What is the Net Force on B in an Equilateral Triangle of Point Charges?

[axxon]

Homework Statement

Three identical point charges A, B, and C are located in the shape of an equilateral triangle with sides of length 15 cm. What is the net force on B if each charge has a magnitude of -5.0 x 10-3 C?

Homework Equations

coulombs law equation

The Attempt at a Solution

(please see the attached scanned page and attached diagram)
I think that i am doing something wrong the angle...hope you guys can help.

 

[LowlyPion]

Homework Statement

Three identical point charges A, B, and C are located in the shape of an equilateral triangle with sides of length 15 cm. What is the net force on B if each charge has a magnitude of -5.0 x 10-3 C?

Homework Equations

coulombs law equation

The Attempt at a Solution

(please see the attached scanned page and attached diagram)
I think that i am doing something wrong the angle...hope you guys can help.

Without seeing your diagram, I would hope that you are treating the E-field as vectors. The best way to calculate the Sum is to separate them into the x,y components and then add and then report the resultant.

 

[axxon]

did you see my work shown? Is it not opening up?
I trying to solve the triangle to find the net force. I think that is where I may be getting the angle wrong. It is not suppose to be 60 degrees.

 

[LowlyPion]

did you see my work shown? Is it not opening up?
I trying to solve the triangle to find the net force. I think that is where I may be getting the angle wrong. It is not suppose to be 60 degrees.

Until images are approved, they are not generally available.

You of course do have 60° angles.

You may want to choose your axis as the bisector of the opposite side if you did not already. That way you are likely only interested in the components of the 2 other charges directed along this bisector. That should reduce your calculation to simply be 2*cos30*|F| calculated from one of them alone.

 

[axxon]

okay, so i know that the angle for the original triangle is 60 degrees. And by using coulombs law to find Fbc and Fba we can create another triangle to find the net force. I wish there was a way i could draw it out in here, but i guess I will have to wait for the approval. Basically the net force on B i get is 1.0 x 10^7 N.

 

[LowlyPion]

okay, so i know that the angle for the original triangle is 60 degrees. And by using coulombs law to find Fbc and Fba we can create another triangle to find the net force. I wish there was a way i could draw it out in here, but i guess I will have to wait for the approval. Basically the net force on B i get is 1.0 x 10^7 N.

That looks only like |F| between either of the Fab or Fcb

When you add them as vectors I think you should get something more like

2*Cos30*|F| = 2*.866*1*10⁷ = 1.732 * 10⁷

directed away from the other 2 charges along the line that is the ⊥ bisector of the line AC.

 

[axxon]

But don't you have to add the vectors head to tail? and then using that we can use cosine law to solve or vector components like you said earlier.

 

[LowlyPion]

But don't you have to add the vectors head to tail? and then using that we can use cosine law to solve or vector components like you said earlier.

That's one way to do it.

I didn't use the law of cosines to solve however.

I simply exploited the symmetry of the charges noting that since they are equal and equidistant, and on opposite sides of the bisector, that their contributions ⊥ TO the bisector cancel, and along the bisector they will add. Since Cos30 is the side in the direction of the bisector, then it is Cos30*|F| for each or 2*Cos30*|F|.

 

[axxon]

when will the attachments be approved? I think it will help me if you maybe see what I did for my solution.
Thanks!