Two point charges of `+3muC and +4muC` repel each other with a force of 10 N. If each is given an additional charge of `-6muC`, the new force is

Options

A. `2 N`
B. `4 N`
C. `5 N`
D. `7.5 N`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the initial charges and the initial force
Given:
- Charge \( q_1 = +3 \, \mu C = 3 \times 10^{-6} \, C \)
- Charge \( q_2 = +4 \, \mu C = 4 \times 10^{-6} \, C \)
- Initial force \( F = 10 \, N \)

### Step 2: Calculate the new charges after adding the additional charge
Each charge is given an additional charge of \( -6 \, \mu C \):
- New charge \( q_1' = q_1 + (-6 \, \mu C) = 3 \, \mu C - 6 \, \mu C = -3 \, \mu C = -3 \times 10^{-6} \, C \)
- New charge \( q_2' = q_2 + (-6 \, \mu C) = 4 \, \mu C - 6 \, \mu C = -2 \, \mu C = -2 \times 10^{-6} \, C \)

### Step 3: Use Coulomb's Law to find the new force
Coulomb's Law states that the force between two point charges is given by:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
Where:
- \( k \) is Coulomb's constant
- \( r \) is the distance between the charges (which remains constant)

Since we are looking for the new force \( F' \) in terms of the initial force \( F \), we can use the ratio:
\[ \frac{F'}{F} = \frac{q_1' q_2'}{q_1 q_2} \]

### Step 4: Substitute the values into the equation
Substituting the values we have:
\[ F' = F \cdot \frac{q_1' q_2'}{q_1 q_2} \]
Where:
- \( F = 10 \, N \)
- \( q_1' = -3 \times 10^{-6} \, C \)
- \( q_2' = -2 \times 10^{-6} \, C \)
- \( q_1 = 3 \times 10^{-6} \, C \)
- \( q_2 = 4 \times 10^{-6} \, C \)

Calculating \( q_1' q_2' \) and \( q_1 q_2 \):
\[ q_1' q_2' = (-3 \times 10^{-6}) \times (-2 \times 10^{-6}) = 6 \times 10^{-12} \]
\[ q_1 q_2 = (3 \times 10^{-6}) \times (4 \times 10^{-6}) = 12 \times 10^{-12} \]

### Step 5: Calculate the new force
Now substituting back into the equation:
\[ F' = 10 \cdot \frac{6 \times 10^{-12}}{12 \times 10^{-12}} = 10 \cdot \frac{6}{12} = 10 \cdot \frac{1}{2} = 5 \, N \]

### Final Answer
The new force \( F' \) is \( 5 \, N \).

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