What is the net electric force on particle 3 due to the other charges (diagram)?

Question

Please show all work so I can get a full understanding of these questions for the upcoming exam.

http://img406.imageshack.us/img406/6928/... (Diagram)
What is the net electric force on particle 3 due to the other charges?

What is the Electric Field experienced by the -4.0 µC charge due to the other two charges?

http://img72.imageshack.us/img72/1096/examreviewsw8.jpg
Is the URL of the diagram. Sorry, my bad!

Answer 1

First calculate the electric field on particle 3 due to particles 1 and 2 (E = -kq/(r^2)) where q is the charge of the particle causing the electric field (you must add the vector components of the electric fields due to both particle 1 and particle 2), and r is the distance between particle 3 and the source charge (that is particle 1 or 2, respectively). Check to see if the components of your electric field make sense; it appears the electric field should be down and to the left. (Remember k = Coulomb's constant = 9 e9)

Then multiply the charge of particle 3 by the electric field to obtain the force on particle 3 due to the electric field (which is caused by the other two particles). This will also have vector components, and since particle 3 is negatively charged, should have a net force upward and do the right.

To get you started:

The particles are arranged nicely. The electric field due to particle 2 is entirely to the left, or in the -i direction, with magnitude k*(6 e-6)/(.3^2). The electric field due to particle 1 is entirely downward, or in the -j direction, with magnitude k*(3 e-6)/(.2^2). Thus, the net electric field is the respective magnitudes in the -i and -j directions.

Although I left units off of here for the sake of clarity, you should check to make sure they work accordingly.

Answer 2

The force between two charges is given by the formula F = k.q1.q2/ (r.r) F23 (force on 3 by 2) = - k.6µ.4µ/ (0.3x0.3) = - 266.7 (k.µ.µ) newton. (k.µ.µ) = 9x 10^(-3) F23= 2.4 to the right. F13 = k (-3µ) (-4µ)/ (0.3x0.3) = 133.3 (k.µ.µ) newton F13 = 1.2 N in the upward direction. Resultant of these two forces is Sqrt of (2.4) ^2 + (1.2)^2 The force is 2.7 N The direction is given by tan (2.4/ 1.2) 63.4 degree up from the 1.2 newton

Answer 3

You pasted the wrong URL for your picture.

Answer 4

I know how to do it, but the diagram isn't showing up.