Three conducting spheres A, B and C are shown in figure. The radii of the spheres are a, b and c respectively. A and B connected by a conducting wire. The capacity of the system is

Options

A. `4pi epsilon_(0)(a+b+c)`
B. `4pi epsilon_(0)((bc)/(c-b))`
C. `4pi epsilon_(0)((1)/(a)+(1)/(b)+(1)/(c))`
D. `4 pi epsilon_(0)((abc)/(ab+bc+ca))`

Answer

To find the capacitance of the system consisting of three conducting spheres A, B, and C, where A and B are connected by a conducting wire, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have three conducting spheres: A (radius = a), B (radius = b), and C (radius = c).
- Spheres A and B are connected by a conducting wire, which means they will share the same electric potential.

2. **Charge Distribution**:
- When A and B are connected, any excess charge will redistribute between them until they reach the same potential. Since sphere A has no charge on its surface when connected to B, we can ignore it in our calculations for capacitance.

3. **Effective Capacitance**:
- The capacitance of two spheres can be calculated using the formula for capacitance between two spheres:
\[ C = 4\pi \epsilon_0 \frac{b \cdot c}{c - b} \]
- Here, \( \epsilon_0 \) is the permittivity of free space, \( b \) is the radius of sphere B, and \( c \) is the radius of sphere C.

4. **Final Expression**:
- Thus, the capacitance of the system formed by spheres B and C is:
\[ C = 4\pi \epsilon_0 \frac{b \cdot c}{c - b} \]

5. **Conclusion**:
- Therefore, the capacitance of the system is given by the expression above.