1-\sqrt{10}}{\sqrt{5}}\right]$