Let σ be the uniform surface charge density of a thin spherical shell of radius R (Fig.). The situation has obvious spherical symmetry. The field at any point P, outside or inside, can depend only on r (the radial distance from the centre of the shell to the point) and must be radial (i.e., along the radius vector).
(i) Field outside the shell:
Consider a point P outside the shell with radius vector r. To calculate E at P, we take the Gaussian surface to be a sphere of radius r and with centre O, passing through P. All points on this sphere are equivalent relative to the given charged configuration (spherical symmetry). The electric field at each point of the Gaussian surface, therefore, has the same magnitude E and is along the radius vector at each point. Thus, E and ∆S at every point are parallel and the flux through each element is E ∆S. Summing over all ∆S, the flux through the Gaussian surface is E × 4 πr2. The charge enclosed is σ × 4π R2. By Gauss’s law.
where q = 4πR2σ is the total charge on the spherical shell. Vectorially,
The electric field is directed outward if q > 0 and inward if q < 0. This, however, is exactly the field produced by a charge q placed at the centre O. Thus for points outside the shell, the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at its centre.
(ii) Field inside the shell: In Fig. (b), the point P is inside the shell. The Gaussian surface is again a sphere through P centred at O.
The flux through the Gaussian surface, calculated as before, is E × 4π r2. However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives
E × 4π r2 = 0
i.e., E = 0 (r < R ) that is, the field due to a uniformly charged thin shell is zero at all points inside the shell. This important result is a direct consequence of Gauss’s law which follows from Coulomb’s law. The experimental verification of this result confirms the 1/r2 dependence in Coulomb’s law.
