Electric field of a point near a dipole

AI Thread Summary
The discussion revolves around calculating the electric field at a point near a dipole formed by two equal and opposite charges. The key equation derived is E = kqd/x^3 for points very far from the dipole. Participants express confusion about using trigonometric functions, specifically whether to use sine or cosine in their calculations. The Pythagorean theorem is also mentioned as a method to determine the distance r in the context of the dipole. Ultimately, one participant resolves their confusion and expresses gratitude for the assistance received.
jack343
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Homework Statement


Two equal and opposite charges form a dipole, as shown in the figure. What is the magnitude of the electric field at point P, in terms of k,q,d, and x? If point P is VERY far away, show that the result is approximately E = kqd/x^3.

http://imageshack.us/photo/my-images/403/img20111215161448.jpg/

please help :(


Homework Equations



F = kq1q2/r^2
E= F/q

The Attempt at a Solution



I don't know how to start? Does the dipole basically act like a wire?
I did some research on dipoles and nothing was related.
 
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Maybe this will help.
 

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Spinnor said:
Maybe this will help.

Thanks, now I know where the forces are going. But why is sin(theta) = d/2L? shouldn't it be cos(theta) since L is the adjacent side?

Can't we use the Pythagorean theorem to get that r = sqrt(L^2 + (d/2)^2) ?
 
jack343 said:
Thanks, now I know where the forces are going. But why is sin(theta) = d/2L? shouldn't it be cos(theta) since L is the adjacent side?

Can't we use the Pythagorean theorem to get that r = sqrt(L^2 + (d/2)^2) ?

My mistake, should be sin(theta) = d/2r. As r gets large sin(theta) gets small as required, the forces tend to cancel to a greater extent.
 
THANK YOU SO MUCH! I FIGURED IT OUT :)

Unfortunately, I bombed my electrostatics test today... great way to start winter vacation. :(
 
You lost a battle but you can still win the "war".

Good luck!
 
I lost another battle... deffered from my top school :/

anyhow, thanks for the help, in return I will help someone else.
 
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