Q.A thin conducting spherical shell of radius $R$ has a charge $Q$ which is uniformly distributed on its surface. Which of the four given $V$ versus $r$ plots correctly shows the electrostatic potential due to this spherical shell -- constant for $r<R$ and falling as $1/r$ for $r>R$, or one of the three other, incorrect, shapes offered as options (a), (b), (c), (d)?

Inside a uniformly charged conducting shell the field is zero, so the potential stays constant at its surface value all the way to the centre; outside, it falls off smoothly as $1/r$ -- exactly the graph described in option (d).

Working

For a thin conducting spherical shell of radius R carrying charge Q, uniformly distributed on its surface:

Inside ($r<R$): The electric field inside a uniformly charged spherical shell is zero (by Gauss's law, since the enclosed charge for any Gaussian surface with $r<R$ is zero). Since $E=-dV/dr=0$ inside, the potential does not change with $r$ inside the shell -- it remains constant, equal to the value at the surface:

$V_{in} = V(R) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R} \qquad (r \le R)$

Outside ($r>R$): The shell behaves exactly like a point charge Q at the centre for all external points, so

$V_{out}(r) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r} \qquad (r > R)$ …

The electric field inside a hollow charged conducting sphere is zero at every interior point.

For a conductor in electrostatic equilibrium, free charges redistribute themselves so that the electric field inside the body of the conductor is always zero; any excess charge resides entirely on the outer surface. For a hollow charged metal (conducting) sphere, this means all the charge $q$ resides uniformly on the outer surface.

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