The surface charge densities on two square plates of a parallel plate capacitor are to `+ sigma and - sigma` respectively. What is the magnitude of the electric field in the region between the plates?

Options

A. `sigma/(2epsilon_0)`
B. `sigma/(epsilon_0)`
C. 0
D. None of these

Explanation

To find the magnitude of the electric field in the region between two plates of a parallel plate capacitor with surface charge densities of \( +\sigma \) and \( -\sigma \), we can follow these steps:

### Step 1: Understand the Electric Field Due to a Single Plate
The electric field \( E \) due to a single infinite plane 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 Electric Fields from Each Plate
For the positive plate with charge density \( +\sigma \):
- The electric field \( E_1 \) due to this plate is directed away from the plate (upward if we consider the positive plate on the bottom):
\[ E_1 = \frac{\sigma}{2\epsilon_0} \]

For the negative plate with charge density \( -\sigma \):
- The electric field \( E_2 \) due to this plate is directed towards the plate (downward if we consider the negative plate on the top):
\[ E_2 = \frac{\sigma}{2\epsilon_0} \]

### Step 3: Combine the Electric Fields in the Region Between the Plates
In the region between the plates, the electric fields due to both plates add up because they are in the same direction (from the positive plate to the negative plate):
\[ E_{\text{total}} = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \]

### Step 4: Conclusion
Thus, the magnitude of the electric field in the region between the plates is:
\[ E = \frac{\sigma}{\epsilon_0} \]