Two square plates are at potential difference of 100V separated by 2cm. Calculate electric intensity between them

Options

A. `5 xx 10^(-3)`
B. 5000
C. 200
D. 98

Answer

To solve the problem of calculating the electric intensity (electric field strength) between two square plates with a potential difference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Potential difference (V) = 100 V
- Distance between the plates (D) = 2 cm

2. **Convert Distance to Meters:**
- Since the standard unit for distance in the SI system is meters, we need to convert 2 cm to meters.
- \( D = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} \)

3. **Use the Formula for Electric Field Intensity:**
- The electric field intensity (E) between two plates is given by the formula:
\[ E = \frac{V}{D} \]
- Where:
- \( E \) is the electric field intensity in volts per meter (V/m),
- \( V \) is the potential difference in volts (V),
- \( D \) is the distance between the plates in meters (m).

4. **Substitute the Values into the Formula:**
- Substitute \( V = 100 \, \text{V} \) and \( D = 2 \times 10^{-2} \, \text{m} \) into the formula:
\[ E = \frac{100 \, \text{V}}{2 \times 10^{-2} \, \text{m}} \]

5. **Calculate the Electric Field Intensity:**
- Performing the division:
\[ E = \frac{100}{0.02} = 5000 \, \text{V/m} \]

6. **Conclusion:**
- The electric field intensity between the two square plates is \( 5000 \, \text{V/m} \).

### Final Answer:
The electric intensity between the two square plates is **5000 V/m**.

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