Force acting upon charged particle kept between the plates of a charged condenser is `F`. If one of the plates of the condenser is removed, force acting on the same particle will become.

Options

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

Explanation

To solve the problem, we need to understand the concept of electric fields and forces acting on charged particles in a capacitor.

### Step-by-step Solution:

1. **Understanding the Initial Setup**:
- We have a parallel plate capacitor with two plates: one positively charged and one negatively charged.
- The electric field (E) between the plates is uniform and directed from the positive plate to the negative plate.

2. **Calculating the Electric Field**:
- The electric field (E) between two plates of a capacitor is given by:
\[ E = \frac{V}{d} \]
where \( V \) is the voltage across the plates and \( d \) is the separation between them.
- In the case of a parallel plate capacitor, the net electric field between the plates is:
\[ E_{\text{net}} = E + E = 2E \]

3. **Force on the Charged Particle**:
- If a charge \( q \) is placed between the plates, the force \( F \) acting on the charge due to the electric field is given by:
\[ F = q \cdot E_{\text{net}} = q \cdot 2E \]
- Thus, the initial force \( F \) is:
\[ F = 2qE \]

4. **Removing One Plate**:
- When one of the plates is removed, only the electric field from the remaining plate is present.
- The electric field due to a single charged plate is given by:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.
- Therefore, the electric field \( E' \) acting on the charge \( q \) now becomes:
\[ E' = E \]

5. **Calculating the New Force**:
- The new force \( F' \) acting on the charge \( q \) when one plate is removed is:
\[ F' = q \cdot E' = q \cdot E \]

6. **Finding the Relationship Between Forces**:
- Now, we can relate the new force \( F' \) to the original force \( F \):
\[ F' = \frac{F}{2} \]
- This shows that the new force is half of the original force.

### Conclusion:
The force acting on the charged particle after one plate of the capacitor is removed becomes:
\[ F' = \frac{F}{2} \]