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 .