Electrostatics Part 3

Assertion A: Inside a uniformly charged spherical shell the electric field is zero at every point, so the potential is constant throughout the interior and equal to its value on the surface.Since the potential is the same at all interior points, the potential difference between any two points inside the shell is zero.Hence the work done in moving a test charge between them is $W = q\,\Delta V = q \times 0 = 0$, whatever path is chosen.So Assertion $A$ is true.Reason R: For a uniformly charged thin spherical shell of radius $R$ carrying charge $Q$, the potential inside ($r < R$) is $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}$, which is constant and equal to the potential on the surface ($r = R$).Thus Reason $R$ is also true.Moreover, it is this constant potential inside the shell that makes $\Delta V = 0$ between any two interior points, which is exactly why the work done is zero for every path.Hence $R$ is the correct explanation of $A$.Therefore, the correct option is: Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.