Question

A spherical conducting shell of inner radius '$r_1$' and outer radius '$r_2$' has a charge 'Q'. A charge $-q$ is placed at the centre of the shell. The surface charge density on the inner and outer surface of the shell will be

• A. $\frac{q}{4\pi r_1^2}$ and $\frac{Q - q}{4\pi r_2^2}$
• B. $\frac{q}{4\pi r_1^2}$ and $\frac{Q}{4\pi r_2^2}$
• C. $\frac{-q}{4\pi r_1^2}$ and $\frac{Q + q}{4\pi r_2^2}$
• D. zero and $\frac{Q - q}{4\pi r_2^2}$

Hint:

Electrostatic induction inside conductors always acts to mirror charges: a central charge of $-q$ requires an inner surface layout of exactly $+q$. This immediately eliminates option (C) and leaves option (A) as the correct choice!

Answer 1

The Correct Option is A

Let's track the charge distributions across the conducting material layers using electrostatic induction principles: 1. When a charge $-q$ is placed at the hollow center point, it creates an electric field that pulls free electrons away from the inner boundary. This induces an equal and opposite positive charge $+q$ uniformly across the inner surface area ($A_1 = 4\pi r_1^2$). 2. To preserve charge neutrality rules within the bulk metal, this migration leaves behind an equal negative charge $-q$ that moves to the outer boundary. Since the shell initially held a net background charge $Q$, the updated net charge residing on the outer boundary surface becomes $(Q - q)$. Surface charge density ($\sigma$) is defined as total charge divided by surface area ($\sigma = \frac{\text{Charge}}{\text{Area}}$): * Inner surface density: $\sigma_{\text{inner}} = \frac{q}{4\pi r_1^2}$ * Outer surface density: $\sigma_{\text{outer}} = \frac{Q - q}{4\pi r_2^2}$
Final Answer:
The surface charge densities correspond to option (A).