Question

Obtain the formula for the intensity of electric field outside a charged thin spherical shell with the help of Gauss' law.

Hint:

The electric field outside a charged spherical shell behaves as if all the charge were concentrated at the center of the shell. This is a direct consequence of Gauss' law applied to spherical symmetry.

Answer 1

Electric Field Outside a Charged Spherical Shell:
According to Gauss' law, the electric field due to a spherically symmetric charge distribution can be determined by considering a Gaussian surface that is a sphere of radius \(r\) outside the charged spherical shell. The total charge enclosed by the Gaussian surface is the charge on the spherical shell. Gauss's law is expressed as: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}, \] where: - \( Q_{\text{enc}} \) is the total charge enclosed by the Gaussian surface, and - \( \varepsilon_0 \) is the permittivity of free space. Since the electric field is radially symmetric and uniform over the Gaussian surface, the left-hand side of Gauss's law becomes: \[ E \cdot 4\pi r^2, \] where \(E\) is the electric field at the distance \(r\) from the center of the shell. Therefore, Gauss's law gives: \[ E \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\varepsilon_0}. \] Solving for \(E\), we get the formula for the electric field outside the spherical shell: \[ E = \frac{Q_{\text{enc}}}{4\pi \varepsilon_0 r^2}. \] This is the same expression for the electric field as if all the charge \( Q_{\text{enc}} \) were concentrated at the center of the spherical shell. This result holds true outside the shell for any point at a distance \(r\) from the center.