Two equally charged small metal balls placed at a fixed distance experience a force F. A similar unchanged metal ball after touching one of them is placed at the middle point between the two balls. The force experienced by this ball is

Options

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

Answer

To solve the problem step by step, we will analyze the situation involving three metal balls and the forces acting on them.

### Step 1: Understand the Initial Configuration
We have two equally charged small metal balls, each with a charge \( Q \), placed at a fixed distance \( R \) apart. The force experienced between these two balls is given as \( F \).

**Hint:** Recall Coulomb's law, which states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

### Step 2: Determine the Charge of the Third Ball
A third uncharged metal ball touches one of the charged balls. When it touches the ball with charge \( Q \), it will acquire half of the charge. Therefore, the charge on the third ball becomes:
\[ Q_{3} = \frac{Q}{2} \]

**Hint:** When two conductors touch, charge is shared equally if they are identical.

### Step 3: Position the Third Ball
The third ball is then placed at the midpoint between the two original balls. The distance from each of the two charged balls to the third ball is:
\[ d = \frac{R}{2} \]

**Hint:** Visualize the arrangement to understand the distances involved.

### Step 4: Calculate the Forces Acting on the Third Ball
Now, we need to calculate the forces acting on the third ball due to the other two balls.

1. **Force due to the first ball (charge \( Q \)):**
Using Coulomb's law, the force \( F_1 \) on the third ball due to the first ball is given by:
\[ F_1 = k \cdot \frac{Q \cdot \frac{Q}{2}}{\left(\frac{R}{2}\right)^2} = k \cdot \frac{Q^2/2}{R^2/4} = \frac{2kQ^2}{R^2} \]

2. **Force due to the second ball (charge \( Q \)):**
The force \( F_2 \) on the third ball due to the second ball (which also has charge \( Q \)) is calculated similarly:
\[ F_2 = k \cdot \frac{Q \cdot \frac{Q}{2}}{\left(\frac{R}{2}\right)^2} = \frac{2kQ^2}{R^2} \]

**Hint:** Remember that both forces \( F_1 \) and \( F_2 \) are repulsive since all charges are positive.

### Step 5: Determine the Net Force on the Third Ball
Since both forces \( F_1 \) and \( F_2 \) act in opposite directions (away from each other), the net force \( F_{net} \) on the third ball is:
\[ F_{net} = F_1 - F_2 = \frac{2kQ^2}{R^2} - \frac{2kQ^2}{R^2} = 0 \]

**Hint:** When forces are equal and opposite, they cancel each other out.

### Conclusion
The force experienced by the third ball placed at the midpoint between the two charged balls is:
\[ \boxed{0} \]