Each of two large conducting parallel plates has one sided surface area A. If one of the plates is given a charge Q whereas the other is neutral, then the electric field at a point in between the plates is given by

Options

A. `(Q)/(A epsilon_(0))`
B. `(Q)/(2A epsilon_(0))`
C. `(Q)/(4A epsilon_(0))`
D. Zero

Answer

To find the electric field at a point between two large conducting parallel plates, where one plate has a charge \( Q \) and the other is neutral, we can follow these steps:

### Step 1: Understand the Configuration
We have two large parallel plates:
- Plate 1: Charged with a charge \( +Q \)
- Plate 2: Neutral

### Step 2: Determine the Electric Field Due to Each Plate
1. **Electric Field Due to the Charged Plate**:
- The electric field \( E \) due to an infinite sheet of charge is given by the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
- Where \( \sigma \) is the surface charge density, defined as:
\[ \sigma = \frac{Q}{A} \]
- Substituting \( \sigma \) into the electric field formula gives:
\[ E = \frac{Q}{2A\epsilon_0} \]

2. **Electric Field Due to the Neutral Plate**:
- A neutral conducting plate does not create an electric field in the region between the plates. Therefore, the electric field due to the neutral plate is:
\[ E = 0 \]

### Step 3: Calculate the Total Electric Field Between the Plates
- The total electric field \( E_{total} \) in the region between the plates is the vector sum of the electric fields due to each plate. Since the neutral plate contributes no electric field, we have:
\[ E_{total} = E_{charged} + E_{neutral} = \frac{Q}{2A\epsilon_0} + 0 = \frac{Q}{2A\epsilon_0} \]

### Conclusion
The electric field at a point in between the plates is given by:
\[ E = \frac{Q}{2A\epsilon_0} \]