A test charge exeperiences force F when placed in between a PPC , if one plate is removed then the force on this test charge will be :–

Options

A. 0
B. F/2
C. F
D. 2F

Answer

To solve the problem, we need to analyze the situation of a test charge placed between the plates of a parallel plate capacitor and how the force acting on it changes when one plate is removed.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- When the test charge \( Q \) is placed between two parallel plates of a capacitor, it experiences a force \( F \).
- The electric field \( E \) between the plates is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \( \sigma \) is the surface charge density on the plates and \( \epsilon_0 \) is the permittivity of free space.

2. **Calculating the Initial Force**:
- The force \( F \) experienced by the test charge \( Q \) is given by:
\[ F = Q \cdot E = Q \cdot \frac{\sigma}{\epsilon_0} \]

3. **Analyzing the Condition After Removing One Plate**:
- When one plate is removed, the configuration changes. Now, the electric field \( E' \) due to a single plate (the remaining plate) is:
\[ E' = \frac{\sigma}{2\epsilon_0} \]
- This is because the electric field due to a single infinite plane sheet of charge is half that of two sheets.

4. **Calculating the New Force**:
- The new force \( F' \) acting on the test charge \( Q \) after removing one plate is:
\[ F' = Q \cdot E' = Q \cdot \frac{\sigma}{2\epsilon_0} \]
- Substituting the expression for \( E' \):
\[ F' = \frac{1}{2} \left( Q \cdot \frac{\sigma}{\epsilon_0} \right) = \frac{1}{2} F \]

5. **Conclusion**:
- Therefore, the force on the test charge after removing one plate will be:
\[ F' = \frac{F}{2} \]

### Final Answer:
The force on the test charge after one plate is removed will be \( \frac{F}{2} \).

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