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(D) The electric potential $dV$ due to a small charge element $dq$ at the centre of the ring is given by $dV = \frac{1}{4 \pi \epsilon_0} \cdot \frac{

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$\frac{\sigma}{4 \epsilon_0}$

Vedclass · Answer

(D) The electric potential $dV$ due to a small charge element $dq$ at the centre of the ring is given by $dV = \frac{1}{4 \pi \epsilon_0} \cdot \frac{dq}{R}$.Since the ring is a half-ring,the total charge $q$ is the product of the linear charge density $\sigma$ and the length of the half-ring $(L = \pi R)$.Thus,$q = \sigma \cdot \pi R$.The total potential $V$ at the centre is the integral of $dV$ over the entire half-ring:$V = \int dV = \int \frac{1}{4 \pi \epsilon_0} \cdot \frac{dq}{R} = \frac{1}{4 \pi \epsilon_0 R} \int dq$.Substituting the total charge $q = \sigma \pi R$:$V = \frac{1}{4 \pi \epsilon_0 R} \cdot (\sigma \pi R) = \frac{\sigma}{4 \epsilon_0}$.

$A$ uniformly charged half-ring of radius $R$ has a linear charge density $\sigma$. The electric potential at the centre of the half-ring is ($\epsilon_0$ = permittivity of free space).

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