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: Identify the given values
- 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 around a conducting sphere
The electric field (E) at the surface of a conducting sphere is given by the formula:
\[ E = \frac{kQ}{r^2} \]
where:
- \( E \) is the electric field,
- \( k \) is Coulomb's constant (\( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( Q \) is the total charge on the surface of the sphere,
- \( r \) is the radius of the sphere.

### Step 3: Rearrange the formula to solve for Q
We can rearrange the formula to find \( Q \):
\[ Q = \frac{E \cdot r^2}{k} \]

### Step 4: Substitute the known values into the formula
Now, substitute the values of \( E \), \( r \), and \( k \):
- \( E = 150 \, \text{V/m} \)
- \( r = 6.4 \times 10^6 \, \text{m} \)
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

So,
\[ Q = \frac{150 \cdot (6.4 \times 10^6)^2}{9 \times 10^9} \]

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

### Step 6: Substitute \( r^2 \) back into the equation for Q
Now substitute \( r^2 \) back into the equation:
\[ Q = \frac{150 \cdot 40.96 \times 10^{12}}{9 \times 10^9} \]

### Step 7: Calculate the numerator
Calculate the numerator:
\[ 150 \cdot 40.96 \times 10^{12} = 6144 \times 10^{12} \]

### Step 8: Divide by the denominator
Now divide by \( 9 \times 10^9 \):
\[ Q = \frac{6144 \times 10^{12}}{9 \times 10^9} = \frac{6144}{9} \times 10^{3} \]

### Step 9: Calculate the final value of Q
Calculating \( \frac{6144}{9} \):
\[ \frac{6144}{9} \approx 684.89 \]

So,
\[ Q \approx 684.89 \times 10^{3} \, \text{C} = 6.849 \times 10^5 \, \text{C} \]

### Final Answer
The total charge on the Earth's surface is approximately:
\[ Q \approx 6.85 \times 10^5 \, \text{C} \]

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