Question

Three charges, each of charge 5.0 ""C are located at the vertices of an equilateral triangle whose sides are 10 cm in length. a) What is the total force, magnitude and direction, on charge I? b) What is-the total potential energy of the three charges? c) Charges 2 and 3 remain fixed, but charge one is now allowed to move. If it starts from rest, what i.s its velocity when it is very far away from charges 2 and 3? Charge 1 has a mass of 10 g.

Answer 1

given

three charges

q1 = q2 = q3 = q = 5 C, at vertices of an equilateral triangle

sides length, l = 0.1 m

a. so for the charge onthe top most vertex

from symmetry the force is in upward direction

magnitude of force = F = 2*kq^2*cos(30)/l^2

F = 2*8.98*10^9*25*cos(30)/0.1^2 = 3.888454*10^13 N

b. total potential energy = U

U = 3kq^2/l = 673.5*10^10 J

c. for m = 10 g

from conservation of energy

initial PE = final PE + final KE

2kq^2/l = 0.5mv^2

2*8.98*10^9*25/0.1 = 0.5*10*v^2/1000

v = 0.2996664*10^8 m/s