Two metal plates having charges `Q and -Q` face each other at some separation and are dipped into an oil tank .If the oil is pumped out, the eletric field between the plates will

Options

A. increase
B. decrease
C. remain unchangeed
D. become zero

Answer

To solve the problem, we need to analyze the effect of removing the dielectric medium (oil) between two charged plates on the electric field between them.

### Step-by-Step Solution:

1. **Understanding the Electric Field in a Capacitor:**
The electric field \( E \) between two parallel plates with surface charge densities \( +\sigma \) and \( -\sigma \) is given by the formula:
\[ E = \frac{\sigma}{\epsilon} \]
where \( \epsilon \) is the permittivity of the medium between the plates.

2. **Identifying the Medium:**
Initially, the plates are dipped in an oil tank, which acts as a dielectric medium. The permittivity of the oil can be expressed as:
\[ \epsilon = k \epsilon_0 \]
where \( k \) is the dielectric constant of the oil, and \( \epsilon_0 \) is the permittivity of free space.

3. **Electric Field with Dielectric:**
When the plates are submerged in oil, the electric field can be expressed as:
\[ E = \frac{\sigma}{k \epsilon_0} \]
Here, \( k > 1 \) since oil is a dielectric material.

4. **Effect of Pumping Out the Oil:**
When the oil is pumped out, the dielectric constant \( k \) approaches 1 (the plates are now in a vacuum or air). Therefore, the electric field becomes:
\[ E = \frac{\sigma}{\epsilon_0} \]

5. **Comparing Electric Fields:**
Since \( k \) was greater than 1 when oil was present, we can conclude that:
\[ E \text{ (in oil)} = \frac{\sigma}{k \epsilon_0} < \frac{\sigma}{\epsilon_0} = E \text{ (in vacuum)} \]
This indicates that the electric field increases when the oil is removed.

6. **Conclusion:**
Therefore, when the oil is pumped out, the electric field between the plates will **increase**.

### Final Answer:
The electric field between the plates will **increase** when the oil is pumped out.

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