Two metal plates having a.p.d 600 volts are 2 cm apart. It is found that a particle of mass `1.96xx10^(-12)` g remains suspended in the electric field. The intensity of elelctric is

Options

A. `1.96xx10^(-120xx600` volts/m
B. `3xx10^(4)` volts /
C. `3xx10^(2)` volts /m
D. `12xx10^(4)`volts/m

Explanation

To find the intensity of the electric field between two metal plates with a potential difference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Potential difference (V) = 600 volts
- Distance between the plates (d) = 2 cm

2. **Convert the Distance to Meters:**
- Since 1 cm = 0.01 m, we convert 2 cm to meters:
\[ d = 2 \, \text{cm} = 2 \times 0.01 \, \text{m} = 0.02 \, \text{m} \]

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

4. **Substitute the Values into the Formula:**
- Now, substitute the values of V and d into the formula:
\[ E = \frac{600 \, \text{V}}{0.02 \, \text{m}} \]

5. **Calculate the Electric Field Intensity:**
- Performing the calculation:
\[ E = \frac{600}{0.02} = 30000 \, \text{V/m} = 3 \times 10^4 \, \text{V/m} \]

6. **Final Answer:**
- The intensity of the electric field between the plates is:
\[ E = 3 \times 10^4 \, \text{V/m} \]