Two plates are `2cm` apart, a potential difference of `10` volt is applied between them, the electric field between the plates is

Options

A. `20 N//C`
B. `500 N//C`
C. `5 N//C`
D. `250 N//C`

Explanation

To find the electric field between two plates when a potential difference is applied, we can use the formula:

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

where:
- \( E \) is the electric field (in volts per meter or V/m),
- \( V \) is the potential difference (in volts),
- \( d \) is the distance between the plates (in meters).

### Step-by-Step Solution:

1. **Identify the given values:**
- Potential difference \( V = 10 \) volts
- Distance between the plates \( d = 2 \) cm

2. **Convert the distance from centimeters to meters:**
\[ d = 2 \text{ cm} = 2 \times 10^{-2} \text{ m} = 0.02 \text{ m} \]

3. **Substitute the values into the formula for electric field:**
\[ E = \frac{V}{d} = \frac{10 \text{ volts}}{0.02 \text{ m}} \]

4. **Calculate the electric field:**
\[ E = \frac{10}{0.02} = 500 \text{ V/m} \]

5. **Convert the unit of electric field to Newton per coulomb (N/C):**
Since \( 1 \text{ V/m} = 1 \text{ N/C} \), we have:
\[ E = 500 \text{ N/C} \]

### Final Answer:
The electric field between the plates is \( 500 \text{ N/C} \).

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