Three charged particles are placed at each of three corners of an equilateral tr

Question

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.4 cm. Two of the particles have a negative charge: q1 = -6.5 nC and q2 = -13.0 nC. The remaining particle has a positive charge, q3 = 8.0 nC. What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Find the net force acting on particle 3 due to the presence of the other two particles. Report your answer as a magnitude and a direction measured from the positive x axis.

Explanation / Answer

force by particle 1: F1=k(-6.5*8)*10^-18/(2.4*10^-2)^2=-8.125*10^-4 N can be written in vector form as=-F1 *cos60 i - F1 *sin60 j =-4.0625*10^-4 i - 7.036 *10^-4 j force by particle 2 :F2=k(-13*8)*10^-18/(2.4*10^-2)^2=-1.625*10^-3 N can be written in vector form as=F2*cos60 i + F2 *sin60 j =8.125*10^-4 i - 1.407*10^-4 j net force acting on particle F= F1 + F2=4.0625*10^-4 i - 8.443*10^-4 j |F|=sqrt(4.0625^2 +8.443^2 )*10^-4 N =9.37 *10^-4 N direction tan@=-8.443/4.0625 @=-64.3 degree from x axis or 295.69 dgre from x axis

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