Two large metal plates are placed parallel to each other The inner surfaces of plates are charge by `+sigmaand-sigma(Cm^(-2))` The outer surfaces are neutral .The electric field in the negion between the plates and outside the plates is

Options

A. `(2sigma)/(epsilon_(0)),(sigma)/(epsilon_(0))`
B. `(sigma)/(epsilon_(0))`, zero
C. `(2sigma)/(epsilon_(0))`, zero
D. zero , `(2sigma)/(epsilon_(0))`

Explanation

To solve the problem of finding the electric field between and outside two parallel metal plates with charges +σ and -σ, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**: We have two large parallel metal plates. The inner surfaces of the plates are charged with +σ (for Plate A) and -σ (for Plate B). The outer surfaces of both plates are neutral.

2. **Drawing the Diagram**:
- Draw two parallel plates, labeling one as Plate A (with charge +σ) and the other as Plate B (with charge -σ).
- Indicate the direction of the electric field due to each plate. The electric field due to a positively charged plate points away from the plate, while the electric field due to a negatively charged plate points towards the plate.

3. **Electric Field due to Each Plate**:
- The electric field (E) due to a single infinite plane sheet of charge is given by the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
- Therefore, for Plate A (positive charge +σ), the electric field \(E_A\) in the region between the plates is:
\[ E_A = \frac{\sigma}{2\epsilon_0} \]
- For Plate B (negative charge -σ), the electric field \(E_B\) in the region between the plates is:
\[ E_B = -\frac{\sigma}{2\epsilon_0} \]
- The negative sign indicates that the field is directed towards Plate B.

4. **Calculating the Net Electric Field Between the Plates**:
- Since both electric fields are in the same direction between the plates (from Plate A to Plate B), we can add them:
\[ E_{between} = E_A + E_B = \frac{\sigma}{2\epsilon_0} + \left(-\frac{\sigma}{2\epsilon_0}\right) = \frac{\sigma}{\epsilon_0} \]

5. **Electric Field Outside the Plates**:
- Outside the plates, the electric fields due to each plate cancel each other out because the outer surfaces are neutral. Therefore, the electric field outside the plates is:
\[ E_{outside} = 0 \]

6. **Final Result**:
- The electric field in the region between the plates is:
\[ E_{between} = \frac{\sigma}{\epsilon_0} \]
- The electric field outside the plates is:
\[ E_{outside} = 0 \]

### Summary:
- Electric field between the plates: \( \frac{\sigma}{\epsilon_0} \)
- Electric field outside the plates: \( 0 \)