Deriving Electric Field of a Dipole in Cylindrical Polars: Can You Help?

AI Thread Summary
The discussion focuses on deriving the electric field around a dipole without resolving the dipole moment into components. The approach involves considering the charge separation (2a) as a vector between the negative and positive charges, and defining vector "r" from the midpoint of the dipole to the observation point. Participants suggest rewriting the vectors in terms of "r" and "a," and using the relation r=(r.r)^1/2 to find expressions for 1/r+ and 1/r-. The standard expression for electric potential is then used, followed by taking the negative gradient to obtain the electric field in cylindrical polar coordinates. This method provides a structured way to achieve the desired derivation.
Ashu2912
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Can someone help me with the derivation of the electric field at any point around a dipole. I DO NOT want the dipole moment to be resolved into components and then the field found out. I want it in terms of charge q (+q and -q constitute the dipole), a (2a being the charge separation), theta (angle between the line joining the point to the center of the line joining the 2 charges and the line joining the 2 charges) and of course k (= 1/(4*pi*e0)). Please help!
 
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Hi,

Iv'e been struggling on the same kind of question but think I can do it up to the stage you're asking about.

It's useful to consider 2a as a vector going from the -ve dipole to the +ve one. Also take vector "r" to go from the half way point of "a" to the point of observation. You should now be able to rewrite the vectors from each end of the dipole to the point of observation in terms of "r" and "a".

Using the relation r=(r.r)^1/2 get terms for both 1/r+ and 1/r- then sub into the standard expression for electric potential.

Finally, take the negative gradient of the potential in cylindrical polars to get the electric field in the terms that you asked.
 
Hi,

Iv'e been struggling on the same kind of question but think I can do it up to the stage you're asking about.

It's useful to consider 2a as a vector going from the -ve dipole to the +ve one. Also take vector "r" to go from the half way point of "a" to the point of observation. You should now be able to rewrite the vectors from each end of the dipole to the point of observation in terms of "r" and "a".

Using the relation r=(r.r)^1/2 get terms for both 1/r+ and 1/r- then sub into the standard expression for electric potential.

Finally, take the negative gradient of the potential in cylindrical polars to get the electric field in the terms that you asked.
 
It may be shown from the equations of electromagnetism, by James Clerk Maxwell in the 1860’s, that the speed of light in the vacuum of free space is related to electric permittivity (ϵ) and magnetic permeability (μ) by the equation: c=1/√( μ ϵ ) . This value is a constant for the vacuum of free space and is independent of the motion of the observer. It was this fact, in part, that led Albert Einstein to Special Relativity.
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