The electric potential at the center O of two | vedclass

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(B) The electric potential $V$ due to a small charge element $dq$ at a distance $r$ is given by $dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{r}$.For a hal

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

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(B) The electric potential $V$ due to a small charge element $dq$ at a distance $r$ is given by $dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{r}$.For a half ring of radius $R$ and linear charge density $\lambda$, the total charge is $q = \lambda 	imes (\pi R)$.Since all points on the half ring are at the same distance $R$ from the center, the potential at the center is $V = \frac{1}{4\pi\epsilon_0} \frac{q}{R} = \frac{1}{4\pi\epsilon_0} \frac{\lambda \pi R}{R} = \frac{\lambda}{4\epsilon_0}$.For two concentric half rings of radii $R_1$ and $R_2$ with the same linear charge density $\lambda$, the total potential at the center $O$ is the sum of the potentials due to each half ring:$V_{total} = V_1 + V_2 = \frac{\lambda}{4\epsilon_0} + \frac{\lambda}{4\epsilon_0} = \frac{2\lambda}{4\epsilon_0} = \frac{\lambda}{2\epsilon_0}$.

The electric potential at the center $O$ of two concentric half rings of radii $R_1$ and $R_2$, having the same linear charge density $\lambda$, is $(\epsilon_0 = \text{permittivity of free space})$:

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