Question

Three charges, Q1=4.5 μC, Q2=-3.8 μC, and Q3=2.2 μC, are placed on each corner of a right triangle as shown in Figure. The distance between Q1 and Q2 is 38 cm and that between Q1 and Q3 is 51 cm. Find the magnitude of the electric force exerted on Q1 by Q2 and Q3.

Answer 1

u have to apply the law of superpositioning of electric fiels
f12=force on 1 due to 2 ,like wise f13
now f12=kq1.q2/38^2
f13=kq1.q3/51^2
the direction of f12 is a straight line in the direction f the perpendicular of the triangle
the direction of f13 will be at an angle of theta from the f12
now resove the f13 force as f13 cos(theta) and f13 sin(theta)
total force in upward direction will be f12 +f13 cos(theta)
total force perpendicular to upward direction will be f13sin(theta)
afterdoing all this just get the resultant of both the directions
the resultant will be the total force acting due on q1 due to q2 and q3