Question

A charge \( +Q \) is placed on a thin conducting spherical shell of radius \( r \). Derive an expression for the electric field at a point lying (i) inside and (ii) outside the shell.

Hint:

Remember that the double electric field strength near a conducting plate compared to a nonconducting sheet is a result of the conductive property, which causes charges to redistribute and maximize the field on the exposed side. This principle is fundamental in designing effective electromagnetic shields and capacitors, where surface charge distribution plays a critical role in performance.

Answer 1

Step 1: Electric field inside the shell. According to the properties of conductors in electrostatic equilibrium, the electric field inside a conducting shell is zero. This is because the charges reside on the surface and symmetrical distribution of charge ensures no net electric field points inside the shell. 

Step 2: Electric field outside the shell. For points outside the spherical shell, the shell can be considered as a point charge at the center for the purpose of calculating the electric field. The electric field \( E \) at a distance \( x \) from the center (where \( x>r \)) is given by Coulomb's Law: \[ E = \frac{1}{4\pi \epsilon_0} \frac{Q}{x^2} \] where \( \epsilon_0 \) is the permittivity of free space. This formula indicates that the electric field behaves as if all the charge \( Q \) were concentrated at the center of the sphere.