Calculating the electric field


by aliaze1
Tags: electric, field
aliaze1
aliaze1 is offline
#1
Oct5-07, 09:45 PM
P: 177
1. The problem statement, all variables and given/known data

What are the strength and direction of the electric field at the position indicated by the dot in the figure?



2. Relevant equations

Edipole = ~ [1/(4πε0)] * [2p/r3 ]
on the axis of an electric dipole

Edipole = ~ [-1/(4πε0)] * [p/r3 ]
in the plane perpendicular to an electric dipole

3. The attempt at a solution

Which equation should I use??

Thanks
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Avodyne
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#2
Oct5-07, 11:23 PM
Sci Advisor
P: 1,185
Neither. Your equations only apply if the two charges in the dipole are much closer together than the distance r from the dipole.
aliaze1
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#3
Oct6-07, 05:36 PM
P: 177
Quote Quote by Avodyne View Post
Neither. Your equations only apply if the two charges in the dipole are much closer together than the distance r from the dipole.
what formula would i use?

aliaze1
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#4
Oct6-07, 06:21 PM
P: 177

Calculating the electric field


my textbook uses these forumulas...so i tried the problem using them

so here is my process:

p = qs

s=0.1
q=1*10-9
p=1*10-10

1/(4πε0) = 9*109

r=0.05m

plugging everything into the second equation {[-1/(4πε0)] * [p/r3 ]}, i get -7200, but this is incorrect
nicksauce
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#5
Oct6-07, 06:23 PM
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Quote Quote by aliaze1 View Post
what formula would i use?
Why not just use coulomb's law, and superimpose the two electric fields?
aliaze1
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#6
Oct6-07, 06:39 PM
P: 177
Quote Quote by nicksauce View Post
Why not just use coulomb's law, and superimpose the two electric fields?
good idea..this was my original approach, which didn't work for some reason

E=[1/(4??0)]*[q/r2]

using the two charges:

q1=1*109
q2=-1*109

and their respective distances:

r1=0.05
r2=0.01250.5 (square root)

and doing all calculations, and then adding the two charges (3600 and -720) gives me 2880, which is incorrect...
aliaze1
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#7
Oct7-07, 12:00 PM
P: 177
lol so yea....umm...any help?


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