A and B similarly charged bodies which repel with force F. Another uncharged sphere C of same size and material is touched with B and removed. It is kept at mid-point of distance between A and B. Force on C is

Options

A. `(F)/(2)`
B. 2F
C. F
D. `(2)/(3)F`

Explanation

To solve the problem, we need to analyze the forces acting on the uncharged sphere C after it has been touched with charged body B and placed at the midpoint between A and B.

### Step-by-step Solution:

1. **Identify the Charges**:
- Let the charge on bodies A and B be \( Q \).
- When sphere C is touched with B, it acquires a charge of \( \frac{Q}{2} \) because it is made of the same material and is of the same size as B.

2. **Positioning the Bodies**:
- Let the distance between A and B be \( R \).
- The midpoint between A and B is at a distance of \( \frac{R}{2} \) from both A and B.

3. **Calculate the Force on C due to A**:
- The force \( F_{AC} \) on sphere C due to sphere A can be calculated using Coulomb's law:
\[ F_{AC} = k \frac{Q \cdot \frac{Q}{2}}{\left(\frac{R}{2}\right)^2} \]
- Simplifying this:
\[ F_{AC} = k \frac{Q^2/2}{R^2/4} = k \frac{2Q^2}{R^2} = 2F \]
- Here, \( F \) is the force between A and B when they are at distance \( R \).

4. **Calculate the Force on C due to B**:
- The force \( F_{BC} \) on sphere C due to sphere B is:
\[ F_{BC} = k \frac{\frac{Q}{2} \cdot \frac{Q}{2}}{\left(\frac{R}{2}\right)^2} \]
- Simplifying this:
\[ F_{BC} = k \frac{Q^2/4}{R^2/4} = k \frac{Q^2}{R^2} = F \]

5. **Determine the Resultant Force on C**:
- Since both forces \( F_{AC} \) and \( F_{BC} \) act in the same direction (repulsive), the resultant force \( F_R \) on sphere C is:
\[ F_R = F_{AC} - F_{BC} = 2F - F = F \]

### Final Answer:
The force on sphere C is \( F \).