A parallel plate capacitor of plate separation d = 1 mm and area 50 `cm^2` is connected to a battery of 50 V.
The magnitude of electric field between plates is:

Options

A. `5xx10^3V//m`
B. `5xx10^4V//m`
C. `5xx10^2V//m`
D. `5xx10^5V//m`

Explanation

To find the magnitude of the electric field (E) between the plates of a parallel plate capacitor, we can use the formula:

\[ E = \frac{V}{d} \]

where:
- \( V \) is the potential difference (voltage) across the plates,
- \( d \) is the separation between the plates.

### Step-by-step Solution:

1. **Identify the given values**:
- Plate separation, \( d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
- Voltage, \( V = 50 \text{ V} \)

2. **Convert the plate separation to SI units**:
- Since \( d \) is given as 1 mm, we convert it to meters:
\[ d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \]

3. **Substitute the values into the formula**:
- Now, we can substitute the values of \( V \) and \( d \) into the electric field formula:
\[ E = \frac{V}{d} = \frac{50 \text{ V}}{1 \times 10^{-3} \text{ m}} \]

4. **Calculate the electric field**:
- Performing the division:
\[ E = 50 \times 10^{3} \text{ V/m} = 5 \times 10^{4} \text{ V/m} \]

5. **Conclusion**:
- The magnitude of the electric field between the plates is:
\[ E = 5 \times 10^{4} \text{ V/m} \]

### Final Answer:
The magnitude of the electric field between the plates is \( 5 \times 10^{4} \text{ V/m} \).