If the field near the earth's surface is 300V `m^(-1)` directed downwards , what is the surface density of change on the surface of the earth?

Options

A. `3.0 xx 10^(-9) C//m^(2)`
B. `5.0 xx 10^(-9) C//m^(2)`
C. `2.6 xx 10^(-9) C//m^(2)`
D. `7.0 xx 10^(-9) C//m^(2)`

Answer

To find the surface charge density on the surface of the Earth given the electric field strength, we can follow these steps:

### Step 1: Understand the relationship between electric field and surface charge density
The electric field \( E \) due to a surface charge density \( \sigma \) can be expressed using the formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space, approximately equal to \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \).

### Step 2: Rearrange the formula to find surface charge density
Rearranging the formula gives us:
\[ \sigma = 2 \epsilon_0 E \]

### Step 3: Substitute the known values
Given that \( E = 300 \, \text{V/m} \) and \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \), we can substitute these values into the equation:
\[ \sigma = 2 \times (8.85 \times 10^{-12}) \times 300 \]

### Step 4: Calculate the surface charge density
Now we can perform the calculation:
\[ \sigma = 2 \times 8.85 \times 10^{-12} \times 300 \]
\[ \sigma = 2 \times 8.85 \times 300 \times 10^{-12} \]
\[ \sigma = 2 \times 2655 \times 10^{-12} \]
\[ \sigma = 5310 \times 10^{-12} \]
\[ \sigma = 5.31 \times 10^{-9} \, \text{C/m}^2 \]

### Conclusion
The surface charge density on the surface of the Earth is approximately:
\[ \sigma \approx 5.31 \times 10^{-9} \, \text{C/m}^2 \]