Consider a charged thin spherical shell of radius R. Let the charge +q distribute uniformly on the outer surface of the conductor, so that the surface charge density be ‘σ’.
Case 1: Outside the spherical shell:
Let ‘P’ be a point outside the conductor at a distance r from the centre of the spherical shell.
With ‘r’ as the radius imagine a Gaussian surface. The electric field at every point on the Gaussian surface is same and radial (θ = 0).
The flux through the Gaussian surface of radius r is
Φ = E.S cosθS – Surface
area of the Gaussian surface of radius r.
∴ Φ = E.(4πr2) ……(1)
From Gauss’s Law Φ = \(\frac{q}{\varepsilon_0}\) …….(2)
Equating (1) and (2), we get
\(\hat n\) = Unit radial vector
Case 2: On the surface of the spherical shell:
The electric field on the surface of the conductor r = R, then
Case 3: Inside the spherical shell: Since the charges enclosed by the Gaussian surface of radius ('r') is zero, the electric field inside the spherical shell is zero.
