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Dipole moment, given E-field

  • Thread starter satchmo05
  • Start date
  • #1
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Homework Statement


E-field at (2[m], 30⁰, 90⁰) is E = 4aѳ [V/m]. Find the magnitude and direction of the dipole moment p.

Homework Equations


I know that I need I need to solve for the equation: p = Qd, where Q is the point charge at the given point and the distance between +Q and -Q, which should be 4[m].
The other equation I know of that may lead me in the right direction is:
E = [(Qd)/(4∏єor3)]*(2cosѳ*ar + sinѳ*aѳ).


The Attempt at a Solution


The only thing wrong with the E equation above is that in my text it states that this equation is only valid when the point charges (+Q and -Q) are aligned with the z-axis. My question is what formula (or what variation of the formula above) do I use to determine what I need? It seems as if I need to solve for Q and d to solve for dipole moment. I know that the d vector is directed from -Q --> +Q. How would I determine the vector from this?

Thank you for all of the help in advance! I greatly appreciate it! :biggrin:
 
Last edited:

Answers and Replies

  • #2
674
2
Looking at the electric field equation you put up, what angle will the dipole have to make with your point of interest to not have an radial vector?
 
  • #3
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Looking at the electric field equation you put up, what angle will the dipole have to make with your point of interest to not have an radial vector?
It appears as if I would have to shift my perspective +30 degrees to make it appear as if the charges were along the "z-axis." I assume this is what you mean by your question?!
 
  • #4
674
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There is no charge at 30 degrees. The 30 degrees is just telling you the point in space where you measure the electric field.

I am asking in your electric field equation, at what angle theta will there be no radial vector in your electric field.
 
  • #5
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There is no charge at 30 degrees. The 30 degrees is just telling you the point in space where you measure the electric field.

I am asking in your electric field equation, at what angle theta will there be no radial vector in your electric field.

Oh, I see what you're saying. Well, the radial component in my E equation is 2cos(theta)*ar. Therefore, cos(90) = 0.
 
  • #6
674
2
Exactly, so the dipole must be at a 90 degree angle with respect to the point at which you are measuring the electric field. So once you solve for the proper angle, you will then have your dipole direction. Then all you need to worry about is the magnitude.
 
  • #7
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So if the dipole is located at the origin, the (theta) angle I need to use for the E equation is 90 degrees? I believe this is what you're saying...
 
  • #8
674
2
Yes, the theta angle in your electric field equation will remain 90 degrees. Knowing it is 90 degrees, you get the direction of the dipole. Also, plugging it into your equation you can easily solve for the magnitude of the dipole.
 
  • #9
Yes, the theta angle in your electric field equation will remain 90 degrees. Knowing it is 90 degrees, you get the direction of the dipole. Also, plugging it into your equation you can easily solve for the magnitude of the dipole.

so is the direction vector the same as the E-field vector since they are both directed from -Q to +Q?!
 
  • #10
674
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There is a dipole vector, a direction vector, and an electric field vector. Let's not get confused here... To get the given electric field vector we solved that the dipole vector must be perpendicular to the direction vector. Not sure how else to explain this without drawing a picture.
 
  • #11
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alright, i think that makes sense. I'll write back if I have any more questions for you. thank you very much for spelling this out for me. I greatly appreciate it!
 
  • #12
114
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I have another question for you nickjer. If the dipole needs to be perpendicular to the E-field direction, wouldn't the theta angle need to be 120 degrees?
 
  • #13
674
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You have to be more specific on what theta you are talking about.
 
  • #14
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Well, let me rephrase what we have been talking about. I have the equation for the E-field in my original post. From what you're saying, the dipole needs to be perpindicular to the E-field (this makes sense). From the E-field equation, the theta angle is 30 degrees. So, in order for the dipole to be perpindicular, I would represent the E-field equation with a theta angle of 120 degrees, correct?
 
  • #15
674
2
No, the dipole needs to be perpendicular to the direction vector pointing to the point in space where you measure the E-field. That means the theta in the E-field equation you posted must be 90 degrees.

Lets call the dipole frame the z'-frame where it is aligned in the positive z'-axis. Since your direction vector is 30 degrees with respect to the positive z-axis, and your direction vector is 90 degrees with respect to the positive z'-axis (dipole rotated frame). Going back to the z-frame, the dipole will be -60 degrees with respect to the positive z-axis.

Although you don't usually say negative degrees, you will have to convert it using the phi angle.
 
  • #16
114
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Yes, the theta angle in your electric field equation will remain 90 degrees. Knowing it is 90 degrees, you get the direction of the dipole. Also, plugging it into your equation you can easily solve for the magnitude of the dipole.
But here, I am not getting a magnitude of the dipole if I do not know what Q or d is. After plugging in 90 degrees, the result is (Q*d)/constant*a(theta). How does this give me the magnitude of the E-field?
 
  • #17
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I apologize for the confusion, I thought it made sense last night until I looked at it again a little bit ago... :D
 
  • #18
674
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The magnitude of the dipole is p=Q*d. You don't need to solve for Q and d separately.
 
  • #19
114
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so the magnitude of the dipole moment is (1/constant) and the direction is (1/constant)a(theta)?
 
  • #20
674
2
You just solved for the magnitude using the electric field equation. I already explained how to get the direction in a previous post.

Since you are just saying 1/constant, I will assume you did all the math correct.
 
  • #21
Ok, so what you two have been talking about has been a bit over my head. Nickjer, I'm kinda with satchmo here, you kinda lost me on the direction part. Can you break it down to a simpler visual?

I dont mean to waste your time; I had posted this problem on the introductory level forum page to see if someone could break it down to easier terms and the dude locked it saying that it was already here.
 
  • #22
674
2
Unfortunately I can't explain it in better without a picture. So here is a picture I drew up to help explain it better.

The red vector at the origin is the dipole. The black vector is your direction vector. Remember you solved for the angle between these two vectors as 90 degrees.

The blue vector is the measured electric field vector. It has no radial component and only a theta component.

I drew a 2d plot along the y-z axis, since phi=90 degrees in spherical coords. So nothing really of any importance happens along the x-axis.

I also (horribly) drew what an electric field line would look like for a dipole (dashed line), so that you can see it is plausible that the electric field vector (blue vector) is caused by a dipole in that direction.
 

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  • #23
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Visually, your picture makes a whole lot of sense. I'm still confused about how to find the direction of the dipole mathematically. Somehow, I must have to incorporate the phi angle of 60 degrees. I also calculated the magnitude of the dipole from what you told satchmo, and I received an answer of ~ 1/(8.901e-10) [m]. Obviously, this cannot be correct! haha. any suggestions there, as well?
 
  • #24
674
2
You solved for the direction in one of the previous posts. Using the electric field equation you listed in the first post, you said that the angle between the dipole and the direction vector must be 90 degrees for there to be no radial component of the electric field.
 
  • #25
114
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Thank you much for your help and the time you put into it. I think I understand now. I'm still new to the concept :D
 

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