Two balls carrying charges `+ 7 muC` and `-5 mu C` attract each other with a force F. If a charge `-2 mu C` is added to both, the force between them will be :-

Options

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

Answer

To solve the problem, we will use Coulomb's Law, which states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

### Step-by-Step Solution:

1. **Identify the Initial Charges:**
- Let the first charge \( Q_1 = +7 \, \mu C = +7 \times 10^{-6} \, C \)
- Let the second charge \( Q_2 = -5 \, \mu C = -5 \times 10^{-6} \, C \)

2. **Calculate the Initial Force (F):**
- According to Coulomb's Law, the force \( F \) between the two charges is given by:
\[ F = k \frac{|Q_1 Q_2|}{r^2} \]
- Here, \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, N \, m^2/C^2 \)), and \( r \) is the distance between the charges.

3. **Adding Charges to Both Balls:**
- Now, we add a charge of \( -2 \, \mu C \) to both charges:
- New charge for the first ball:
\[ Q_1' = Q_1 + (-2 \, \mu C) = 7 \, \mu C - 2 \, \mu C = 5 \, \mu C \]
- New charge for the second ball:
\[ Q_2' = Q_2 + (-2 \, \mu C) = -5 \, \mu C - 2 \, \mu C = -7 \, \mu C \]

4. **Calculate the New Force (F'):**
- Now we calculate the new force \( F' \) between the modified charges:
\[ F' = k \frac{|Q_1' Q_2'|}{r^2} \]
- Substitute the new charges:
\[ F' = k \frac{|5 \, \mu C \cdot (-7 \, \mu C)|}{r^2} \]
- This simplifies to:
\[ F' = k \frac{35 \, \mu C^2}{r^2} \]

5. **Compare the New Force with the Initial Force:**
- The initial force \( F \) was:
\[ F = k \frac{|7 \, \mu C \cdot (-5 \, \mu C)|}{r^2} = k \frac{35 \, \mu C^2}{r^2} \]
- Thus, we find that:
\[ F' = F \]

### Conclusion:
The force between the two balls after adding the charges will remain the same, i.e., \( F' = F \).