Question

A solid metal sphere of radius R having charge q is enclosed inside the concentric spherical shell of inner radius a and outer radius b as shown in figure. The approximate variation electric field \(\vec{E}\) as a function of distance r from centre O is given by : 

• A. A
• B. B
• C. C
• D. D

Hint:

Electrostatic shielding: The electric field is always zero inside the cavity of a conductor or within the body of the conductor itself.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a metallic (conducting) sphere or shell in electrostatic equilibrium, the electric field inside the material of the conductor is always zero. Outside the conductor, the field follows the inverse square law (\(E \propto 1/r^2\)).
Step 2: Key Formula or Approach:
1. \(E = 0\) for \(r<R\) (inside solid metal sphere).
2. \(E = \frac{kq}{r^2}\) for \(R<r<a\) (air gap).
3. \(E = 0\) for \(a<r<b\) (inside the thickness of the shell).
4. \(E = \frac{kq_{total}}{r^2}\) for \(r>b\) (outside).
Step 3: Detailed Explanation:
The solid sphere is a conductor, so the field is zero from \(0\) to \(R\).
In the region between the sphere and the shell (\(R<r<a\)), the field decreases as \(1/r^2\).
The shell is also metallic, so the field inside its thickness (\(a<r<b\)) is zero.
Outside the entire assembly (\(r>b\)), the field again decreases as \(1/r^2\).
Looking at the graphs:
Graph (A) correctly shows zero field in \([0, R]\) and \([a, b]\), with \(1/r^2\) decays in between.
Step 4: Final Answer:
The correct variation is represented by the graph in Option (A).