Homework Help

Three charged particles are placed at the corners of an equilateral triangle of side d...

user profile pic

zlemisch | Student, Grade 11 | eNotes Newbie

Posted April 13, 2010 at 11:57 PM via web

dislike 0 like

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

user profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted April 14, 2010 at 5:49 AM (Answer #1)

dislike 1 like

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