A sphere of 10 cm diameter is suspended within a hollow sphere of 12 cm diameter. If the inner sphere be charged to a potential of 15,000 V and the outer sphere be earthed, the charge on the inner sphere is

Options

A. `5 xx 10^(-7) C`
B. `6 xx 10^(-7) C`
C. `7 xx 10^(-7) C`
D. `8 xx 10^(-7) C`

Explanation

To solve the problem, we need to find the charge on the inner sphere when it is charged to a potential of 15,000 V and the outer sphere is earthed (grounded). Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the Radii of the Spheres**:
- The diameter of the inner sphere is 10 cm, so its radius \( r_1 = \frac{10}{2} = 5 \) cm = 0.05 m.
- The diameter of the outer sphere is 12 cm, so its radius \( r_2 = \frac{12}{2} = 6 \) cm = 0.06 m.

2. **Understand the Potential Difference**:
- The inner sphere is at a potential of 15,000 V.
- The outer sphere is earthed, which means its potential is 0 V.
- Therefore, the potential difference \( \Delta V = V_{\text{inner}} - V_{\text{outer}} = 15000 \, \text{V} - 0 \, \text{V} = 15000 \, \text{V} \).

3. **Calculate the Capacitance of the System**:
- The formula for the capacitance \( C \) of a system of two concentric spheres is given by:
\[ C = \frac{4 \pi \epsilon_0 r_1 r_2}{r_2 - r_1} \]
- Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).
- Plugging in the values:
\[ C = \frac{4 \pi (8.85 \times 10^{-12}) (0.05)(0.06)}{0.06 - 0.05} \]

4. **Calculate the Capacitance**:
- First, calculate the denominator:
\[ r_2 - r_1 = 0.06 - 0.05 = 0.01 \, \text{m} \]
- Now substituting the values:
\[ C = \frac{4 \pi (8.85 \times 10^{-12}) (0.05)(0.06)}{0.01} \]
- Calculate:
\[ C = \frac{4 \pi (8.85 \times 10^{-12}) (0.003)}{0.01} = \frac{4 \pi (2.655 \times 10^{-14})}{0.01} \]
\[ C \approx \frac{3.34 \times 10^{-13}}{0.01} = 3.34 \times 10^{-11} \, \text{F} \]

5. **Calculate the Charge on the Inner Sphere**:
- The charge \( Q \) on the inner sphere can be calculated using the formula:
\[ Q = C \Delta V \]
- Substituting the values:
\[ Q = (3.34 \times 10^{-11} \, \text{F})(15000 \, \text{V}) = 5.01 \times 10^{-7} \, \text{C} \]

6. **Final Result**:
- The charge on the inner sphere is approximately \( 5 \times 10^{-7} \, \text{C} \).

### Final Answer:
The charge on the inner sphere is \( 5 \times 10^{-7} \, \text{C} \).