The Poisson Equation for Electrostatics · 2014. 12. 20. · Derivation from Maxwell’s Equations Example: Laplace Equation in Rectangular Coordinates Uniqueness Theorems Bibliography

Derivation from Maxwell’s EquationsExample: Laplace Equation in Rectangular Coordinates

Uniqueness TheoremsBibliography

The Poisson Equation for Electrostatics

Yesé J. Felipe

University of Puerto Rico - Mayagüez

Yesé J. Felipe The Poisson Equation for Electrostatics



Table of Contents

1 Derivation from Maxwell’s Equations

2 Example: Laplace Equation in Rectangular Coordinates

3 Uniqueness Theorems

4 Bibliography




Derivation from Maxwell’s Equations

Maxwell’s Equations for Electrodynamics in differential form are:

∇ · ~E = 1�0ρ(~r , t) (Gauss’s Law for Electricity) (1a)

∇ · ~B = 0 (Gauss’s Law for Magnetism)(1b)

∇× ~E = −∂~B

∂t(Faraday’s Law of Induction) (1c)

∇× ~B = µ0~J(~r , t) + µ0�0∂~E

∂t(Ampère-Maxwell’s Law) .(1d)




Using the Helmholtz Theorem and that ~B is divergenceless, themagnetic field can be expressed in terms of a vector potential, ~A:

~B = ∇× ~A (2)

From this and Faraday’s Law, Eq. (1c), the electric field can beexpressed as:

~E = −∇V − ∂~A

∂t(3)




Substiting into Gauss’ Law, Eq. (1a), results in:

∇2V + ∂∂t

(∇ · ~A) = − 1�0ρ (4)

In the electrostatic case, it reduces to

∆V ≡ ∇2V = − 1�0ρ(~r), (5)

which is Poisson’s equation. For a region of space in which there isno charge, we obtain Laplace’s equation:

∆V ≡ ∇2V = 0 (6)




Example: Laplace Equation in Rectangular Coordinates

The Laplace equation in rectangular coordinates is

∂2V

∂x2+∂2V

∂y2+∂2V

∂z2= 0 (7)

To solve by separation of variables we assume that:

V (x , y , z) = X (x)Y (y)Z (z) (8)

After substituting and diving, this results in:

1

X (x)

d2X

dx2+

1

Y (y)

d2Y

dy2+

1

Z (z)

d2Z

dz2= 0 (9)




In order for the result to hold for arbitrary values of thecoordinates, each of the terms must be individually constant:

1

X (x)

d2X

dx2= −α2 (10a)

1

Y (y)

d2Y

dy2= −β2 (10b)

1

Z (z)

d2Z

dz2= γ2, (10c)

where α2 + β2 = γ2.




If α2 and β2 are arbitrarily chosen to be positive, the solutions tothe set of ODEs are then:

X (x) = e±iαx (11a)

Y (y) = e±iβy (11b)

Z (z) = e±√α2+β2z . (11c)




In order to determine α and β, we impose the boundary conditionson the potential. An example is shown in the figure.




Since V = 0, for x = 0, y = 0, z = 0, we obtain that

X (x) = sin(αx) (12a)

Y (y) = sin(βy) (12b)

Z (z) = sinh(√α2 + β2z), (12c)

From V = 0, for x = a, and y = b, we must have that α(a) = nπ,and β(b) = mπ. Therefore,

αn =nπ

a(13a)

βm =mπ

b(13b)

γnm = π

√n2

α2+

m2

β2, (13c)




So, the partial potential that satisfies the above boundaryconditions is

Vnm = sin(αnx)sin(βmy)sinh(γnmz) (14)

The potential can then be expanded in terms of Vnm with arbitrarycoefficients that will be chosen to fulfill the final boundarycondition:

V (x , y , z) =∞∑

n,m=1

Anmsin(αnx)sin(βmy)sinh(γnmz) (15)




Evaluating the final boundary condition, V (x , y , c) = V0(x , y) atz = c :

V0(x , y) =∞∑

n,m=1

Anmsin(αnx)sin(βmy)sinh(γnmc) (16)

Which is a double Fourier series for the function V0(x , y). Herethe coefficients Anm are given by

Anm =4

ab sinh γnmc

∫ a0

dx

∫ b0

dyV0(x , y)sin(αnx)sin(βmy) (17)




If the box has potentials different from zero on all six sides, thesolution for the potential inside the box can be obtained by linearsuperposition of six solutions, one for each side, equivalent toEqs.(17) and (15). In the case of the potential inside the box witha charge distribution inside, Poisson’s equation with prescribedboundary conditions on the surface, requires the construction ofthe appropiate Green function, whose discussion shall be ommited.




Uniqueness Theorems

First uniqueness theorem: The solution to Laplace’s equation insome volume τ is uniquely determined if V is specified on theboundary surface S .This theorem guarantees that the solution found for the previousexample is unique.Corollary: The potential in a volume τ is uniquely determined if(a) the charge density throughout the region and (b) the value ofV on all boundaries, are specified.This corollary is crucial for the validity of the solutions obtainedwith another method used for finding the electric potential, knownas The Method of Images.




Second uniqueness theorem: In a volume τ surrounded byconductors and containing a specified charge density ρ, the electricfield is uniquely determined if the total charge, Qtot =

∑i Qi , on

each conductor is given (as ilustrated below).




Bibliography

J.D. JacksonClassical Electrodynamics.John Wiley and Sons, Inc., 3rd edition, 1998.

D.J. GriffithsIntroduction to Electrodynamics.Prentice-Hall, 3rd edition, 1999.


Derivation from Maxwell's EquationsExample: Laplace Equation in Rectangular CoordinatesUniqueness TheoremsBibliography

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The Poisson Equation for Electrostatics · 2014. 12. 20. · Derivation from Maxwell’s Equations Example: Laplace Equation in Rectangular Coordinates Uniqueness Theorems Bibliography