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(B) The electric potential $V$ due to a charge $q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$.For a half ring of radiu

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$\frac{\lambda}{2 \epsilon_0}$

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(B) The electric potential $V$ due to a charge $q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$.For a half ring of radius $R$ and linear charge density $\lambda$,the total charge $q$ is $\lambda 	imes (\pi R)$.The potential at the centre due to the first half ring of radius $R_1$ is $V_1 = \frac{1}{4 \pi \epsilon_0} \frac{\lambda \pi R_1}{R_1} = \frac{\lambda}{4 \epsilon_0}$.The potential at the centre due to the second half ring of radius $R_2$ is $V_2 = \frac{1}{4 \pi \epsilon_0} \frac{\lambda \pi R_2}{R_2} = \frac{\lambda}{4 \epsilon_0}$.The total potential at the centre is $V = V_1 + V_2 = \frac{\lambda}{4 \epsilon_0} + \frac{\lambda}{4 \epsilon_0} = \frac{2 \lambda}{4 \epsilon_0} = \frac{\lambda}{2 \epsilon_0}$.

The electric potential at the centre of two concentric half rings of radii $R_1$ and $R_2$,having same linear charge density $\lambda$ is

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