Two parallel metal plates having charges `+Q` and `-Q` face each other at a certain distance between them.If the plates are now dipped in kerosene oil tank ,the electric field between the plates will

Options

A. became zero
B. increase
C. decrease
D. remain same

Answer

To solve the problem, we need to analyze the effect of dipping two parallel charged plates into kerosene oil on the electric field between them.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- We have two parallel metal plates with charges +Q and -Q.
- The electric field (E) between two parallel plates is given by the formula:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the permittivity of free space.

2. **Effect of a Dielectric**:
- When the plates are dipped into a dielectric medium (in this case, kerosene oil), the electric field is modified. The new electric field \(E'\) in the presence of a dielectric is given by:
\[ E' = \frac{\sigma}{k \epsilon_0} \]
where \(k\) is the dielectric constant of the medium.

3. **Understanding the Dielectric Constant**:
- The dielectric constant \(k\) for kerosene oil is greater than 1 (typically around 2.2 to 2.5). This means that the electric field will be reduced when the dielectric is introduced.

4. **Comparing the Electric Fields**:
- Since \(k > 1\), it follows that:
\[ E' < E \]
This indicates that the electric field between the plates decreases when they are submerged in kerosene oil.

5. **Conclusion**:
- Therefore, the electric field between the plates decreases when they are dipped into kerosene oil.

### Final Answer:
The electric field between the plates will **decrease** when they are dipped in kerosene oil.