Two large parallel metal carry charges +Q and –Q respectively . A test charge `q_0` placed between them experiences a force F. If the separation between the plants is doubled, then the force on the test charge will be

Options

A. F
B. 2F
C. `F//2`
D. `F//4`

Answer

To solve the problem, we need to understand the relationship between the electric field created by two parallel plates and the force experienced by a test charge placed between them.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two large parallel metal plates, one carrying a charge of +Q and the other -Q.
- A test charge \( q_0 \) is placed between these plates.

2. **Electric Field Between the Plates**:
- The electric field \( E \) between two parallel plates with surface charge densities \( \sigma \) and \( -\sigma \) is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
- Here, \( \sigma = \frac{Q}{A} \), where \( A \) is the area of the plates.
- Therefore, the electric field \( E \) can be expressed as:
\[ E = \frac{Q}{A \epsilon_0} \]

3. **Force on the Test Charge**:
- The force \( F \) experienced by the test charge \( q_0 \) in an electric field \( E \) is given by:
\[ F = q_0 E \]
- Substituting the expression for \( E \):
\[ F = q_0 \left(\frac{Q}{A \epsilon_0}\right) \]

4. **Doubling the Separation**:
- If the separation between the plates is doubled, the electric field \( E \) between the plates remains unchanged because it depends only on the surface charge density and not on the distance between the plates.
- Thus, the electric field remains:
\[ E = \frac{Q}{A \epsilon_0} \]

5. **New Force on the Test Charge**:
- Since the electric field does not change, the force on the test charge \( q_0 \) also remains the same:
\[ F' = q_0 E = q_0 \left(\frac{Q}{A \epsilon_0}\right) = F \]

### Conclusion:
The force on the test charge \( q_0 \) remains \( F \) even after the separation between the plates is doubled.

### Final Answer:
The force on the test charge will be \( F \).

---