For two infinitely long charged parallel sheets, the electric feild at P will be

Options

A. `(sigma)/(2x)-(sigma)/(2(r-x))`
B. `(sigma)/(2epsilon_(0)x)+(sigma)/(2pi(r-x)epsilon_(0))`
C. `(sigma)/(epsilon_(0))`
D. Zero

Answer

To solve the problem of finding the electric field at point P between two infinitely long charged parallel sheets, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have two infinitely long parallel sheets, one positively charged and the other negatively charged. Let's denote the charge density of the positive sheet as \( \sigma \) and that of the negative sheet as \( -\sigma \).

2. **Electric Field Due to a Single Sheet**:
- The electric field \( E \) due to an infinitely long charged sheet is given by the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
- This electric field points away from the positively charged sheet and towards the negatively charged sheet.

3. **Determine the Direction of the Electric Fields**:
- For the positively charged sheet, the electric field \( E_1 \) at point P (which is between the two sheets) will point away from the sheet (towards the negative sheet).
- For the negatively charged sheet, the electric field \( E_2 \) at point P will point towards the sheet (also towards the negative sheet).

4. **Calculate the Total Electric Field at Point P**:
- Since both electric fields \( E_1 \) and \( E_2 \) point in the same direction (towards the negative sheet), we can add their magnitudes:
\[ E_{\text{total}} = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \]

5. **Conclusion**:
- The electric field at point P between the two sheets is:
\[ E_{\text{total}} = \frac{\sigma}{\epsilon_0} \]
- Therefore, the electric field at point P is not zero; it has a magnitude of \( \frac{\sigma}{\epsilon_0} \) directed from the positive sheet to the negative sheet.