Two infinite parallel,non- conducting sheets carry equal positive charge density `sigma` ,One is placed in the `yz` plane other at distance `x=a` Take potential `V=0` at `x=0` then

Options

A. for `0 le x le a`, potential `V = 0`
B. for `x ge a`, potential `V = (sigma)/(epsilon_0)(x - a)`
C. for `x ge a`, potential `V = (sigma)/(epsilon_0)(x - a)`
D. for `x le 0`, potential `V = (sigma)/(epsilon_0) x`

Explanation

To solve the problem of determining the electric potential due to two infinite parallel non-conducting sheets with equal positive charge density \( \sigma \), we will analyze the electric field and potential in different regions defined by the positions of the sheets.

### Step 1: Understand the Configuration
We have two infinite parallel sheets:
- The first sheet is located in the \( yz \)-plane (at \( x = 0 \)).
- The second sheet is at a distance \( x = a \).

Both sheets have a uniform positive charge density \( \sigma \).

### Step 2: Determine the Electric Field
The electric field due to an infinite sheet of charge with charge density \( \sigma \) is given by:
\[ E = \frac{\sigma}{2\epsilon_0} \]
This electric field points away from the sheet if the charge is positive.

**Regions to consider:**
1. **Region 1:** \( x < 0 \) (to the left of the first sheet)
2. **Region 2:** \( 0 < x < a \) (between the two sheets)
3. **Region 3:** \( x > a \) (to the right of the second sheet)

**Electric Field in Each Region:**
- **For \( x < 0 \):** The electric field due to the first sheet is directed to the left (negative x-direction), and there is no contribution from the second sheet. Thus:
\[ E = -\frac{\sigma}{2\epsilon_0} \]

- **For \( 0 < x < a \):** The electric fields from both sheets add up:
\[ E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \]

- **For \( x > a \):** The electric field due to both sheets is directed to the right:
\[ E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \]

### Step 3: Calculate the Potential
The potential difference \( V \) between two points is given by:
\[ V = -\int E \, dx \]

**Potential in Each Region:**

1. **For \( x < 0 \):**
- Taking \( V = 0 \) at \( x = 0 \):
\[ V(x) = V(0) + \int_{0}^{x} -\frac{\sigma}{2\epsilon_0} \, dx = 0 - \left(-\frac{\sigma}{2\epsilon_0} \cdot x\right) = \frac{\sigma}{2\epsilon_0} x \]

2. **For \( 0 < x < a \):**
- Since the electric field is constant:
\[ V(x) = V(0) + \int_{0}^{x} \frac{\sigma}{\epsilon_0} \, dx = 0 + \frac{\sigma}{\epsilon_0} \cdot x = \frac{\sigma}{\epsilon_0} x \]

3. **For \( x > a \):**
- The potential at \( x = a \) is:
\[ V(a) = \frac{\sigma}{\epsilon_0} a \]
- Now, for \( x > a \):
\[ V(x) = V(a) - \int_{a}^{x} \frac{\sigma}{\epsilon_0} \, dx = \frac{\sigma}{\epsilon_0} a - \frac{\sigma}{\epsilon_0}(x - a) = \frac{\sigma}{\epsilon_0} (a - x) \]

### Summary of Potential in Different Regions:
- For \( x < 0 \): \( V(x) = \frac{\sigma}{2\epsilon_0} x \)
- For \( 0 < x < a \): \( V(x) = \frac{\sigma}{\epsilon_0} x \)
- For \( x > a \): \( V(x) = \frac{\sigma}{\epsilon_0} (a - x) \)

### Final Answer:
- **Option 1:** For \( 0 < x < a \), \( V = 0 \) is incorrect.
- **Option 2:** For \( x > a \), \( V = -\frac{\sigma}{\epsilon_0}(x - a) \) is correct.
- **Option 3:** Incorrect.
- **Option 4:** For \( x < 0 \), \( V = \frac{\sigma}{2\epsilon_0} x \) is correct.