Electric charges of `1mu C, -1 mu C` and `2mu C` 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. `1.8 N`
C. `2.7 N`
D. `3.6 N`

Explanation

To find the resultant force on the charge at point C due to the charges at points A and B, we will follow these steps:

### Step 1: Identify the Charges and Their Positions
- Charge at A (Q_A) = +1 µC = +1 × 10^(-6) C
- Charge at B (Q_B) = -1 µC = -1 × 10^(-6) C
- Charge at C (Q_C) = +2 µC = +2 × 10^(-6) C
- The distance between each pair of charges (sides of the triangle) is 10 cm = 0.1 m.

### Step 2: Calculate the Force on Charge C due to Charge A (F_A)
Using Coulomb's Law:
\[ F_A = k \cdot \frac{|Q_A \cdot Q_C|}{r^2} \]
Where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( r = 0.1 \, \text{m} \)

Substituting the values:
\[ F_A = 9 \times 10^9 \cdot \frac{(1 \times 10^{-6}) \cdot (2 \times 10^{-6})}{(0.1)^2} \]
\[ F_A = 9 \times 10^9 \cdot \frac{2 \times 10^{-12}}{0.01} \]
\[ F_A = 9 \times 10^9 \cdot 2 \times 10^{-10} \]
\[ F_A = 1.8 \, \text{N} \]

### Step 3: Calculate the Force on Charge C due to Charge B (F_B)
Using Coulomb's Law again:
\[ F_B = k \cdot \frac{|Q_B \cdot Q_C|}{r^2} \]
Substituting the values:
\[ F_B = 9 \times 10^9 \cdot \frac{|-1 \times 10^{-6} \cdot 2 \times 10^{-6}|}{(0.1)^2} \]
\[ F_B = 9 \times 10^9 \cdot \frac{2 \times 10^{-12}}{0.01} \]
\[ F_B = 9 \times 10^9 \cdot 2 \times 10^{-10} \]
\[ F_B = 1.8 \, \text{N} \]

### Step 4: Determine the Directions of the Forces
- The force \( F_A \) (due to charge A) will be repulsive and directed away from charge A towards C.
- The force \( F_B \) (due to charge B) will be attractive and directed towards charge B from charge C.

### Step 5: Calculate the Resultant Force
Since both forces have the same magnitude (1.8 N) and are at an angle of 120° to each other (in an equilateral triangle), we can use the formula for the resultant of two forces:
\[ R = \sqrt{F_A^2 + F_B^2 + 2 F_A F_B \cos(120°)} \]

Substituting the values:
\[ R = \sqrt{(1.8)^2 + (1.8)^2 + 2 \cdot 1.8 \cdot 1.8 \cdot (-0.5)} \]
\[ R = \sqrt{3.24 + 3.24 - 3.24} \]
\[ R = \sqrt{3.24} \]
\[ R = 1.8 \, \text{N} \]

### Conclusion
The resultant force on the charge at point C is 1.8 N.

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