Two uniform sheets of charge with surface charge densities `+sigma` and `-sigma` are intersecting at right angles to each other. Electric field intensity due to this system is

Options

A. `sigma/(2epsilon_0)`
B. `sigma/(2sqrt2epsilon_0)`
C. `sigma/(epsilon_0sqrt2)`
D. None

Explanation

To find the electric field intensity due to two uniform sheets of charge with surface charge densities \( +\sigma \) and \( -\sigma \) that intersect at right angles, we can follow these steps:

### Step 1: Understand the Electric Field due to a Single Sheet
The electric field \( E \) due to an infinite sheet of charge with surface charge density \( \sigma \) is given by the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Determine the Direction of Electric Fields
- For the positively charged sheet with surface charge density \( +\sigma \), the electric field \( E_1 \) points away from the sheet.
- For the negatively charged sheet with surface charge density \( -\sigma \), the electric field \( E_2 \) points towards the sheet.

### Step 3: Analyze the Configuration
Since the two sheets intersect at right angles, we can consider the electric fields \( E_1 \) and \( E_2 \) to be perpendicular to each other.

### Step 4: Calculate the Magnitudes of the Electric Fields
From Step 1, we know:
\[ E_1 = \frac{\sigma}{2\epsilon_0} \quad \text{(from the +\sigma sheet)} \]
\[ E_2 = \frac{\sigma}{2\epsilon_0} \quad \text{(from the -\sigma sheet)} \]

### Step 5: Use the Pythagorean Theorem to Find the Resultant Electric Field
Since \( E_1 \) and \( E_2 \) are perpendicular, the net electric field \( E_{net} \) can be calculated using the Pythagorean theorem:
\[ E_{net} = \sqrt{E_1^2 + E_2^2} \]

### Step 6: Substitute the Values
Substituting the values of \( E_1 \) and \( E_2 \):
\[ E_{net} = \sqrt{\left(\frac{\sigma}{2\epsilon_0}\right)^2 + \left(\frac{\sigma}{2\epsilon_0}\right)^2} \]
\[ E_{net} = \sqrt{2 \left(\frac{\sigma}{2\epsilon_0}\right)^2} \]
\[ E_{net} = \frac{\sigma}{2\epsilon_0} \sqrt{2} \]

### Step 7: Final Result
Thus, the electric field intensity due to the system is:
\[ E_{net} = \frac{\sigma \sqrt{2}}{2\epsilon_0} \]