# Proof on a uniqueness theorem in electrostatics

• ELB27
In summary, the field is uniquely determined when the charge density ##\rho## is given and either ##V## or the normal derivative ##\partial V/\partial n## is specified on each boundary surface, without assuming the boundaries are conductors or that V is constant over any given surface. This can be proven by showing that the potential difference ##V_3## satisfies Laplace's equation and must have its minima or maxima on the boundary, and that when ##\partial V/\partial n## is specified but ##V## is not, the divergence theorem can be used to obtain a condition on the boundary, leading to the standard proof.
ELB27

## Homework Statement

Prove that the field is uniquely determined when the charge density ##\rho## is given and either ##V## or the normal derivative ##\partial V/\partial n## is specified on each boundary surface. Do not assume the boundaries are conductors, or that V is constant over any given surface.

## Homework Equations

Poisson's equation which the potential ##V## must satisfy:
$$\nabla^2V = -\frac{\rho}{\epsilon_0}$$
Gauss' law in differential form:
$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$
The definition of the potential:
$$\vec{E} = -\nabla V$$
Perhaps various product rules of vector calculus will also be useful.

## The Attempt at a Solution

I must admit that I'm not sure about the strategy I should use in this problem so I just played around with the equations trying to get some insight and using the proofs given for the two classical uniqueness theorems as models.
First, suppose there are two fields that satisfy the conditions of the problem:
$$\vec{E_1} = \frac{\rho}{\epsilon_0} ; \vec{E_2} = \frac{\rho}{\epsilon_0}$$
Looking at their difference ##\vec{E_3} = \vec{E_1} - \vec{E_2}##, it must satisfy ##\nabla \cdot \vec{E_3} = 0## or in integral form: ##\oint \vec{E_3}\cdot d\vec{a} = 0## over each boundary surface. Looking at the potential difference ##V_3 = V_1 - V_2##, it must satisfy ##V_3 = 0## on the boundary or ##\partial V_3/\partial n = 0## on the boundary.
In addition, ##\nabla^2V = 0## thus ##V_3## satisfies Laplace's equation and must have it's minima or maxima on the boundary.
Now I don't have any idea on how to proceed from here. In particular, I feel that the normal derivative can be exploited somehow but since I can't assume that the surfaces are conductors, I don't know how. I would highly appreciate any insight on this problem as well as general tips for going about such a problem in the future.

EDIT: After thinking a bit more, if ##V## is given then it follows from the above that ##V_3 = 0## in the region (since it is zero on the boundary and has both maxima and minima zero there). If this reasoning is correct, the only thing left is to prove the case where ##\partial V/\partial n## is specified but ##V## is not.

Last edited:
The usual proof goes by noting that ## \Delta \cdot \left(V \Delta V\right) = (\Delta V)^2 + V \Delta^2 V = (\Delta V)^2 ## and then taking the integral of the equation over the entire region within the surface, and making use of the divergence theorem to obtain a condition on the boundary.

1 person
voko said:
The usual proof goes by noting that ## \Delta \cdot \left(V \Delta V\right) = (\Delta V)^2 + V \Delta^2 V = (\Delta V)^2 ## and then taking the integral of the equation over the entire region within the surface, and making use of the divergence theorem to obtain a condition on the boundary.
Thank you very much!
I figured out how to use ##\partial V/\partial n = 0##. This turns the surface integral that follows from the divergence theorem into zero leading to the standard proof. So the question is solved.

Note to self: write ##\hat{n}da## instead of ##d\vec{a}##. The former gives much more insight.

## 1. What is a uniqueness theorem in electrostatics?

A uniqueness theorem in electrostatics states that the solution to an electrostatic problem is unique, meaning there can only be one possible solution. This means that given the boundary conditions and charge distribution, there can only be one electric potential and one electric field in a given region.

## 2. Why is a uniqueness theorem important in electrostatics?

A uniqueness theorem is important because it allows us to confidently and accurately solve electrostatic problems. It ensures that there is only one possible solution, which helps us avoid any errors or discrepancies in our calculations.

## 3. How is a uniqueness theorem proved in electrostatics?

A uniqueness theorem in electrostatics can be proved using mathematical techniques such as the method of images or the method of separation of variables. These methods involve using the boundary conditions and the governing equations of electrostatics to show that there can only be one possible solution.

## 4. Are there any limitations to the uniqueness theorem in electrostatics?

Yes, there are some limitations to the uniqueness theorem in electrostatics. It assumes that the boundary conditions and charge distribution are well-defined and do not change over time. It also assumes that the electric fields are static and do not vary with time. Additionally, the uniqueness theorem may not hold in cases where there are overlapping or intersecting conductors.

## 5. How does the uniqueness theorem relate to other principles in electrostatics?

The uniqueness theorem is closely related to other principles in electrostatics, such as Gauss's law and the principle of superposition. These principles all work together to help us understand and solve electrostatic problems. Gauss's law helps us determine the electric field from a given charge distribution, and the principle of superposition allows us to find the total electric field by summing the individual contributions of each charge. The uniqueness theorem then ensures that this total electric field is the only possible solution.

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