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)/(2)CV^(2)`
C. `CV^(2)`
D. `2 CV^(2)`

Answer

To solve the problem of energy loss when two spherical conductors are connected by a fine wire, we can follow these steps:

### Step 1: Understand the Initial Conditions
We have two spherical conductors, each with a capacitance \( C \). One is charged to a potential \( V \) and the other to \( -V \).

### Step 2: Calculate Initial Energy
The initial energy stored in each capacitor can be calculated using the formula for energy stored in a capacitor:
\[ U = \frac{1}{2} C V^2 \]
- For the first conductor (charged to \( V \)):
\[ U_1 = \frac{1}{2} C V^2 \]
- For the second conductor (charged to \( -V \)):
\[ U_2 = \frac{1}{2} C (-V)^2 = \frac{1}{2} C V^2 \]

### Step 3: Total Initial Energy
The total initial energy \( U_{\text{initial}} \) is the sum of the energies of both conductors:
\[ U_{\text{initial}} = U_1 + U_2 = \frac{1}{2} C V^2 + \frac{1}{2} C V^2 = C V^2 \]

### Step 4: Connect the Conductors
When the two conductors are connected by a fine wire, charge will redistribute until they reach the same potential. The final potential \( V_f \) can be calculated as follows:
- The total charge \( Q \) on the system before connecting is:
\[ Q = CV + C(-V) = 0 \]
Since the total charge is zero, when connected, the final potential \( V_f \) will be:
\[ V_f = 0 \]

### Step 5: Calculate Final Energy
After connecting, the energy stored in the system is:
\[ U_{\text{final}} = \frac{1}{2} C V_f^2 = \frac{1}{2} C (0)^2 = 0 \]

### Step 6: Calculate Loss of Energy
The loss of energy \( \Delta U \) is given by the difference between the initial and final energies:
\[ \Delta U = U_{\text{initial}} - U_{\text{final}} = C V^2 - 0 = C V^2 \]

### Final Answer
The loss of energy when the two spherical conductors are connected by a fine wire is:
\[ \Delta U = C V^2 \]

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