The force between two similar charges of magnitude 2C each separated by a distance 2km

Options

A. 9N
B. `9 xx 10^(3)N`
C. 300N
D. 50N

Answer

To find the force between two similar charges of magnitude 2C each, separated by a distance of 2 km, we can use Coulomb's Law. The formula for the electrostatic force \( F \) between two point charges is given by:

\[ F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r^2} \]

Where:
- \( F \) is the force between the charges,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \),
- \( \frac{1}{4 \pi \epsilon_0} \) is approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step-by-Step Solution:

1. **Identify the values**:
- Both charges \( q_1 \) and \( q_2 \) are \( 2 \, \text{C} \).
- The distance \( r \) is \( 2 \, \text{km} = 2000 \, \text{m} \).

2. **Convert the distance to meters**:
- \( r = 2 \, \text{km} = 2 \times 10^3 \, \text{m} \).

3. **Substitute the values into Coulomb's Law**:
\[ F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r^2} \]
\[ F = 9 \times 10^9 \cdot \frac{(2 \, \text{C}) \cdot (2 \, \text{C})}{(2000 \, \text{m})^2} \]

4. **Calculate \( r^2 \)**:
\[ r^2 = (2000 \, \text{m})^2 = 4 \times 10^6 \, \text{m}^2 \]

5. **Substitute \( r^2 \) back into the equation**:
\[ F = 9 \times 10^9 \cdot \frac{4}{4 \times 10^6} \]

6. **Simplify the equation**:
\[ F = 9 \times 10^9 \cdot 1 = 9 \times 10^3 \, \text{N} \]

7. **Conclusion**:
- The force between the two charges is \( 9 \times 10^3 \, \text{N} \).

### Final Answer:
The force between the two similar charges is \( 9000 \, \text{N} \) (or \( 9 \times 10^3 \, \text{N} \)).