Two equal and like charges when placed 5 cm apart experience a repuslive force of 0.144 newton. The magnitude of the charge in micro-coulomb will be

Options

A. `0.2`
B. 2
C. 20
D. 12

Answer

To solve the problem, we can use Coulomb's law, which states that the electrostatic force (F) between two point charges (Q1 and Q2) separated by a distance (r) is given by the formula:

\[ F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1 \cdot Q_2}{r^2} \]

Since the charges are equal (let's denote them as Q), the formula simplifies to:

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

### Step-by-step Solution:

1. **Identify the given values:**
- The repulsive force \( F = 0.144 \, \text{N} \)
- The distance \( r = 5 \, \text{cm} = 0.05 \, \text{m} \) (convert cm to m)
- The value of \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

2. **Substitute the values into the formula:**
\[ 0.144 = \frac{9 \times 10^9 \cdot Q^2}{(0.05)^2} \]

3. **Calculate \( (0.05)^2 \):**
\[ (0.05)^2 = 0.0025 \, \text{m}^2 \]

4. **Rearrange the equation to solve for \( Q^2 \):**
\[ Q^2 = \frac{0.144 \cdot 0.0025}{9 \times 10^9} \]

5. **Calculate the numerator:**
\[ 0.144 \cdot 0.0025 = 0.00036 \]

6. **Now substitute this back into the equation:**
\[ Q^2 = \frac{0.00036}{9 \times 10^9} \]

7. **Calculate \( Q^2 \):**
\[ Q^2 = 4 \times 10^{-14} \, \text{C}^2 \]

8. **Take the square root to find \( Q \):**
\[ Q = \sqrt{4 \times 10^{-14}} = 2 \times 10^{-7} \, \text{C} \]

9. **Convert \( Q \) to micro-coulombs:**
\[ Q = 2 \times 10^{-7} \, \text{C} = 0.2 \, \mu\text{C} \]

### Final Answer:
The magnitude of the charge is \( 0.2 \, \mu\text{C} \).