Physics - Electrostatics Question with Solution | TestHub

The potential at the centre is${\mathrm{V}}_0=\frac{2\mathrm{k}\mathrm{Q}}{\mathrm{a}}$

Potential on the$\mathrm{x}$-axis is given by,${\mathrm{V}}_{\mathrm{A}}=\frac{2\mathrm{k}\mathrm{Q}}{\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}}$

At$\mathrm{x}=\mathrm{\infty },{\mathrm{V}}_{\mathrm{\infty }}=0$

$\therefore$The graph will tend to$0$at$\mathrm{x}=\pm \mathrm{\infty }$and will be maximum at$\mathrm{x}=0$