I'm looking for a quick and dirty was to calculate an electric field.


I know this shit should be easy, I just have to work it out.



Problem) Calculate the electric field between the 2 charges on the x axis:

       2
       /\
      /  \
     /    \
    0----1

9/5/2007 7:19:48 PM

http://www.physics247.com/physics-homework-help/electric-field.php



that's not the kind of help i need.


it might as well be written in Arabic.

9/5/2007 7:23:05 PM

DH Hill, 6th floor, North Stack, QC21 is where you need to look lol

It's right behind me so I'll go see what I can find

9/5/2007 8:13:01 PM

that would be lovely.






[Edited on September 5, 2007 at 8:13 PM. Reason : [nohomo]]

9/5/2007 8:13:43 PM

Do you at least have this equation?



This will give you the force the particle is creating at the origin (push or pull). Treat that similar to the magnitude of a vector and the sum them together.

[Edited on September 5, 2007 at 8:28 PM. Reason : k]

9/5/2007 8:25:32 PM

uhhh.




my equation doesnt look like that.





fuck

9/5/2007 8:28:04 PM

im rollin with F=(ke) ( |8.00nC||-5.00nC| / .500^2)




...

[Edited on September 5, 2007 at 8:29 PM. Reason : word]

9/5/2007 8:29:35 PM

Is it a dipole?

[Edited on September 5, 2007 at 8:30 PM. Reason : nvm]

9/5/2007 8:29:51 PM

meaning?

9/5/2007 8:30:18 PM

I think you have the right formula, you need to sum the charges and then COS(60) to give you the magnitude (along X).

ke = 1/(4PIe0)

It's been a while since I've done these. They're pretty easy.

What book are you working out of?

[Edited on September 5, 2007 at 8:40 PM. Reason : .]

9/5/2007 8:36:33 PM

i was multiplying charges



and not cos(shit)



lol.


so just add 8 + -5, divide by .5 for the first part of the answer.....


and the second part of the answer is cos(60)


?

9/5/2007 8:39:35 PM

What book are you working out of?

9/5/2007 8:42:32 PM

aim



for the love of god.

9/5/2007 8:44:03 PM

ALL HAIL DHCPME

9/5/2007 9:56:02 PM