An electric charge 10 - 6 μ C is placed at origin ( 0 , 0 ) m of X - Y co-ordinate system. Two points P and…

The formula to calculate the potential difference between $P$ and $Q$ is given by

$∆V=kq\left(\frac{1}{r_1}-\frac{1}{r_2}\right) ...\left(1\right)$

Now,

$\begin{array}{rcl}{r}_1& =& \sqrt{(\sqrt{3}{)}^2+(\sqrt{3}{)}^2} \mathrm{m}\\ & =& \sqrt{6} \mathrm{m}\end{array}$

And,

$\begin{array}{rcl}{r}_2& =& \sqrt{(\sqrt{6}{)}^2+0} \mathrm{m}\\ & =& \sqrt{6} \mathrm{m}\end{array}$

As $r_1=r_2=\sqrt{6} \mathrm{m}$ and the value of the charge remains constant, the potential difference between the points is zero.