Two equally charged identical small balls kept some fixed disttance apart exert a repulsive force F on each other. A similar uncharged ball, after touching one of them is placed at the mid-point of line joining the two balls. Force experienced by the third ball is

Options

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

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understanding the Initial Setup
We have two identical small balls, each with a charge \( Q \), placed a fixed distance \( D \) apart. They exert a repulsive force \( F \) on each other.

### Step 2: Introducing the Third Ball
A third ball, which is initially uncharged, touches one of the charged balls. When it touches the charged ball, it acquires a charge due to the redistribution of charge.

### Step 3: Charge Distribution
Since the two balls are identical and one is charged with \( Q \) while the other is neutral, when the uncharged ball touches the charged ball, the charge will redistribute equally. Thus, the uncharged ball will acquire a charge of \( \frac{Q}{2} \) from the charged ball.

- After touching, the charges will be:
- The first ball (initially charged) will have a charge of \( Q - \frac{Q}{2} = \frac{Q}{2} \).
- The second ball (initially uncharged) will now have a charge of \( \frac{Q}{2} \).

### Step 4: Positioning the Third Ball
The third ball, now with a charge of \( \frac{Q}{2} \), is placed at the midpoint between the two charged balls.

### Step 5: Calculating Forces on the Third Ball
Now we need to calculate the net force acting on the third ball due to the other two balls.

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

2. **Force due to the second ball (also with charge \( \frac{Q}{2} \))**:
- The distance from the second ball to the third ball is also \( \frac{D}{2} \).
- The force \( F_2 \) on the third ball due to the second ball is:
\[ F_2 = k \frac{\frac{Q}{2} \cdot \frac{Q}{2}}{\left(\frac{D}{2}\right)^2} = k \frac{\frac{Q^2}{4}}{\frac{D^2}{4}} = \frac{kQ^2}{D^2} \]

### Step 6: Net Force on the Third Ball
Since both forces \( F_1 \) and \( F_2 \) are repulsive and act in the same direction, the net force \( F_{net} \) on the third ball is:
\[ F_{net} = F_1 - F_2 = \frac{2kQ^2}{D^2} - \frac{kQ^2}{D^2} = \frac{kQ^2}{D^2} \]

### Step 7: Relating to the Original Force \( F \)
We know that the original force \( F \) between the two charged balls was given by:
\[ F = k \frac{Q^2}{D^2} \]
Thus, we find that the force experienced by the third ball is:
\[ F_{net} = F \]

### Conclusion
The force experienced by the third ball is equal to the original force \( F \) exerted by the two charged balls on each other.

### Final Answer
The force experienced by the third ball is \( F \).

---