Two parallel large thin metal sheets have equal surface charge densities `(sigma = 26.4 xx 10^(-12) C//m^(2))` of opposite signs. The electric field between these sheets is

Options

A. `1.5 N//C`
B. `1.5xx10^(-10)N//C`
C. `3.N//C`
D. `3xx10^(-6)N//C`

Explanation

To solve the problem of finding the electric field between two parallel large thin metal sheets with equal surface charge densities of opposite signs, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two parallel sheets: one with a positive surface charge density \( \sigma = 26.4 \times 10^{-12} \, \text{C/m}^2 \) and the other with a negative surface charge density \( -\sigma \).
- The electric field produced by a single infinite sheet of charge is given by the formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

2. **Calculating the Electric Field Due to Each Sheet**:
- For the positively charged sheet, the electric field \( E_+ \) is directed away from the sheet:
\[ E_+ = \frac{\sigma}{2 \epsilon_0} \]
- For the negatively charged sheet, the electric field \( E_- \) is directed towards the sheet:
\[ E_- = \frac{-\sigma}{2 \epsilon_0} \]
- However, since we are interested in the region between the sheets, both electric fields will add up in that region.

3. **Total Electric Field Between the Sheets**:
- The total electric field \( E \) between the sheets is the sum of the magnitudes of the electric fields due to both sheets:
\[ E = E_+ + |E_-| = \frac{\sigma}{2 \epsilon_0} + \frac{\sigma}{2 \epsilon_0} = \frac{\sigma}{\epsilon_0} \]

4. **Substituting the Values**:
- Now, substituting the value of \( \sigma \) and \( \epsilon_0 \):
\[ E = \frac{26.4 \times 10^{-12}}{8.85 \times 10^{-12}} \]

5. **Calculating the Electric Field**:
- Performing the division:
\[ E \approx 2.99 \, \text{N/C} \approx 3 \, \text{N/C} \]

6. **Final Result**:
- The electric field between the two sheets is approximately \( 3 \, \text{N/C} \).