Electric charge of `1 muC, -1 muC` and `2 muC` are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is

Options

A. 0.9 N
B. 1. 8N
C. 2.7 N
D. 3.6 N

Answer

To find the resultant force on the charge at point C in the given configuration, we will follow these steps:

### Step 1: Identify the Charges and Their Positions
We have three charges:
- Charge at A: \( Q_A = 1 \, \mu C = 1 \times 10^{-6} \, C \)
- Charge at B: \( Q_B = -1 \, \mu C = -1 \times 10^{-6} \, C \)
- Charge at C: \( Q_C = 2 \, \mu C = 2 \times 10^{-6} \, C \)

These charges are placed at the corners of an equilateral triangle ABC with each side measuring \( 10 \, cm = 0.1 \, m \).

### Step 2: Calculate the Forces on Charge C
The forces acting on charge C due to charges A and B need to be calculated.

#### Force due to Charge A on Charge C (\( F_{AC} \))
The force between two charges is given by Coulomb's Law:
\[ F = k \frac{|Q_1 Q_2|}{r^2} \]
Where:
- \( k \) (Coulomb's constant) \( = 8.99 \times 10^9 \, N \, m^2/C^2 \)
- \( Q_1 = Q_A = 1 \times 10^{-6} \, C \)
- \( Q_2 = Q_C = 2 \times 10^{-6} \, C \)
- \( r = 0.1 \, m \)

Calculating \( F_{AC} \):
\[ F_{AC} = k \frac{|Q_A Q_C|}{r^2} = 8.99 \times 10^9 \frac{(1 \times 10^{-6})(2 \times 10^{-6})}{(0.1)^2} \]
\[ = 8.99 \times 10^9 \frac{2 \times 10^{-12}}{0.01} = 8.99 \times 10^9 \times 2 \times 10^{-10} = 1.798 \, N \]

The direction of \( F_{AC} \) is away from charge A (repulsive force).

#### Force due to Charge B on Charge C (\( F_{BC} \))
Calculating \( F_{BC} \):
\[ F_{BC} = k \frac{|Q_B Q_C|}{r^2} = 8.99 \times 10^9 \frac{|-1 \times 10^{-6} \cdot 2 \times 10^{-6}|}{(0.1)^2} \]
\[ = 8.99 \times 10^9 \frac{2 \times 10^{-12}}{0.01} = 8.99 \times 10^9 \times 2 \times 10^{-10} = 1.798 \, N \]

The direction of \( F_{BC} \) is towards charge B (attractive force).

### Step 3: Determine the Angle Between the Forces
Since the triangle is equilateral, the angle between the forces \( F_{AC} \) and \( F_{BC} \) is \( 120^\circ \).

### Step 4: Calculate the Resultant Force
Using the formula for the resultant of two forces:
\[ F_R = \sqrt{F_{AC}^2 + F_{BC}^2 + 2 F_{AC} F_{BC} \cos(120^\circ)} \]
Substituting the values:
\[ F_R = \sqrt{(1.798)^2 + (1.798)^2 + 2 \cdot (1.798) \cdot (1.798) \cdot (-0.5)} \]
\[ = \sqrt{2 \cdot (1.798)^2 - (1.798)^2} = \sqrt{(1.798)^2} \]
\[ = 1.798 \, N \]

### Final Result
The resultant force on the charge at C is approximately:
\[ F_R \approx 1.798 \, N \]