If atmospheric electric field is approximately 150 volt/m and radius of the earth is 6400 km, then the total charge on the earth's surface is

Options

A. `6.8xx10^(5)` coulomb
B. `6.8xx10^(6)` coulomb
C. `6.8xx10^(4)` coulomb
D. `6.8xx10^(9)` coulomb

Answer

To find the total charge on the Earth's surface given the atmospheric electric field and the radius of the Earth, we can follow these steps:

### Step 1: Understand the Given Values
- Atmospheric electric field (E) = 150 V/m
- Radius of the Earth (R) = 6400 km = 6400 × 10^3 m = 6.4 × 10^6 m

### Step 2: Use the Formula for Electric Field
The electric field (E) due to a charged sphere at its surface can be expressed using the formula:
\[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2} \]
where:
- \( Q \) is the total charge on the surface of the sphere (Earth in this case),
- \( R \) is the radius of the sphere,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 3: Rearranging the Formula
We can rearrange the formula to solve for \( Q \):
\[ Q = E \cdot 4 \pi \epsilon_0 R^2 \]

### Step 4: Substitute the Known Values
Now we can substitute the known values into the rearranged formula:
- \( E = 150 \, \text{V/m} \)
- \( R = 6.4 \times 10^6 \, \text{m} \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

Substituting these values:
\[ Q = 150 \cdot 4 \pi (8.85 \times 10^{-12}) (6.4 \times 10^6)^2 \]

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

### Step 6: Calculate \( Q \)
Now plug in the values:
\[ Q = 150 \cdot 4 \pi (8.85 \times 10^{-12}) \cdot (40.96 \times 10^{12}) \]

Calculating \( 4 \pi (8.85 \times 10^{-12}) \):
\[ 4 \pi (8.85 \times 10^{-12}) \approx 1.11 \times 10^{-10} \, \text{C}^2/\text{N m}^2 \]

Now substituting this back into the equation for \( Q \):
\[ Q \approx 150 \cdot (1.11 \times 10^{-10}) \cdot (40.96 \times 10^{12}) \]

Calculating this gives:
\[ Q \approx 150 \cdot 4.54 \times 10^{2} \]
\[ Q \approx 68100 \, \text{C} \]

### Step 7: Final Answer
Thus, the total charge on the Earth's surface is approximately:
\[ Q \approx 6.81 \times 10^4 \, \text{C} \]