Question

A charge $Q$ is placed at the center of an uncharged conducting spherical shell of inner radius $R_1$ and outer radius $R_2$. The surface charge density on the outer surface of the shell is:

• A. \( \frac{Q}{(4\pi R_1^2)} \)
• B. \( \frac{Q}{(4\pi R_2^2)} \)
• C. \( \frac{-Q}{(4\pi R_1^2)} \)
• D. \( \text{Zero} \)

Hint:

By Gauss's Law, the total charge enclosed within any Gaussian surface drawn inside the conducting material must be zero, forcing the inner surface charge to be completely opposite to the central cavity charge.

Answer 1

The Correct Option is B

Concept: Inside a conductor under electrostatic equilibrium, the electric field is zero everywhere. When a charge is placed inside the cavity of a conducting shell, electrostatic induction occurs:
• An equal and opposite charge is induced on the inner surface of the shell.
• An equal and similar charge is induced on the outer surface of the shell to maintain its initial net charge state. The surface charge density (\(\sigma\)) is given by the formula: \[ \sigma = \frac{\text{Charge on the surface}}{\text{Surface area}} = \frac{q}{4\pi R^2} \] Step 1: Determining the induced charges on the inner and outer surfaces.
The initial net charge on the conducting spherical shell is zero (uncharged). When a charge \(+Q\) is placed at its center: \[ \text{Induced charge on the inner surface (at radius } R_1) = -Q \] To keep the net charge of the shell zero, a balancing positive charge must appear on its exterior boundary: \[ \text{Induced charge on the outer surface (at radius } R_2) = +Q \]

Step 2: Calculating the surface charge density on the outer surface.
The surface area of the outer sphere with radius \(R_2\) is \(4\pi R_2^2\). Substituting the outer charge and surface area into the density formula gives: \[ \sigma_{\text{outer}} = \frac{\text{Charge on outer surface}}{\text{Outer surface area}} = \frac{Q}{4\pi R_2^2} \]