Physics - Electrostatics Question with Solution | TestHub

The electric potential $V$ at the center of a uniformly charged circular arc of radius $R$ and total charge $Q$ is given by:

$V=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R}$

For a half ring with linear charge density $\lambda$, the length of the arc is $\pi R$. Therefore, the total charge on the half ring is:

$Q=\lambda (\pi R)$

Substituting this into the potential formula, the potential at the center due to one half ring is:

$V=\frac{1}{4\pi {ϵ}_0}\frac{\lambda \pi R}{R}=\frac{\lambda }{4{ϵ}_0}$

Notice that the potential at the center of a half ring depends only on the linear charge density $\lambda$ and is independent of its radius $R$.

Since electric potential is a scalar quantity, the total potential at the center $O$ due to the two concentric half rings is the algebraic sum of their individual potentials:

$V_{net}=V_1+V_2$

$V_{net}=\frac{\lambda }{4{ϵ}_0}+\frac{\lambda }{4{ϵ}_0}=\frac{\lambda }{2{ϵ}_0}$

Answer: $\lambda /2{ϵ}_0$