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. `4.55xx10^(7)` C
B. `4.55xx10^(8)` C
C. `4.55xx10^(5)` C
D. `4.55xx10^(6)` 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 due to a charged sphere. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Electric field intensity (E) = 100 V/m
- Radius of the Earth (R) = 6400 km = 6400 × 10^3 m = 6.4 × 10^6 m

2. **Use the Formula for Electric Field**:
The electric field intensity (E) at the surface of a charged sphere is given by the formula:
\[ E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{R^2} \]
where:
- \( E \) is the electric field intensity,
- \( Q \) is the total charge on the surface,
- \( R \) is the radius of the sphere,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

3. **Rearrange the Formula to Solve for Q**:
Rearranging the formula to find \( Q \):
\[ Q = E \cdot 4 \pi \epsilon_0 R^2 \]

4. **Substitute the Values**:
Substitute \( E = 100 \, \text{V/m} \), \( R = 6.4 \times 10^6 \, \text{m} \), and \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \):
\[ Q = 100 \cdot 4 \pi (8.85 \times 10^{-12}) (6.4 \times 10^6)^2 \]

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

6. **Calculate the Charge Q**:
Now substituting \( R^2 \) into the equation:
\[ Q = 100 \cdot 4 \pi (8.85 \times 10^{-12}) (40.96 \times 10^{12}) \]
\[ Q = 100 \cdot 4 \cdot 3.14 \cdot 8.85 \times 10^{-12} \cdot 40.96 \times 10^{12} \]
\[ Q \approx 100 \cdot 4 \cdot 3.14 \cdot 8.85 \cdot 40.96 \times 10^{0} \]
\[ Q \approx 4.55 \times 10^5 \, \text{C} \]

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