MHTCET 13 Apr 2026 Shift 2 Paper

Concept:The potential at the centre of a charged circular arc depends only on the total charge and the radius, and for a half ring, the radius cancels out.Explanation:For any charged arc of radius $R$, every small charge element is at the same distance $R$ from the centre.So, the potential at the centre due to the whole arc is:$$V = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$Each ring is a half ring.The length of a half ring of radius $R$ is $\pi R$.Given the linear charge density $\lambda$, the charge on one half ring is:$$Q = \lambda (\pi R)$$Substituting this $Q$ into the potential formula gives:$$V = \frac{1}{4\pi\epsilon_0}\frac{\lambda \pi R}{R} = \frac{\lambda}{4\epsilon_0}$$Thus, the potential at the centre due to one half ring is $\frac{\lambda}{4\epsilon_0}$, independent of its radius.The two concentric half rings have radii $R_1$ and $R_2$, but both give the same potential $\frac{\lambda}{4\epsilon_0}$.Since electric potential is a scalar quantity, the net potential is the sum of the individual potentials:$$V_{\text{net}} = \frac{\lambda}{4\epsilon_0} + \frac{\lambda}{4\epsilon_0} = \frac{\lambda}{2\epsilon_0}$$Answer:$$V_{\text{net}} = \frac{\lambda}{2\epsilon_0}$$So, the correct option is B.