Let A B C is a right angled triangle in which A B=3 cm and B C=4 cm and angle A B C=90°. The three charges +15,+12 and -20 esu are placed on A, B and C respectively. The force acting on B will be

Question

Let A B C is a right angled triangle in which A B=3 cm and B C=4 cm and angle A B C=90°. The three charges +15,+12 and -20 esu are placed on A, B and C respectively. The force acting on B will be (A) Zero (B) 25 dyne (C) 30 dyne (D) 150 dyne

Tardigrade Admin · Accepted Answer

The given condition can be shown as <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3f65b46a17b34d0245975a89e331ed18-.png" /> Net force on B, F text net =√FA2+FC2 ∴ FA=(5 × 12/(3)2)=20 dyne and FC=(12 × 20/(4)2)=15 dyne So, F text net =√(20)2+(15)2=25 dyne.

Q. Let $A BC$ is a right angled triangle in which $A B = 3 c m$ and $BC = 4 c m$ and $∠ A BC = 9 0^{\circ}$ . The three charges $+ 15, + 12$ and $- 20$ esu are placed on $A, B$ and $C$ respectively. The force acting on $B$ will be

Zero

25 dyne

30 dyne

150 dyne

The given condition can be shown as

Net force on $B, F_{net} = F_{A}^{2} + F_{C}^{2}$
$∴ F_{A} = \frac{5 \times 12}{( 3 ) ^{2}} = 20$ dyne
and $F_{C} = \frac{12 \times 20}{( 4 ) ^{2}} = 15$ dyne
So, $F_{net} = (20)^{2} + (15)^{2} = 25$ dyne.

Q. Let ABC is a right angled triangle in which AB=3cm and BC=4cm and ∠ABC=90∘. The three charges +15,+12 and −20 esu are placed on A,B and C respectively. The force acting on B will be