If the electric field intensity in a fair weather atmosphere is 100 V/m, then total charge on the earth's surface is (radius of the earth is 6400 km)

Options

A. `4.55 xx 10^(7) C`
B. ` 4.55 xx 10^(8) C`
C. ` 4.55 xx 10^(9) C`
D. ` 4.55 xx 10^(5) C`

Answer

To find the total charge on the Earth's surface given the electric field intensity, we can use the formula for the electric field (E) around a conducting sphere. The formula is:

\[ E = \frac{KQ}{R^2} \]

Where:
- \(E\) is the electric field intensity (in V/m),
- \(K\) is Coulomb's constant (\(9 \times 10^9 \, \text{N m}^2/\text{C}^2\)),
- \(Q\) is the total charge (in C),
- \(R\) is the radius of the sphere (in meters).

### Step-by-Step Solution:

1. **Identify the given values**:
- Electric field intensity, \(E = 100 \, \text{V/m}\)
- Radius of the Earth, \(R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} = 6.4 \times 10^6 \, \text{m}\)
- Coulomb's constant, \(K = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\)

2. **Rearrange the formula to solve for \(Q\)**:
\[ Q = \frac{E R^2}{K} \]

3. **Substitute the known values into the equation**:
\[ Q = \frac{100 \, \text{V/m} \times (6.4 \times 10^6 \, \text{m})^2}{9 \times 10^9 \, \text{N m}^2/\text{C}^2} \]

4. **Calculate \(R^2\)**:
\[ R^2 = (6.4 \times 10^6)^2 = 40.96 \times 10^{12} \, \text{m}^2 \]

5. **Substitute \(R^2\) back into the equation for \(Q\)**:
\[ Q = \frac{100 \times 40.96 \times 10^{12}}{9 \times 10^9} \]

6. **Calculate the numerator**:
\[ 100 \times 40.96 \times 10^{12} = 4096 \times 10^{12} \]

7. **Now divide by \(9 \times 10^9\)**:
\[ Q = \frac{4096 \times 10^{12}}{9 \times 10^9} = \frac{4096}{9} \times 10^{3} \, \text{C} \]

8. **Calculate \(\frac{4096}{9}\)**:
\[ \frac{4096}{9} \approx 455.11 \]

9. **Final calculation for \(Q\)**:
\[ Q \approx 455.11 \times 10^{3} \, \text{C} = 4.55 \times 10^{5} \, \text{C} \]

### Final Answer:
The total charge on the Earth's surface is approximately \(4.55 \times 10^5 \, \text{C}\).