Two thin conducting concentric shells of radii \( R_1 \) and \( R_2 \) are shown in the figure. The inner shell carries a charge \( +Q \) and the outer shell is neutral. Then the correct statement is:
Show Hint
When dealing with concentric spherical conductors, remember that the potential inside a conductor is constant and can be manipulated by placing the conductor at a specific potential, such as zero.
(a) When the switch is closed, the potential on the inner shell becomes zero. This is because when the switch is closed, the outer shell is neutral and does not influence the inner shell. The potential on the inner shell is grounded and becomes zero.
(b) With the switch closed, the charge on the outer shell is \( -Q \), and it is uniformly distributed over the surface of the outer shell. This happens because of the influence of the charge \( +Q \) on the inner shell, which induces a charge \( -Q \) on the inner surface of the outer shell.
Was this answer helpful?
0
0
Hide Solution
Verified By Collegedunia
Approach Solution -2
Step 1: Understand the system.
Two concentric conducting spherical shells are given:
- Inner shell of radius \( R \) carries charge \( +Q \)
- Outer shell of radius \( 2R \) is neutral (initially no net charge)
- Outer shell is grounded
Step 2: Determine the induced charges.
Since the inner shell carries a charge of \( +Q \), it induces a charge \( -Q \) on the inner surface of the outer shell to cancel the electric field inside the conductor.
To keep the outer shell neutral overall, a charge \( +Q \) appears on its outer surface.
Step 3: Analyze grounding effect.
The outer shell is connected to the ground. This means its potential becomes zero.
So the \( +Q \) on the outer surface will flow to the ground, making the net charge on the outer shell \(-Q\) (since only the \(-Q\) induced on the inner surface remains).
Step 4: Evaluate options (as per standard choices in such questions).
Based on the usual multiple-choice options for such problems, these are likely the correct statements:
(a) Inner surface of outer shell has charge \( -Q \)
(b) Outer surface of outer shell is at zero potential (due to grounding)
Hence, both (a) and (b) are correct.
Final Answer: \( \boxed{(a) \text{ and } (b) \text{ are correct}} \)