The force between two small charged spheres having charges of `1xx10^(-7)C and 2xx10^(-7)C` placed 20 cm apart in air is

Options

A. `4.5xx10^(-2)N`
B. `4.5xx10^(-3)N`
C. `5.4xx10^(-2)N`
D. `5.4xx10^(-3)N`

Explanation

To find the electrostatic force between two small charged spheres, we can use Coulomb's Law, which is given by the formula:

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

Where:
- \( F \) is the electrostatic force between the charges,
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the charges,
- \( r \) is the distance between the charges.

### Step-by-Step Solution:

1. **Identify the charges and distance:**
- \( q_1 = 1 \times 10^{-7} \, \text{C} \)
- \( q_2 = 2 \times 10^{-7} \, \text{C} \)
- Distance \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)

2. **Convert the distance to meters:**
- Since \( r \) is given in centimeters, we convert it to meters:
\[ r = 20 \, \text{cm} = 20 \times 10^{-2} \, \text{m} = 0.2 \, \text{m} \]

3. **Substitute the values into Coulomb's Law:**
\[ F = k \frac{|q_1 q_2|}{r^2} \]
\[ F = 9 \times 10^9 \frac{(1 \times 10^{-7})(2 \times 10^{-7})}{(0.2)^2} \]

4. **Calculate \( r^2 \):**
\[ r^2 = (0.2)^2 = 0.04 \, \text{m}^2 \]

5. **Calculate the product of the charges:**
\[ q_1 q_2 = (1 \times 10^{-7})(2 \times 10^{-7}) = 2 \times 10^{-14} \, \text{C}^2 \]

6. **Substitute \( r^2 \) and \( q_1 q_2 \) back into the equation:**
\[ F = 9 \times 10^9 \frac{2 \times 10^{-14}}{0.04} \]

7. **Calculate the fraction:**
\[ \frac{2 \times 10^{-14}}{0.04} = 5 \times 10^{-13} \]

8. **Now substitute this back into the equation:**
\[ F = 9 \times 10^9 \times 5 \times 10^{-13} \]

9. **Perform the multiplication:**
\[ F = 45 \times 10^{-4} \]

10. **Express in standard form:**
\[ F = 4.5 \times 10^{-3} \, \text{N} \]

### Final Answer:
The force between the two small charged spheres is \( 4.5 \times 10^{-3} \, \text{N} \).