Two spherical conductors each of capacity `C` are charged to potetnial `V` and `-V`. These are then conneted by means of a fine wire. The loss of energy will be

Options

A. zero
B. `1/2CV^(2)`
C. `CV^(2)`
D. `2CV^(2)`

Explanation

To solve the problem of finding the loss of energy when two spherical conductors, each with a capacity \( C \), charged to potentials \( V \) and \( -V \) respectively, are connected by a fine wire, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Charges**:
Each conductor has a capacitance \( C \). The charge on the first conductor (charged to potential \( V \)) is given by:
\[ Q_1 = CV \]
The charge on the second conductor (charged to potential \( -V \)) is:
\[ Q_2 = -CV \]

2. **Connect the Conductors**:
When the two conductors are connected by a wire, the potential of both conductors will equalize. Let the final potential of both conductors be \( V_f \).

3. **Conservation of Charge**:
The total initial charge before connecting the conductors is:
\[ Q_{\text{initial}} = Q_1 + Q_2 = CV - CV = 0 \]
After connecting, since the total charge is conserved, the final charges on both conductors will be equal:
\[ Q_{1f} + Q_{2f} = 0 \]
Let \( Q_{1f} = Q_{2f} = Q_f \). Therefore:
\[ Q_f + Q_f = 0 \implies 2Q_f = 0 \implies Q_f = 0 \]

4. **Calculate Initial Potential Energy**:
The initial potential energy \( U_i \) of the system is the sum of the potential energies of both conductors:
\[ U_i = \frac{1}{2} C V^2 + \frac{1}{2} C (-V)^2 = \frac{1}{2} C V^2 + \frac{1}{2} C V^2 = CV^2 \]

5. **Calculate Final Potential Energy**:
After connecting the conductors, since both have zero charge, the final potential energy \( U_f \) is:
\[ U_f = 0 \]

6. **Calculate Loss of Energy**:
The loss of energy \( \Delta U \) is given by:
\[ \Delta U = U_i - U_f = CV^2 - 0 = CV^2 \]

### Conclusion:
The loss of energy when the two spherical conductors are connected is:
\[ \Delta U = CV^2 \]