Question

A ring of radius R is having two charges q and 2q on its two half parts. The electric potential at a distance2 root 2Rfromthecentrealongtheaxiallineis

Answer 1

Lets solve this using principle of superposition. From every point on the ring the distance of the point at $2\sqrt{2}R$ from the centre of the half rings are exactly same.
So electric potential due to the two half rings will be,
$$\begin{gathered} V_E=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{\sqrt{R^2+(2\sqrt{2}R{)}^2}}+\frac{q}{\sqrt{R^2+(2\sqrt{2}R{)}^2}}\right) \\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{3R}+\frac{q}{3R}\right)=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R} \end{gathered}$$