Question

Two conducting shells of radius \(a\) and \(b\) are connected by conducting wire as shown in figure. The capacity of system is :

• A. $4 \pi \varepsilon_{0} \frac{a b}{b-a}$
• B. $4 \pi \varepsilon_{0}(a+b)$
• C. zero
• D. infinite

Answer 1

The Correct Option is D

Explanation:
When two conducting shells are connected by a conducting wire, they form an equipotential system. This means that the potential difference (\(V\)) between the two shells is zero because the wire ensures that both shells are at the same electric potential.
Let's denote the radii of the inner and outer shells as \(a\) and \(b\) respectively, with \(a < b\).
Step-by-Step Analysis:
1. Potential on the shells:
  Since the shells are connected by a wire, the potential on both shells must be the same, say \(V\).
2. Charge Distribution:
  - Let's assume that the inner shell has a charge \(+Q\).
  - By electrostatic induction, the inner surface of the outer shell will have a charge \(-Q\), and the outer surface of the outer shell will have a charge \(+Q\) to maintain neutrality.
3. Electric Potential Calculation:
  - The potential on the inner shell of radius \(a\) due to its own charge is:
    \[ V_{\text{inner}} = \frac{Q}{4\pi \epsilon_0 a} \]
  - The potential on the outer shell of radius \(b\) due to the charge \(Q\) on it is:
    \[  V_{\text{outer}} = \frac{Q}{4\pi \epsilon_0 b}\]
  Since the potentials are the same (\(V_{\text{inner}} = V_{\text{outer}}\)), we have:
    \[ \frac{Q}{4\pi \epsilon_0 a} = \frac{Q}{4\pi \epsilon_0 b}\]
  This equation implies that for \(a \neq b\), \(Q = 0\) to maintain equality, since there cannot be any potential difference.
4. Capacitance Calculation:
  - Capacitance \(C\) is defined as the charge \(Q\) stored per unit potential difference \(V\):
    \[C = \frac{Q}{V}\]
  - Here, \(V = 0\) as the shells are at the same potential. In electrostatics, the capacitance of such a system is effectively infinite because the shells can theoretically hold an unlimited amount of charge without creating any potential difference.
    \[C = \frac{Q}{0} \rightarrow \infty\]
Conclusion:
The capacity (capacitance) of the system, when two conducting shells are connected by a wire and made to be at the same potential, is theoretically infinite. This is because the potential difference \(V\) is zero, leading to an infinite capacitance value mathematically.
So, the final impressive and detailed answer is:
When two conducting shells of radii \(a\) and \(b\) are connected by a conducting wire, they form an equipotential system. The potential difference between them is zero (\(V = 0\)). Thus, the capacitance \(C\) of the system is given by:
\[C = \frac{Q}{V} \rightarrow \infty\]
Therefore, the capacitance of the system is theoretically infinite i.e correct Answer is Option 4 .