The charged particles A, B, and C, occupy fixed positions at the vertices of a r

Question

The charged particles A, B, and C, occupy fixed positions at the vertices of a right triangle, as shown. The charges on the particles are all equal in magnitude. Consider only the electrostatic forces between the particles. Draw a force diagram for each particle showing all of the forces acting on it. Also draw the net force on each particle. For each force indicate the object exerting the force. Identify all Newton's Third Law pairs. If the magnitude of the charges is 2.6 Times 10^-6C, and the distance between the charges A and B and B and C is 2 Times 10^-2 m. what is the magnitude of the net force on particle A? Show your work.

Explanation / Answer

b) Fnet = FA+ FB

FA=FAy= kq^2/d^2 = (9*10^9*(2.6*10^-6)^2)/(2*1^-2)^2 = 0.01521 N

FB=kq^2/r^2 = (9*10^9*(2.6*10^-6)^2)/(sqrt2*2*1^-2)^2 = 7.6*10^-3 N

FBx = FBcos45 = 7.6*10^-3cos45 = 5.37*10^-3 N

FBy = FBsin45 = -7.6*10^-3cos45 =- 5.37*10^-3 N

Fnetx = FAx + FBx = 0 + 7.6*10^-3 = 7.6*10^-3 N

Fnety = FAy + FBy = 0.01521 + 7.6*10^-3 = 0.02281 N

Fnet = sqrt( 0.0076^2 + 0.02281^2) = 0.024 N

= tan^-1(0.02281/0.0076) = 71.57 deg

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