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Electric Field Due to Dipole Problem

by Renaldo
Tags: dipole, electric, field
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Renaldo
#1
Feb8-13, 06:52 PM
P: 57
1. The problem statement, all variables and given/known data

Consider the following figure.



For the electric dipole shown in the figure, express the magnitude of the resulting electric field as a function of the perpendicular distance x from the center of the dipole axis. (Use the following as necessary: k, q for the charges, x, and d.)

2. Relevant equations

Et= E1 + E2

E = k|q|/r2



3. The attempt at a solution

The x component of the electric fields will cancel out, leaving only the y-components. Adding the two vector fields in the y-direction:

[k|q|/r2]sinθ + [k|q|/r2]sinθ = 2[k|q|/r2]sinθ

r^2 = [d/2 + x^2]1/2

My final answer:

2[kq/[d2/4 + x2]1/2]sinθ

This is not correct.
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Dick
#2
Feb8-13, 07:05 PM
Sci Advisor
HW Helper
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P: 25,250
Quote Quote by Renaldo View Post
1. The problem statement, all variables and given/known data

Consider the following figure.



For the electric dipole shown in the figure, express the magnitude of the resulting electric field as a function of the perpendicular distance x from the center of the dipole axis. (Use the following as necessary: k, q for the charges, x, and d.)

2. Relevant equations

Et= E1 + E2

E = k|q|/r2



3. The attempt at a solution

The x component of the electric fields will cancel out, leaving only the y-components. Adding the two vector fields in the y-direction:

[k|q|/r2]sinθ + [k|q|/r2]sinθ = 2[k|q|/r2]sinθ

r^2 = [d/2 + x^2]1/2

My final answer:

2[kq/[d2/4 + x2]1/2]sinθ

This is not correct.
For one thing r^2=x^2+(d/2)^2. You've got an extra square root. For another, you should be able to express sinθ in terms of x and d as well.
Renaldo
#3
Feb8-13, 07:24 PM
P: 57
Quote Quote by Dick View Post
For one thing r^2=x^2+(d/2)^2. You've got an extra square root. For another, you should be able to express sinθ in terms of x and d as well.
Yes, I noticed that extra square root. When I fix things up, and set sinθ equal to d/2r:

It's a lot of algebra, but it comes out to kqd/[(d2/4)+x2)]3/2. Does that look right?

Dick
#4
Feb8-13, 07:29 PM
Sci Advisor
HW Helper
Thanks
P: 25,250
Electric Field Due to Dipole Problem

Quote Quote by Renaldo View Post
Yes, I noticed that extra square root. When I fix things up, and set sinθ equal to d/2r:

It's a lot of algebra, but it comes out to kqd/[(d2/4)+x2)]3/2. Does that look right?
Looks ok to me.
Renaldo
#5
Feb8-13, 07:31 PM
P: 57
Quote Quote by Dick View Post
Looks ok to me.
All right. Thanks. Webassign doesn't like the answer but I feel confident in the math. Could be a problem with Webassign.


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