Dipole moment, given E-field

1. Mar 22, 2010

satchmo05

1. The problem statement, all variables and given/known data
E-field at (2[m], 30⁰, 90⁰) is E = 4aѳ [V/m]. Find the magnitude and direction of the dipole moment p.

2. Relevant 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).

3. 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!

Last edited: Mar 22, 2010
2. Mar 22, 2010

nickjer

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. Mar 22, 2010

satchmo05

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. Mar 22, 2010

nickjer

There is no charge at 30 degrees. The 30 degrees is just telling you the point in space where you measure the electric field.

5. Mar 22, 2010

satchmo05

Oh, I see what you're saying. Well, the radial component in my E equation is 2cos(theta)*ar. Therefore, cos(90) = 0.

6. Mar 22, 2010

nickjer

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. Mar 22, 2010

satchmo05

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. Mar 22, 2010

nickjer

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. Mar 22, 2010

Porsche911NFS

so is the direction vector the same as the E-field vector since they are both directed from -Q to +Q?!

10. Mar 22, 2010

nickjer

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. Mar 22, 2010

satchmo05

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. Mar 23, 2010

satchmo05

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. Mar 23, 2010

nickjer

You have to be more specific on what theta you are talking about.

14. Mar 23, 2010

satchmo05

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. Mar 23, 2010

nickjer

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. Mar 23, 2010

satchmo05

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. Mar 23, 2010

satchmo05

I apologize for the confusion, I thought it made sense last night until I looked at it again a little bit ago... :D

18. Mar 23, 2010

nickjer

The magnitude of the dipole is p=Q*d. You don't need to solve for Q and d separately.

19. Mar 23, 2010

satchmo05

so the magnitude of the dipole moment is (1/constant) and the direction is (1/constant)a(theta)?

20. Mar 23, 2010

nickjer

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.