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. increase
B. decrease
C. remain same
D. become zero

Answer

To solve the problem regarding the effect of dipping charged parallel plates into kerosene oil on the electric field between them, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- We have two parallel metal plates with charges +Q and -Q facing each other.
- The area of the plates is denoted as A.
- The surface charge density (σ) on the plates is given by:
\[ \sigma = \frac{Q}{A} \]

2. **Electric Field in Air**:
- The electric field (E) between the plates in air (or vacuum) can be expressed using the formula:
\[ E = \frac{\sigma}{\epsilon_0} \]
- Substituting the expression for σ, we get:
\[ E = \frac{Q}{A \epsilon_0} \]
- Here, ε₀ is the permittivity of free space.

3. **Introducing Kerosene Oil**:
- When the plates are dipped in kerosene oil, the medium between the plates changes.
- The relative permittivity (dielectric constant) of kerosene oil is denoted as ε_r.

4. **Electric Field in Kerosene Oil**:
- The electric field between the plates when kerosene oil is present can be expressed as:
\[ E' = \frac{\sigma}{\epsilon_0 \epsilon_r} \]
- Substituting for σ, we have:
\[ E' = \frac{Q}{A \epsilon_0 \epsilon_r} \]

5. **Comparing Electric Fields**:
- Since ε_r (the dielectric constant of kerosene oil) is greater than 1, the electric field in kerosene oil (E') will be:
\[ E' = \frac{Q}{A \epsilon_0 \epsilon_r} < \frac{Q}{A \epsilon_0} = E \]
- This indicates that the electric field decreases when the plates are submerged in kerosene oil.

6. **Conclusion**:
- Therefore, the electric field between the plates will **decrease** when they are dipped in kerosene oil.

### Final Answer:
The electric field between the plates will **decrease**.