ABC is a right triangle in which AB = 3 cm, BC = 4 cm and right angle is at B. Three charges `+ 15mu C + 12 mu C` and `-20 mu C` are placed respectively at A, B and C. The force acting on the charge at B is

Options

A. `1250 N`
B. `3500 N`
C. `1200 N`
D. `2250N`

Answer

To find the force acting on the charge at point B in the right triangle ABC, we will use Coulomb's law. The steps are as follows:

### Step 1: Identify the charges and their positions
- Charge at A (QA) = +15 µC
- Charge at B (QB) = +12 µC
- Charge at C (QC) = -20 µC
- Distance AB = 3 cm = 0.03 m
- Distance BC = 4 cm = 0.04 m

### Step 2: Calculate the force acting on charge B due to charge A (FAB)
Using Coulomb's law:
\[ F_{AB} = k \cdot \frac{|Q_A \cdot Q_B|}{r_{AB}^2} \]
Where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( r_{AB} = 0.03 \, \text{m} \)

Substituting the values:
\[ F_{AB} = 9 \times 10^9 \cdot \frac{(15 \times 10^{-6}) \cdot (12 \times 10^{-6})}{(0.03)^2} \]
\[ F_{AB} = 9 \times 10^9 \cdot \frac{180 \times 10^{-12}}{0.0009} \]
\[ F_{AB} = 9 \times 10^9 \cdot 200 \times 10^{-6} \]
\[ F_{AB} = 1800 \, \text{N} \]

### Step 3: Calculate the force acting on charge B due to charge C (FBC)
Using Coulomb's law:
\[ F_{BC} = k \cdot \frac{|Q_B \cdot Q_C|}{r_{BC}^2} \]
Where:
- \( r_{BC} = 0.04 \, \text{m} \)

Substituting the values:
\[ F_{BC} = 9 \times 10^9 \cdot \frac{(12 \times 10^{-6}) \cdot (20 \times 10^{-6})}{(0.04)^2} \]
\[ F_{BC} = 9 \times 10^9 \cdot \frac{240 \times 10^{-12}}{0.0016} \]
\[ F_{BC} = 9 \times 10^9 \cdot 150 \times 10^{-6} \]
\[ F_{BC} = 1350 \, \text{N} \]

### Step 4: Determine the direction of the forces
- \( F_{AB} \) is a repulsive force (since both charges at A and B are positive), directed away from A along AB.
- \( F_{BC} \) is an attractive force (since charge B is positive and charge C is negative), directed towards C along BC.

### Step 5: Calculate the resultant force on charge B
Since the forces are perpendicular to each other, we can use the Pythagorean theorem to find the resultant force (FR):
\[ F_R = \sqrt{F_{AB}^2 + F_{BC}^2} \]
\[ F_R = \sqrt{(1800)^2 + (1350)^2} \]
\[ F_R = \sqrt{3240000 + 1822500} \]
\[ F_R = \sqrt{5062500} \]
\[ F_R = 2250 \, \text{N} \]

### Final Answer
The force acting on the charge at B is **2250 N**.

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