Two point charges `+3 muC` and `+8muC` repel each other with a force of `40 N`. If a charge of `-5 muC` is added to each of them, then the force between them will become

Options

A. `-10 N`
B. `+10 N`
C. `+20 N`
D. `- 20 N`

Explanation

To solve the problem step by step, we will use Coulomb's law, which states that the force 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 1: Identify the initial charges and force
We have two point charges:
- Charge \( q_1 = +3 \, \mu C = 3 \times 10^{-6} \, C \)
- Charge \( q_2 = +8 \, \mu C = 8 \times 10^{-6} \, C \)

The initial force \( F \) between them is given as \( 40 \, N \).

### Step 2: Write the equation for the initial force
Using Coulomb's law, we can express the initial force as:

\[ 40 = k \frac{(3 \times 10^{-6})(8 \times 10^{-6})}{r^2} \]

### Step 3: Calculate the product of the charges
Calculate the product of the charges:

\[ q_1 q_2 = (3 \times 10^{-6})(8 \times 10^{-6}) = 24 \times 10^{-12} \]

### Step 4: Rearrange the equation to find \( r^2 \)
Rearranging the equation gives:

\[ r^2 = k \frac{(3 \times 10^{-6})(8 \times 10^{-6})}{40} \]

### Step 5: Introduce the new charges
Now, we add a charge of \( -5 \, \mu C \) to each charge:

- New charge \( q_1' = 3 \, \mu C - 5 \, \mu C = -2 \, \mu C = -2 \times 10^{-6} \, C \)
- New charge \( q_2' = 8 \, \mu C - 5 \, \mu C = 3 \, \mu C = 3 \times 10^{-6} \, C \)

### Step 6: Write the equation for the new force
Using Coulomb's law again for the new charges, we have:

\[ F' = k \frac{|q_1' q_2'|}{r^2} \]

Substituting the new charges:

\[ F' = k \frac{|-2 \times 10^{-6} \cdot 3 \times 10^{-6}|}{r^2} \]

### Step 7: Calculate the product of the new charges
Calculate the product of the new charges:

\[ |q_1' q_2'| = |-2 \times 10^{-6} \cdot 3 \times 10^{-6}| = 6 \times 10^{-12} \]

### Step 8: Relate the new force to the initial force
Now we can relate the new force to the initial force using the ratio:

\[ \frac{F'}{40} = \frac{6 \times 10^{-12}}{24 \times 10^{-12}} \]

### Step 9: Simplify the ratio
This simplifies to:

\[ \frac{F'}{40} = \frac{1}{4} \]

### Step 10: Calculate the new force
Thus, we find:

\[ F' = 40 \times \frac{1}{4} = 10 \, N \]

Since the charges are of opposite signs, the force will be attractive, and we can denote it as \( -10 \, N \) to indicate the direction.

### Final Answer:
The new force between the charges after adding \( -5 \, \mu C \) to each is \( -10 \, N \).

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