Question

Write down the formula and unit of surface charge density. A charge \( Q \) is distributed over two concentric hollow spheres of radii \( r_1 \) and \( r_2 (r_1 > r_2) \). If their surface charge densities are equal, find the electric potential at their common centre.

Hint:

The electric potential at the centre of a spherical shell depends only on the charge on the shell and the distance from the centre. For concentric spheres, the total potential at the centre is the sum of the potentials from each sphere.

Answer 1

The surface charge density \( \sigma \) is the charge per unit area on the surface of a conductor. It is given by the formula:
\[ \sigma = \frac{Q}{A} \] Where:
- \( Q \) is the charge distributed over the surface,
- \( A \) is the surface area of the sphere.
For a sphere, the surface area is given by:
\[ A = 4 \pi r^2 \] Thus, the surface charge density for a sphere of radius \( r \) is:
\[ \sigma = \frac{Q}{4 \pi r^2} \] The unit of surface charge density is:
\[ \text{Unit of } \sigma = \frac{\text{Coulomb}}{\text{meter}^2} = \text{C/m}^2 \] Now, we are given two concentric hollow spheres with radii \( r_1 \) and \( r_2 \), and their surface charge densities are equal. This implies:
\[ \sigma_1 = \sigma_2 \] Thus, we have the following equations for the surface charge densities of both spheres:
\[ \sigma_1 = \frac{Q}{4 \pi r_1^2}, \sigma_2 = \frac{Q}{4 \pi r_2^2} \] Since \( \sigma_1 = \sigma_2 \), we can equate these expressions:
\[ \frac{Q}{4 \pi r_1^2} = \frac{Q}{4 \pi r_2^2} \] This simplifies to:
\[ r_1^2 = r_2^2 \] However, this is not possible if \( r_1 \neq r_2 \), implying that the surface charge densities cannot be the same unless \( r_1 = r_2 \). Electric Potential at the Common Centre Now, let's calculate the electric potential at the common centre (which is the centre of both spheres). The electric potential at the centre of a spherical shell due to a charge \( Q \) distributed uniformly on its surface is given by:
\[ V = \frac{1}{4 \pi \epsilon_0} \times \frac{Q}{r} \] For a concentric configuration of two spheres, the potential at the common centre will be the sum of the potentials due to both spheres. Therefore, the total potential at the common centre is:
\[ V_{\text{total}} = V_1 + V_2 = \frac{1}{4 \pi \epsilon_0} \left( \frac{Q}{r_1} + \frac{Q}{r_2} \right) \] Thus, the electric potential at the common centre of the spheres is:
\[ V_{\text{total}} = \frac{Q}{4 \pi \epsilon_0} \left( \frac{1}{r_1} + \frac{1}{r_2} \right) \] Where: - \( Q \) is the charge distributed on the spheres,
- \( r_1 \) and \( r_2 \) are the radii of the spheres,
- \( \epsilon_0 \) is the permittivity of free space.