Laplace's equation in presence of a dipole (perfect or physical)

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SUMMARY

Laplace's equation is not applicable at the location of a dipole; instead, Poisson's equation must be used to account for the charge distribution. For an ideal dipole, the charge density is represented as $$\rho = \vec p \cdot \nabla \delta^{(3)}(\vec x)$$, indicating that the potential satisfies Laplace's equation everywhere except at the dipole itself. To solve problems involving a dipole embedded in a dielectric sphere, one must apply Laplace's equation outside the sphere and Poisson's equation inside, while ensuring boundary conditions are met and avoiding singularities as distance approaches infinity.

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  • Understanding of Laplace's equation and Poisson's equation
  • Familiarity with electrostatic potential and dipole moments
  • Knowledge of charge distributions, specifically delta functions
  • Basic principles of dielectric materials and boundary conditions
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  • Learn about delta function representations in charge distributions
  • Explore the superposition principle in solving electrostatic problems
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Ahmed1029
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TL;DR
I'm wondering if the laplacian of the electrostatic potential function will still be zero at the location of a dipole.
Does Laplace's equation hold true for electrostatic potential at the location of a dipole? Or should poisson's equation be used?
 
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You need to use Poisson’s equation. However, just as for a point charge, you need to be wary of what charge distribution you put in. In the case of a point charge, the charge distribution is a three-dimensional delta distribution. In the case of an idealised dipole it is a delta distribution in two directions and a derivative of a delta distribution in the direction of the dipole:
$$
\rho = \vec p \cdot \nabla \delta^{(3)}(\vec x).
$$

This means the potential will satisfy the Laplace equation everywhere except at the dipole.
 
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Orodruin said:
You need to use Poisson’s equation. However, just as for a point charge, you need to be wary of what charge distribution you put in. In the case of a point charge, the charge distribution is a three-dimensional delta distribution. In the case of an idealised dipole it is a delta distribution in two directions and a derivative of a delta distribution in the direction of the dipole:
$$
\rho = \vec p \cdot \nabla \delta^{(3)}(\vec x).
$$

This means the potential will satisfy the Laplace equation everywhere except at the dipole.
Shouldn't we have a minus sign here?
 
LCSphysicist said:
Shouldn't we have a minus sign here?
Possibly, I did not think too much about signs.

(As Feynman allegedly said: Factors of 2, pi, and i are only for publication purposes 😉)
 
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so consider the following problem :

"A point dipole p is imbedded at the center of a sphere of linear
dielectric material (with radius R and dielectric constant e). Find the electric po-
tential inside and outside the sphere."

How can I solve it only using Laplace's equation? Do I use the superposition principle ?
 
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Ahmed1029 said:
so consider the following problem :

"A point dipole p is imbedded at the center of a sphere of linear
dielectric material (with radius R and dielectric constant e). Find the electric po-
tential inside and outside the sphere."

How can I solve it only using Laplace's equation? Do I use the superposition principle ?
Solve Laplace equation outside.
Solve Poisson equation inside.
Reject terms that blows up as r goes to infinity.
Check the boundary conditiions at the sphere.
Realize that you have made an algebric mistake.
Returns to step 1
Realize you have made another algebraic mistake.
Returns to step 1.
Get the right answer.

That't the recipe
 
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Orodruin said:
This means the potential will satisfy the Laplace equation everywhere except at the dipole.
I think it is much better to say that the potential will satisfy the Poisson equation everywhere, with source charge density the one you defined.
 
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Delta2 said:
I think it is much better to say that the potential will satisfy the Poisson equation everywhere, with source charge density the one you defined.
Got it, thanks!
 
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