The insulation property of air breaks down at `E=3xx10^(6) "volt"//meter`. The maximum charge that can be given to a sphere of diameter `5m` is approximately (in coulombs)

Options

A. (a) `2 xx 10^(-2)`
B. (b) `2 xx 10^(-3)`
C. (c) `2 xx 10^(-4)`
D. (d) `2 xx 10^(-5)`

Answer

To solve the problem, we need to determine the maximum charge that can be given to a sphere of diameter 5 meters, given that the breakdown electric field of air is \( E = 3 \times 10^6 \, \text{V/m} \).

### Step-by-Step Solution:

1. **Identify the radius of the sphere:**
The diameter of the sphere is given as 5 meters. Therefore, the radius \( r \) can be calculated as:
\[ r = \frac{\text{diameter}}{2} = \frac{5 \, \text{m}}{2} = 2.5 \, \text{m} \]

**Hint:** Remember that the radius is half of the diameter.

2. **Use the formula for electric field due to a charged sphere:**
The electric field \( E \) at the surface of a charged sphere is given by the formula:
\[ E = \frac{k \cdot Q}{r^2} \]
where \( k \) is Coulomb's constant \( (k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2) \), \( Q \) is the charge on the sphere, and \( r \) is the radius of the sphere.

3. **Set up the equation with the breakdown electric field:**
Since the maximum electric field before breakdown is \( E = 3 \times 10^6 \, \text{V/m} \), we can set up the equation:
\[ 3 \times 10^6 = \frac{9 \times 10^9 \cdot Q}{(2.5)^2} \]

4. **Calculate \( (2.5)^2 \):**
\[ (2.5)^2 = 6.25 \]

5. **Substitute \( r^2 \) into the equation:**
The equation now becomes:
\[ 3 \times 10^6 = \frac{9 \times 10^9 \cdot Q}{6.25} \]

6. **Rearrange to solve for \( Q \):**
Multiply both sides by \( 6.25 \):
\[ 3 \times 10^6 \cdot 6.25 = 9 \times 10^9 \cdot Q \]
\[ 18.75 \times 10^6 = 9 \times 10^9 \cdot Q \]
Now, divide both sides by \( 9 \times 10^9 \):
\[ Q = \frac{18.75 \times 10^6}{9 \times 10^9} \]

7. **Calculate \( Q \):**
\[ Q = \frac{18.75}{9} \times 10^{-3} = 2.0833 \times 10^{-3} \, \text{C} \]
Rounding this to two significant figures gives:
\[ Q \approx 2 \times 10^{-3} \, \text{C} \]

### Final Answer:
The maximum charge that can be given to the sphere is approximately \( 2 \times 10^{-3} \, \text{C} \).