Three charged particles are placed at the corners of an equilateral triangle of side d = 1.00 m.
The charges are Q1 = +4.0 µC, Q2 = -7.0 µC, and Q3 = -6.0 µC. Calculate the magnitude and direction of the net force on each due to the other two.
1 Answer | Add Yours
The force of attraction between Q1and Q2 is Q1*Q2/[(4peo)d^2] = 9*10^9*4*7*10^-(12) /1^2 N=0.252N
The force of attraction between Q1 and Q3 = 9*10^8(4*6/1^2) N = 0.192 N.
So the net force of 0.252N and 0.192N with 60 degree between is sqrt(0.252^2+0.192^2+ 2*0.252*0*192cos60) = 0.3856839N
The angle between 0.252N and the resultant 0.3857N is
cos inverse (0.38565839^2+0.252^2-0.192^2)/(2*0.38565839*0.252) = 27.08 degree.
Similarly you can proceed to find the magnitude and direction of the resultant force exerted on other two charges.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes