A charge Q exerts a 12 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q?

Options

A. 3 N
B. 6 N
C. 24 N
D. 36 N

Answer

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

\[ F = k \frac{|Q \cdot q|}{r^2} \]

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

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The initial force \( F \) is given as 12 N.
- The initial distance between the charges is \( r \).

2. **Apply Coulomb's Law:**
- According to Coulomb's Law, we have:
\[ F = k \frac{|Q \cdot q|}{r^2} = 12 \, \text{N} \]

3. **Change the Distance:**
- The problem states that the distance between the charges is doubled. Therefore, the new distance \( r' \) is:
\[ r' = 2r \]

4. **Calculate the New Force:**
- Substitute the new distance into Coulomb's Law:
\[ F' = k \frac{|Q \cdot q|}{(r')^2} = k \frac{|Q \cdot q|}{(2r)^2} = k \frac{|Q \cdot q|}{4r^2} \]
- Since \( F = k \frac{|Q \cdot q|}{r^2} \), we can express \( F' \) in terms of \( F \):
\[ F' = \frac{F}{4} \]

5. **Substitute the Known Force:**
- Now substitute \( F = 12 \, \text{N} \) into the equation:
\[ F' = \frac{12 \, \text{N}}{4} = 3 \, \text{N} \]

### Conclusion:
The magnitude of the force exerted on charge \( Q \) by charge \( q \) when the distance is doubled is \( 3 \, \text{N} \).