Question

A solid spherical conductor of radius \(R\) has a spherical cavity of radius \(a\) (\(a<R\)) at its centre. A charge \(+Q\) is kept at the centre. The charge at the inner surface, outer surface and at a position \(r\) (\(a<r<R\)) are respectively

• A. \(+Q,-Q,0\)
• B. \(-Q,+Q,0\)
• C. \(0,-Q,0\)
• D. \(+Q,0,Q\)

Hint:

Charge placed in cavity induces equal and opposite charge on inner surface, and remaining charge appears on outer surface.

Answer 1

The Correct Option is B

Step 1: Understand electrostatic equilibrium in conductor.
Inside a conductor (in static condition), electric field must be zero and charge resides only on surfaces.
Step 2: Induced charge on inner surface.
A charge \(+Q\) at the centre induces \(-Q\) uniformly on the inner cavity surface to cancel electric field inside conductor material.
Step 3: Charge on outer surface.
Since the conductor was initially neutral, total induced charge must remain zero.
If inner surface has \(-Q\), outer surface must have \(+Q\).
Step 4: Charge inside the metal region \(a<r<R\).
Within the conducting material, no net charge exists in the volume (only surfaces carry charge).
So at any point \(a<r<R\), net charge enclosed in that volume element is \(0\).
Final Answer:
\[ \boxed{(-Q,\,+Q,\,0)} \]