Three charged particles are placed at the corners of an equilateral triangle of side *d* = 1.00 m.

The charges are *Q*_{1} = +4.0 µC, *Q*_{2} = -7.0 µC, and *Q*_{3} = -6.0 µC. Calculate the magnitude and direction of the net force on each due to the other two.

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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.

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