Two point charges repel each other with a force of 100N. One of the charges is increased by 10% and the other is reduced by 10% . The new force of repulsion at the same distance would be

Options

A. 100N
B. 121N
C. 99N
D. none of these

Explanation

To solve the problem, we will use Coulomb's Law, which states that the force \( F \) between two point charges is given by:

\[ F = k \frac{Q_1 Q_2}{r^2} \]

where:
- \( F \) is the force between the charges,
- \( k \) is Coulomb's constant,
- \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

### Step-by-step Solution:

1. **Identify Initial Conditions:**
- The initial force \( F \) is given as 100 N.
- Let the initial charges be \( Q_1 \) and \( Q_2 \).

2. **Modify the Charges:**
- One charge \( Q_1 \) is increased by 10%, so the new charge \( Q_1' \) is:
\[ Q_1' = Q_1 + 0.1Q_1 = 1.1Q_1 \]
- The other charge \( Q_2 \) is reduced by 10%, so the new charge \( Q_2' \) is:
\[ Q_2' = Q_2 - 0.1Q_2 = 0.9Q_2 \]

3. **Calculate the New Force:**
- The new force \( F' \) can be expressed using the modified charges:
\[ F' = k \frac{Q_1' Q_2'}{r^2} = k \frac{(1.1Q_1)(0.9Q_2)}{r^2} \]
- This can be simplified to:
\[ F' = k \frac{1.1 \cdot 0.9 \cdot Q_1 Q_2}{r^2} \]
- Since \( F = k \frac{Q_1 Q_2}{r^2} = 100 \, \text{N} \), we can substitute:
\[ F' = 1.1 \cdot 0.9 \cdot F = 0.99 \cdot 100 \, \text{N} \]

4. **Final Calculation:**
- Now, calculate \( F' \):
\[ F' = 99 \, \text{N} \]

### Conclusion:
The new force of repulsion after the changes in the charges is **99 N**.

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