Topic awaiting preservation: gah! physics problem

If somebody'd be willing to help, this problem has been bugging me and refuses to be solved. Although it should be easy, is easy, and is more emberassing to post than anything, seeing as it's a rudimentary physics problem meant to introduce electric fields, I could seriously use some help here. So if anybody has the time to waste, that'd be great.


Two charges Qx and Qy lie 6cm away from each other on a horizontal line. Directly between them and 4cm up (Qx+3cm, 4cm) (Qy -3cm, 4cm) lies point Z. Determine the magnitude and direction of the electric field at point Z due to the charges at Qx and Qy.

Qx, Qy = + 2.0 x 10^-8 C

I assume I'd calculate the electric field between Qx and Qy using Fe=(k*Qx*Qy) / (d^2), then input that into E = Fe / (r^2) to produce the answer, however nothing comes up with the book answer: 1.2 x 10^5 N/C [up]



(Edited by cfb on 06-10-2005 02:15)

(Edited by cfb on 06-10-2005 02:21)

Electricity... next year for me!

Wow. That was simple; I think I was just momentarily hated by the Gods.

I forgot to treat it as a vector, so:

E = (k*Qx, Qy) / (d^2) = [k(2.0 x 10^-8)] / (.05^2) = 7.2 x 10^4 N/C [53.13 degrees]

and since the two forces are equal (Qx, Qy), they will negate the direction of pull/push, you can just input them into a sine/cosine law to get the answer, 1.2 x 10^5 N/C [90 degrees / up].

Sorry!

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"Abortion clinics are like expressways to heaven."

Ah electrickery no less, glad you solved it cfb. I had the exact same answer
what?

::tao:::: ::cell::

Oh yeah - that's just child stuff. I was doing that in kindergarten...

ZZZZZZZZZZZZZZZZZZzzz....

Yeah, I was looking through the AP Physics and some college physics books. Oh my god! Reminds me of why I'm never, ever, pursuing a career in the sciences.