JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Electrostatics

Every point of a half ring of radius $R$ is the same distance $R$ from the centre, and the ring carries charge $q=\lambda (\pi R)$. The potential it produces at the centre is $V=\frac{1}{4\pi {ϵ}_0}\frac{q}{R}=\frac{1}{4\pi {ϵ}_0}\frac{\lambda \pi R}{R}=\frac{\lambda }{4{ϵ}_0}$, independent of the radius. Since potential is a scalar, the two concentric half rings simply add: $V=\frac{\lambda }{4{ϵ}_0}+\frac{\lambda }{4{ϵ}_0}=\frac{\lambda }{2{ϵ}_0}$.