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 step by step, we will apply Coulomb's law and analyze the changes in the charges.

### Step 1: Identify the initial charges and the force between them.
The initial charges are:
- \( q_1 = +3 \, \mu C = 3 \times 10^{-6} \, C \)
- \( q_2 = +4 \, \mu C = 4 \times 10^{-6} \, C \)

The initial force between them is given as:
- \( F = 10 \, N \)

### Step 2: Write down Coulomb's Law.
Coulomb's law states that the force \( F \) between two point charges is given by:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, N \, m^2/C^2 \)) and \( r \) is the distance between the charges.

### Step 3: Calculate the product of the initial charges.
The product of the initial charges is:
\[ q_1 q_2 = (3 \times 10^{-6})(4 \times 10^{-6}) = 12 \times 10^{-12} \, C^2 \]

### Step 4: Determine the distance between the charges using the initial force.
From Coulomb's law, we can express the distance \( r \) in terms of the known quantities:
\[ 10 = k \frac{12 \times 10^{-12}}{r^2} \]
Rearranging gives:
\[ r^2 = k \frac{12 \times 10^{-12}}{10} \]

### Step 5: Calculate the new charges after adding \(-6 \, \mu C\).
Now we add \(-6 \, \mu C\) to each charge:
- New \( q_1 = 3 \, \mu C - 6 \, \mu C = -3 \, \mu C = -3 \times 10^{-6} \, C \)
- New \( q_2 = 4 \, \mu C - 6 \, \mu C = -2 \, \mu C = -2 \times 10^{-6} \, C \)

### Step 6: Calculate the product of the new charges.
The product of the new charges is:
\[ q_1 q_2 = (-3 \times 10^{-6})(-2 \times 10^{-6}) = 6 \times 10^{-12} \, C^2 \]

### Step 7: Calculate the new force using Coulomb's Law.
Using the distance \( r \) from Step 4, the new force \( F' \) can be calculated:
\[ F' = k \frac{|q_1 q_2|}{r^2} = k \frac{6 \times 10^{-12}}{r^2} \]

### Step 8: Relate the new force to the initial force.
From the previous calculations:
\[ \frac{F'}{F} = \frac{6 \times 10^{-12}}{12 \times 10^{-12}} = \frac{1}{2} \]
Thus:
\[ F' = \frac{F}{2} = \frac{10 \, N}{2} = 5 \, N \]

### Final Answer:
The new force between the charges after adding \(-6 \, \mu C\) to each is \( F' = 5 \, N \).

---