The insulation of air vanishes when the electric field is `5xx10^(5)` V/m .the maximum charge that can be given to a sphere of radius 3m is approximately

Options

A. `5muC`
B. `500muC`
C. `10muC`
D. `1muC`

Explanation

To find the maximum charge that can be given to a sphere of radius 3 m before the insulation of air vanishes, we can use the relationship between electric field (E), charge (Q), and radius (r) of the sphere. The formula we will use is:

\[ E = \frac{k \cdot Q}{r^2} \]

Where:
- \( E \) is the electric field strength (in V/m),
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( Q \) is the charge (in C),
- \( r \) is the radius of the sphere (in m).

### Step-by-Step Solution:

1. **Identify the given values**:
- Electric field strength, \( E = 5 \times 10^5 \, \text{V/m} \)
- Radius of the sphere, \( r = 3 \, \text{m} \)

2. **Rearrange the formula to solve for charge (Q)**:
\[ Q = \frac{E \cdot r^2}{k} \]

3. **Substitute the known values into the equation**:
\[ Q = \frac{(5 \times 10^5) \cdot (3^2)}{8.99 \times 10^9} \]

4. **Calculate \( r^2 \)**:
\[ r^2 = 3^2 = 9 \, \text{m}^2 \]

5. **Substitute \( r^2 \) back into the equation**:
\[ Q = \frac{(5 \times 10^5) \cdot 9}{8.99 \times 10^9} \]

6. **Calculate the numerator**:
\[ 5 \times 10^5 \cdot 9 = 4.5 \times 10^6 \]

7. **Now substitute this back into the equation**:
\[ Q = \frac{4.5 \times 10^6}{8.99 \times 10^9} \]

8. **Perform the division**:
\[ Q \approx 5 \times 10^{-4} \, \text{C} \]

### Final Answer:
The maximum charge that can be given to a sphere of radius 3 m is approximately \( 5 \times 10^{-4} \, \text{C} \).