Two point charges `q_(1)=2xx10^(-3) C` and `q_(2)=-3xx10^(-6) C` are separated by a distance `x = 10` cm. Find the magnitude and nature of the force between the two charges.

Answer

To solve the problem of finding the magnitude and nature of the force between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( x \), we will use Coulomb's Law.

### Step-by-Step Solution:

1. **Identify the Charges and Distance**:
- Given:
- \( q_1 = 2 \times 10^{-3} \, \text{C} \)
- \( q_2 = -3 \times 10^{-6} \, \text{C} \)
- Distance \( x = 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Coulomb's Law**:
- The formula for 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, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( r \) is the distance between the charges.

3. **Substituting Values**:
- Substitute the values of \( q_1 \), \( q_2 \), and \( r \) into the formula:
\[ F = 9 \times 10^9 \frac{|(2 \times 10^{-3})(-3 \times 10^{-6})|}{(0.1)^2} \]

4. **Calculating the Product of Charges**:
- Calculate \( |q_1 q_2| \):
\[ |q_1 q_2| = |(2 \times 10^{-3})(-3 \times 10^{-6})| = 6 \times 10^{-9} \, \text{C}^2 \]

5. **Calculating \( r^2 \)**:
- Calculate \( r^2 \):
\[ r^2 = (0.1)^2 = 0.01 \, \text{m}^2 \]

6. **Substituting Back into the Force Equation**:
- Now substitute \( |q_1 q_2| \) and \( r^2 \) back into the force equation:
\[ F = 9 \times 10^9 \frac{6 \times 10^{-9}}{0.01} \]

7. **Calculating the Force**:
- Simplifying the equation:
\[ F = 9 \times 10^9 \times 6 \times 10^{-9} \times 100 = 54 \, \text{N} \]

8. **Determining the Nature of the Force**:
- Since \( q_1 \) is positive and \( q_2 \) is negative, the force between them is attractive.

### Final Answer:
- The magnitude of the force between the two charges is \( 54 \, \text{N} \) and the nature of the force is attractive.