How Does Dipole Arrangement Affect Electric Field Strength at a Distance?

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Homework Help Overview

The discussion revolves around the calculation of the electric field strength produced by two dipoles arranged back-to-back along an axis. The problem involves applying principles of electrostatics, particularly focusing on the electric potential and electric field equations, while considering the approximation where the distance from the dipoles is much greater than the separation between the charges.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of electric potential and electric field equations to the four charges involved in the dipole arrangement. There are questions about how to properly sum the contributions from each charge and whether to multiply results due to the presence of two dipoles.

Discussion Status

Some participants have proposed using different equations for electric potential and electric field, while others are exploring the implications of the approximation r >> a. There is an ongoing exploration of how to derive the total electric field or potential from the individual contributions of the charges.

Contextual Notes

Participants note the lack of a numeric value for r and question how to handle the derivative of the potential when calculating the electric field. There is also mention of a related question about calculating the electric field of a hexapole, indicating a broader interest in multipole interactions.

scholio
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Homework Statement



assume that two dipoles are placed back -t0 - back along axis as shown:

((-) a (+))(( +) a (-)) ------------------x axisr is oriented between ))(( but above, a distance such that r>>a

compute the strength of the electric field on the axis of the pair at a distance r away from the the center of the dipole pair. use the approximation r>>a to simply your expression. each charge can be assumed to be 1 microcoulomb charge with the sign shown in the diagram

Homework Equations



electric potential V = kq/r where k is 9*10^9 constant, q is charge in coulombs, r is distance the charge is from the point r

electric field E = 2kP/r^3 where P is the dipole moment, given as 10^-8 C-m -->derivative of electric potential eq

The Attempt at a Solution



i am not sure where to draw the diagram as shown with lines connecting each charge (4 of them) and using the V equation to sum them up, but r is not given a numeric value

then do i do the derivative of the summation of V?

or do i just use the electric field equation, but how do i involve all four charges, and what do i put in for r in the denominator?

any help appreciated

cheers...
 
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sorry for double posting, but i think i may have the wrong electric potential V equation.

should it be:

V =kpcosine(theta)/r^2 where p = 2qa, q = charge, a is distance b/w charges of dipole.

but since the problem involves two dipoles, do i multiply V by 2? how about the derivative (electric field E)?

but in order to find the electric field, i would need to take the derivative, so i should get this:

E = 2kpcosine(theta)/r^3 where r is the distance between pt r and the dipole(s)

do i multiply E by two because the problem involves two dipoles? when i put in p = 2qa with q = 1 microcoulomb and r>>a i got

E = 72000a(cosine(theta))/r^3 Newtons/coulomb

any closer?...
 
Last edited:
Hi scholio,

I would use either

[tex] E=\frac{k q}{r^2}[/tex]

or

[tex] V= \frac{kq}{r}[/tex]

and apply it to all four charges. It doesn't make much difference which one you use since everything is along the same line.

Calculate the exact value of the total field or total potential. You'll have parts like (r+a) and (r-a) in some of the denominators. Then you can use the fact that r >> a to find an approximation for those denominators. After a few cancellations you should get the answer. What do you get?
 
yes we can easily calculate it but my question is what we will do if we have to calculate electric field of a hexapole ...can anyone help me.please
 

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