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(D) The potential $V$ at the center of a spherical shell due to a surface charge distribution is given by the integral $V = \oint \frac{k dq}{R}$.Sinc

Vedclass · Accepted Answer

$\frac{Q}{4 \pi \varepsilon_0 R}$

Vedclass · Answer

(D) The potential $V$ at the center of a spherical shell due to a surface charge distribution is given by the integral $V = \oint \frac{k dq}{R}$.Since the radius $R$ is constant for all points on the surface of the sphere,we can take $\frac{k}{R}$ outside the integral.$V = \frac{k}{R} \oint dq$.The integral $\oint dq$ represents the total charge $Q$ on the sphere.Therefore,$V = \frac{kQ}{R}$.Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $V = \frac{Q}{4 \pi \varepsilon_0 R}$.Note that the potential at the center of a charged spherical shell depends only on the total charge $Q$ and the radius $R$,and is independent of the distribution of the charge on the surface.

$A$ charge is spread non-uniformly on the surface of a hollow sphere of radius $R$,such that the charge density is given by $\sigma = \sigma_0(1 - \sin \theta)$,where $\theta$ is the usual polar angle. The potential at the centre of the sphere is

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