Charge of `5 mu C` each are placed at the corners of an equilateral triangle of side 10 cm. Then the force on each charge is :-

Options

A. 78 N
B. 39 N
C. 29 N
D. 22.5 N

Answer

To solve the problem of finding the force on each charge placed at the corners of an equilateral triangle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Geometry**:
- We have three charges \( Q_1, Q_2, Q_3 \) each equal to \( 5 \, \mu C \) placed at the corners of an equilateral triangle.
- The side length of the triangle is \( 10 \, cm = 0.1 \, m \).

2. **Calculate the Force Between Two Charges**:
- The force between any two charges can be calculated using Coulomb's Law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{|Q_1 Q_2|}{r^2} \]
- Here, \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, C^2/N \cdot m^2 \).
- Substituting the values:
\[ F = \frac{1}{4 \pi (8.85 \times 10^{-12})} \frac{(5 \times 10^{-6})^2}{(0.1)^2} \]

3. **Calculate the Numerical Value of the Force**:
- First, calculate \( (5 \times 10^{-6})^2 = 25 \times 10^{-12} \).
- The distance squared \( (0.1)^2 = 0.01 \).
- Now substituting these values into the formula:
\[ F = \frac{9 \times 10^9 \times 25 \times 10^{-12}}{0.01} = 9 \times 10^9 \times 2.5 \times 10^{-10} = 2.25 \times 10^{-1} \, N = 0.225 \, N \]

4. **Determine the Resultant Force on Each Charge**:
- Since the triangle is equilateral, the forces \( F_{12} \) (between \( Q_1 \) and \( Q_2 \)) and \( F_{13} \) (between \( Q_1 \) and \( Q_3 \)) will act at an angle of \( 60^\circ \) to each other.
- The resultant force \( F_{net} \) can be calculated using the formula:
\[ F_{net} = \sqrt{F_{12}^2 + F_{13}^2 + 2 F_{12} F_{13} \cos(60^\circ)} \]
- Since \( F_{12} = F_{13} = F \):
\[ F_{net} = \sqrt{F^2 + F^2 + 2F^2 \cdot \frac{1}{2}} = \sqrt{3F^2} = F \sqrt{3} \]
- Substituting the value of \( F \):
\[ F_{net} = 0.225 \sqrt{3} \approx 0.225 \times 1.732 \approx 0.389 \, N \]

5. **Final Result**:
- The force on each charge is approximately \( 0.389 \, N \).