Explanation :
It is given that,
[tex]q_1=+2\mu C=+2\times 10^{-6}\ C[/tex]
[tex]q_2=-2\mu C=-2\times 10^{-6}\ C[/tex]
Side of square, [tex]x=1.5\ cm=0.015\ m[/tex]
It is given that, the potential at point c is equal to zero.
So, potential at point a due to [tex]q_1[/tex] and [tex]q_2[/tex] is given as:
[tex]V_a=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}[/tex]
where, [tex]r_1=r_2=\dfrac{1}{\sqrt{2} }(0.015\ m)[/tex]
So, [tex]V=\dfrac{kq_1}{r_1}+(-\dfrac{kq_2}{r_2})[/tex] (Since, [tex]q_2[/tex] is negative)
Hence, the potential at a due to [tex]q_1[/tex] and [tex]q_2[/tex] is equal to 0 V.
