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

Topic of the Question:
This question deals with electrostatic induction in conductors and the calculation of surface charge density on a conducting spherical shell.
Step 1 : Understanding the Question:
We are given an uncharged, conducting spherical shell with an inner radius $R_1$ and an outer radius $R_2$. A point charge $Q$ is positioned at the exact center of this shell. We need to find the resulting surface charge density on the outer surface of the conducting shell.
Step 2 : Key Formulas and Approach:

Electrostatic induction: Under electrostatic equilibrium, the electric field inside the metallic bulk of a conductor must be zero.

Gauss's Law states that the net charge enclosed by any Gaussian surface drawn completely inside the conducting material must be zero.

Surface charge density ($\sigma$) represents charge per unit area: $\sigma = \frac{\text{Charge on the surface}}{\text{Total surface area}} = \frac{q}{4\pi R^2}$.

Step 3 : Detailed Explanation:

When a point charge $+Q$ is placed at the center, it produces an electric field pointing outwards. To shield the interior of the conducting material and maintain a zero electric field inside the bulk of the shell, electrostatic induction occurs.

An equal and opposite charge of $-Q$ is drawn to the inner surface of the shell at radius $R_1$. This inner charge completely cancels out the field of the central charge inside the conducting material.

Because the conducting shell was initially uncharged and is electrically isolated, the total net charge on the shell must remain zero by the law of conservation of charge.

Consequently, a charge of $+Q$ must be distributed over the outer boundary of the shell at radius $R_2$ to balance the $-Q$ induced on the inner surface.

The outer boundary of the shell is a sphere of radius $R_2$, which has a total surface area of $4\pi R_2^2$.

We calculate the surface charge density on this outer boundary by dividing the outer surface charge by the outer surface area: $\sigma_{\text{outer}} = \frac{+Q}{4\pi R_2^2}$.

Step 4 : Final Answer:
The induced charge on the outer surface is $+Q$, which gives a surface charge density of $\frac{Q}{4\pi R_2^2}$. This corresponds to Option (B).