Electric Field Due to Dipole Problem

by Renaldo
Tags: dipole, electric, field
 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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P: 25,228
 Quote by Renaldo 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.
P: 57
 Quote by Dick 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?

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Thanks
P: 25,228
Electric Field Due to Dipole Problem

 Quote by Renaldo 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.
P: 57
 Quote by Dick 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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