The contour map of the solution to Laplace's equation inside the disk. (Courtesy http://oak.ucc.nau.edu/jws8/classes/461.2009.7/animations.html and Google).
Having seen the multipole expansion as a method to obtain the approximate potential
far away from a localized charge distribution we now look for other methods to obtian the potential exactly. The most direct method is to solve the Laplace or Poisson equation ∇2 V=0 or ∇2 V= -ρ/ε0 as the case may be to obtain the poetntial V. However its not easy to solve these equations as they are partial differential equations ( PDE) and also they have infinite number of solutions. The solution specific to a problem is obtianed by using boundary conditions ( values of potential at boundaries, which are determined from the physics of the problem. The Uniqueness theoerm pertains to such boundary conditions and the uniqueness of a solution which satisfies a specific set of boundary conditions.
Solutions to the Laplace equations are called harmonic functions and they have certain special properties.
(i) The value of the potential V(r) at a point r is given by the average of the poetntial over a sphere S of radius R with the pint r as its center.
(ii) Follows from (i) There cannot be any local maxima or minima of V, all extremum values of V occur at the boundaries. ( This is obvious because if V had a maxima
its value would be higher than on a sphere of radius R and hence cannot be the average V over that sphere. Average value can neither be larger nor smaller than the
largest and smallest values in the sample over which the average is being taken. It must lie somewhere in between.)
Proof: Take a point charge q outside a spherical surface of radius R. Its easy to show that the average potential over the surface S is equal to the potential produced
at the center. By superposition this can be extended for any arbitrary discrete or continuous charge distributions outside S, For all such configurations the average potential over the sphere is equal to the net potential produced at the center of S
First Uniqueness Theorem: The solution to the Laplace equation in a volume τ is uniquely determined if the potential V is specified at all boundary surfaces
S.
Proof: Assume two solutions V1 and V2 which staisfy the Laplace equation in τ and which have the same value VS at S. Its easy to see that V3=V2 -V1 also satisfies the Laplace equation and its value is ZERO on S. Since for Laplace equation all extrema must occur at the boundary so the value ZERO is both maxima and minima for V3.
Which implies that V3=0 everywhere. So V2-V1=0 so
V2=V1. So V is unique. this can be easily extended to regions with a non zero charge distribution ρ also by using the Poisson equation.
Corollary: The potential in a volume τ is UNIQUELY determined if the charge density is specified throughout and the potential V is specified on all boundaries
S.
The second Uniqueness Theorem is for conductors and is stated as ; In a volume τ surrounded by conductors and having a specified chrage density ρ the electric field E is UNIQUELY determined if the total charge Q on each conductor is given. ( Proof: Read from Griffiths).
Electrostatic boundary conditions: The normal component of the electric field is discontinuous across any boundary with a charge distribution σ by an amount σ / ε0.
The tangential component of the electric field is always continuous across any boundary.
The potential V is always continuous across any boundary but its derivatives are not. In particular the normal derivative ∂ V/∂ n cap =∇ V. cap is discontinuous by -(σ/ε0)
sir i am not convinced, how can uniqueness theorem hold , we say that if we put a charge above a grounded infinite sheet , the image is below it at the same condition but we can have more situations were boundary conditions are fulfilled , suppose we put 2 charges at some distance below the conductor on the z axis and one more charge above the given charge in the same fashion , then also boundary conditions are satisfied but it doesn't work
ReplyDeleteThe Uniqueness Theorem has been proved independently so there should not be any problem with the proof. Now regarding your example which supposedly violates the theorem, in invite you to make a configuration of charges that will satisfy
ReplyDeleteV=0 on the plane conductor exactly and which is different from that. Dont just give me a hand waving argument. Tell me exactly where the charges will be and their strengths such that V=0 on the plane.
There are no such configurations. It should be evident from a study of equipotential surfaces
between charges. The plane equipotential is when the charges are equal and opposite and are exactly the same distance away.
As i explained in the lectures the Image problems solutions always arise from a study of the equipotential surfaces for charge configurations.
I think that you are missing a key point that V=0
ReplyDeleteeverywhere on the infinite plane, not only at the origin z=0.
i can't understand that point(i) what mean by average of pot.over sphere over of radius R and by that avg. how can i cal. v(r)...
ReplyDeleteThanks sir, i forgot that the potential has to be zero everywhere at the plane,not only at z=0
ReplyDeleteAveraging means exactly the same as usual that is
ReplyDeletesum divided by number. So average potential over surface is sum ( integration) of V over the entire surface, that is a surface integral of V and then divide by 4 pi r squared which is area of the surface of the sphere. This is V(r) and it follows from the property of the solutions of Laplace equation which are called Harmonic Functions.
I explained this in detail in the class. I hope that you were attending them. The blog is not a
substitute for the lectures.