Two capacities plates equal and opposite charges. When the space between the plates is evacuated, the electric field between the plates is `5xx10^(5)` v/m. when the space between the plates is filled with a dielectric, the electric field becomes `2xx10^(5)`v/m. what is the dielectric constant os the dielectric material?

Options

A. 2
B. 3
C. 3.5
D. 2.5

Explanation

To find the dielectric constant of the dielectric material, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electric Fields**:
- The electric field between the plates when the space is evacuated (vacuum) is given as \( E_0 = 5 \times 10^5 \, \text{V/m} \).
- The electric field when the space is filled with a dielectric is given as \( E_m = 2 \times 10^5 \, \text{V/m} \).

2. **Using the Relationship Between Electric Fields**:
- The relationship between the electric fields in vacuum and in a dielectric is given by the formula:
\[ E_m = \frac{E_0}{K} \]
where \( K \) is the dielectric constant of the material.

3. **Rearranging the Formula**:
- We can rearrange the formula to solve for the dielectric constant \( K \):
\[ K = \frac{E_0}{E_m} \]

4. **Substituting the Values**:
- Now, substitute the values of \( E_0 \) and \( E_m \):
\[ K = \frac{5 \times 10^5 \, \text{V/m}}{2 \times 10^5 \, \text{V/m}} \]

5. **Calculating the Dielectric Constant**:
- Perform the division:
\[ K = \frac{5}{2} = 2.5 \]

6. **Conclusion**:
- The dielectric constant of the dielectric material is \( K = 2.5 \).

### Final Answer:
The dielectric constant of the dielectric material is **2.5**.