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 find the maximum charge that can be given to a spherical conductor of radius 3 m in an electric field of strength \(3 \times 10^6 \, \text{V/m}\), we will follow these steps:

### Step 1: Calculate the Capacitance of the Spherical Conductor
The capacitance \(C\) of a spherical conductor is given by the formula:
\[ C = 4 \pi \epsilon_0 R \]
where:
- \(R\) is the radius of the sphere (3 m),
- \(\epsilon_0\) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \, \text{F/m}\).

Substituting the values:
\[ C = 4 \pi (8.85 \times 10^{-12}) (3) \]
Calculating this gives:
\[ C \approx 4 \pi (8.85 \times 10^{-12}) \times 3 \approx 3.34 \times 10^{-10} \, \text{F} \]

### Step 2: Calculate the Voltage across the Spherical Conductor
The voltage \(V\) across the conductor can be calculated using the relationship:
\[ V = E \times R \]
where:
- \(E\) is the electric field strength (\(3 \times 10^6 \, \text{V/m}\)),
- \(R\) is the radius of the sphere (3 m).

Substituting the values:
\[ V = (3 \times 10^6) \times 3 = 9 \times 10^6 \, \text{V} \]

### Step 3: Calculate the Maximum Charge \(Q_{\text{max}}\)
The maximum charge that can be stored on the conductor is given by:
\[ Q_{\text{max}} = C \times V \]
Substituting the values we calculated:
\[ Q_{\text{max}} = (3.34 \times 10^{-10}) \times (9 \times 10^6) \]
Calculating this gives:
\[ Q_{\text{max}} \approx 3.006 \times 10^{-3} \, \text{C} \approx 3 \times 10^{-3} \, \text{C} \]

### Final Answer
Thus, the maximum charge that can be given to the spherical conductor is:
\[ Q_{\text{max}} \approx 3 \times 10^{-3} \, \text{C} \]

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