Three identical charges of magnitude `2 mu C `are placed at the corners of right angled triangle ABC whose base BC and height BA respectively 4 cm and 3 cm. Forces on charge at right angled corner B due to charges at 'A' and 'C' are respectively F1 and F2. The angle between their resultant force and F2 is

Options

A. `sin^(-1) ""((3)/(4))`
B. `tan^(-1)""((16)/(9))`
C. `cos^(-1) ((1)/(3))`
D. `45^@`

Answer

To solve the problem, we need to find the angle between the resultant force acting on charge B and the force due to charge C (F2). Here are the steps to arrive at the solution:

### Step 1: Understand the Configuration
We have a right-angled triangle ABC with:
- Charge at A (2 µC)
- Charge at B (2 µC)
- Charge at C (2 µC)

The sides of the triangle are:
- AB = 3 cm (height)
- BC = 4 cm (base)

### Step 2: Calculate the Distances
The distances between the charges are:
- Distance AB = 3 cm
- Distance BC = 4 cm
- Distance AC = √(AB² + BC²) = √(3² + 4²) = √(9 + 16) = √25 = 5 cm

### Step 3: Calculate the Forces F1 and F2
Using Coulomb's Law, the force between two point charges is given by:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

Where:
- \( q_1 = q_2 = 2 \times 10^{-6} \) C (charge at A and C)
- \( r \) is the distance between the charges.

**Calculate F1 (Force on B due to A):**
\[ F_1 = \frac{1}{4 \pi \epsilon_0} \frac{(2 \times 10^{-6})^2}{(3 \times 10^{-2})^2} \]
\[ F_1 = \frac{1}{4 \pi \epsilon_0} \frac{4 \times 10^{-12}}{9 \times 10^{-4}} \]
\[ F_1 = \frac{4}{36 \pi \epsilon_0} \times 10^{-8} \]
\[ F_1 = \frac{1}{9 \pi \epsilon_0} \times 10^{-8} \]

**Calculate F2 (Force on B due to C):**
\[ F_2 = \frac{1}{4 \pi \epsilon_0} \frac{(2 \times 10^{-6})^2}{(4 \times 10^{-2})^2} \]
\[ F_2 = \frac{1}{4 \pi \epsilon_0} \frac{4 \times 10^{-12}}{16 \times 10^{-4}} \]
\[ F_2 = \frac{4}{64 \pi \epsilon_0} \times 10^{-8} \]
\[ F_2 = \frac{1}{16 \pi \epsilon_0} \times 10^{-8} \]

### Step 4: Find the Ratio of Forces
To find the angle between the resultant force and F2, we need the ratio of F1 to F2:
\[ \frac{F_1}{F_2} = \frac{\frac{1}{9 \pi \epsilon_0} \times 10^{-8}}{\frac{1}{16 \pi \epsilon_0} \times 10^{-8}} \]
\[ \frac{F_1}{F_2} = \frac{16}{9} \]

### Step 5: Calculate the Angle
Using the tangent function:
\[ \tan \theta = \frac{F_1}{F_2} = \frac{16}{9} \]
Thus, the angle \( \theta \) can be calculated as:
\[ \theta = \tan^{-1} \left( \frac{16}{9} \right) \]

### Conclusion
The angle between the resultant force and F2 is:
\[ \theta = \tan^{-1} \left( \frac{16}{9} \right) \]