man, I hate electrostatics...

[Semidevil]

I"m just not getting the grasp of this:

attatched is an image...

At each corner of a square of side L, there are point charges of teh magnitude, Q, 2Q, 4Q and 3Q. Determine the force on each charge due to the other force.....

here is all I have:

F12 + F13 + F14 = FNet

k|Q||2Q| k|Q||3Q| k|Q||4Q| 15kQ^2
-------- + ----------- + ---------- = ------
l^2 (sqrt(2)*l)^2 l^2 2l^2


what does that do? I have no idea....

anybody want to help?

link: http://www.geocities.com/hluu410/charge.bmp

 

[morkinva]

If it were just 2 charges between a known distance, would you be able to do it?

Wouldn't one of the charges in the square experience the sum of forces of the other 3 charges?
i.e. F(Q:2Q)+F(Q:3Q)+F(Q:4Q) = the force on charge Q

I forget its been so long

 

[Anubis]

well Quantum mechanics is harder. just had that test. not one of my better showings

 

[arcas]

Break the problem apart into its component pieces instead:

Let charge A = Q, B = 2Q, C = 3Q, D = 4Q

Then,

|F(BA)| = magnitude of force on A due to B = k(Q)(2Q) / (L*L)
|F(CA)| = magnitude of force on A due to C = k(Q)(3Q) / (sqrt(2L^2))
|F(DA)| = magnitude of force on A due to D = k(Q)(4Q) / (L*L)

(I'll let you do the vector part. Hint: F(BA) is in the -X direction, F(DA) is in the +Y direction and F(CA) is 45deg angle so you'll need to break the vector into its X and Y components)

so Force on A = F(BA) + F(CA) + F(DA) [remember, you're doing vector math here...]


When it comes time to calculate the forces on the other charges, use Newton's Law to save yourself some work: F(AB) = -F(BA), etc.