Question

the
ABC is a right angled triangle AB = 3 cm, BC = 4 cm.
charges +15, +12, -12 esu are placed at A, B and C
respectively. The magnitude of the force experienced
by the charge at B in dyne is:
(A) 125
(B) 35
(C) 22
(D) O​

Answer 1

Answer:

the correct answer is option (C) 22 dyne

Explanation:

Given:

Charge at A, $q_A$ = 15 esu

Charge at B $q_B$= 12 esu

Charge at C, $q_C$ = -12 esu

Distance between A and B, r₁ = AB = 3 cm

Distance between B and C, r₂ = BC = 4 cm

Now,

the Force on B due to charge at A = $\frac{kq_Aq_B}{r_1^2}$

here,

k is the coulomb's constant = 1 for CGS system

therefore,

F₁ =  $\frac{1\times15\times12}{3^2}$

or

F₁ = 20 dyne

similarly

the Force on B due to charge at C = $\frac{kq_Cq_B}{r_2^2}$

here,

k is the coulomb's constant = 1 for CGS system

therefore,

F₂ =  $\frac{1\times(-12)\times12}{4^2}$

or

F₂ = -9 dyne

Now,

the resultant force on B due to both the charges is given as:

F = $\sqrt{F_1^2+F_2^2}$

or

F = $\sqrt{20^2+(-9)^2}$

or

F = 21.93 dyne ≈ 22 dyne

Hence, the correct answer is option (C) 22 dyne

Answer 2

Answer:

22 is right answer as according to this question.