Electric Field Due to Dipole Problem

  • Thread starter Renaldo
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  • #1
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Homework Statement



Consider the following figure.

dipole.jpg


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.)

Homework Equations



Et= E1 + E2

E = k|q|/r2



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.
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement



Consider the following figure.

dipole.jpg


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.)

Homework Equations



Et= E1 + E2

E = k|q|/r2



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.
 
  • #3
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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?
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
619
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.
 
  • #5
58
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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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