ABC is a right angled triangle in which AB = 3 cm and BC = 4cm. And `/_ ABC =pi//2`. The three charges +15, + 12 and -20 e.s.u. are placed respectively on A, B and C . The force acting on B is

Options

A. 125 dynes
B. 35 dynes
C. 25 dynes
D. Zero

Answer

To solve the problem, we need to calculate the net force acting on charge B due to the other two charges A and C in the right-angled triangle ABC.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- Charge at A (Q_A) = +15 e.s.u.
- Charge at B (Q_B) = +12 e.s.u.
- Charge at C (Q_C) = -20 e.s.u.
- The distances are: AB = 3 cm, BC = 4 cm.

2. **Calculate the Force on B due to A (F_AB):**
- The distance between A and B (r_AB) = 3 cm.
- Using Coulomb's Law:
\[ F_{AB} = \frac{Q_A \cdot Q_B}{r_{AB}^2} \]
- Substituting the values:
\[ F_{AB} = \frac{15 \cdot 12}{3^2} = \frac{180}{9} = 20 \text{ dyn} \]
- The direction of this force is away from A towards B (repulsive force).

3. **Calculate the Force on B due to C (F_BC):**
- The distance between B and C (r_BC) = 4 cm.
- Using Coulomb's Law:
\[ F_{BC} = \frac{|Q_B \cdot Q_C|}{r_{BC}^2} \]
- Substituting the values:
\[ F_{BC} = \frac{12 \cdot 20}{4^2} = \frac{240}{16} = 15 \text{ dyn} \]
- The direction of this force is towards C (attractive force).

4. **Determine the Resultant Force on B:**
- The forces F_AB and F_BC are perpendicular to each other (since triangle ABC is a right triangle).
- Using the Pythagorean theorem to find the resultant force (F_net):
\[ F_{net} = \sqrt{F_{AB}^2 + F_{BC}^2} \]
- Substituting the values:
\[ F_{net} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \text{ dyn} \]

5. **Final Result:**
- The net force acting on charge B is **25 dyn**.