The electric field strength in air at NTP is `3 xx 10^(6)V//m`. The maximum charge that can be given to a spherical conductor of radius 3m is

Options

A. `3xx 10^(4)C`
B. `3 xx 10^(-3)C`
C. `3 xx 10^(-2)C`
D. `3 xx 10^(-1)C`

Answer

To solve the problem, we need to determine the maximum charge that can be given to a spherical conductor of radius 3 meters, given the electric field strength in air at normal temperature and pressure (NTP) is \(3 \times 10^6 \, \text{V/m}\).

### Step-by-step Solution:

1. **Identify the Given Values:**
- Electric field strength, \(E = 3 \times 10^6 \, \text{V/m}\)
- Radius of the spherical conductor, \(r = 3 \, \text{m}\)

2. **Calculate the Potential (Voltage) on the Surface of the Sphere:**
The potential \(V\) at the surface of the sphere can be calculated using the formula:
\[ V = E \times r \]
Substituting the values:
\[ V = (3 \times 10^6 \, \text{V/m}) \times (3 \, \text{m}) = 9 \times 10^6 \, \text{V} \]

3. **Calculate the Capacitance of the Spherical Conductor:**
The capacitance \(C\) of a spherical conductor is given by:
\[ C = 4 \pi \epsilon_0 r \]
where \(\epsilon_0\) (the permittivity of free space) is approximately \(8.85 \times 10^{-12} \, \text{F/m}\). Thus:
\[ C = 4 \pi (8.85 \times 10^{-12} \, \text{F/m}) \times (3 \, \text{m}) \]
Calculating this:
\[ C \approx 4 \pi \times 8.85 \times 10^{-12} \times 3 \approx 3.34 \times 10^{-10} \, \text{F} \]

4. **Calculate the Maximum Charge (Q_max) that can be Stored:**
The maximum charge \(Q_{\text{max}}\) that can be stored on the conductor is given by:
\[ Q_{\text{max}} = C \times V \]
Substituting the values of capacitance and voltage:
\[ Q_{\text{max}} = (3.34 \times 10^{-10} \, \text{F}) \times (9 \times 10^6 \, \text{V}) \approx 3.006 \times 10^{-3} \, \text{C} \]

5. **Final Result:**
Therefore, the maximum charge that can be given to the spherical conductor is approximately:
\[ Q_{\text{max}} \approx 3.006 \times 10^{-3} \, \text{C} \text{ or } 3 \, \text{mC} \]