I've a doubt on the following situation. Consider a hollow conductor $A$ (of arbitrary shape) containing another conductor $B$ (again of arbitrary shape), with a positive charge $+q$. enter image description here

By the electrostatic induction a total charge $-q$ will appear in the inner surface of $A$ (let's call it $S_{A, int}$) with a density $\sigma_{A,int}$, and a total charge $+q$ will appear on the outer surface $S_{A,ext}$ with a density $\sigma_{A,ext}$.

Let's call the electrostatic fields generated by the charge densities $\sigma_{B}$, $\sigma_{A,int}$ and $\sigma_{A,ext}$ rispectively $\bf{E}_{B}$, $\bf{E}_{\sigma_{A,int}}$ and $\bf{E}_{\sigma_{A,ext}}$.

My question is: how to prove (possibly in a rigorous way) the two following facts?

  1. $\bf{E}_{\sigma_{A,int}}$ is such that $$ \bf{E}_{B}+\bf{E}_{\sigma_{A,int}}=0$$ everywhere outside the cavity (therefore also outside the conductor $A$).

  2. Both $\bf{E}_{\sigma_{A,int}}$ and $\bf{E}_{\sigma_{A,ext}}$ are zero everywhere inside the cavity.


  1. Griffiths- Introduction to Electrodyinamics proposes a similar situation in Chapter 2.5 (Example 2.9) and he states that $\sigma_{A,int}$ is such that "its field cancels that of $B$, for all point exterior to the cavity".

    He explains the statement saying "I cannot give you a satisfactory explanation at the moment" nevertheless I did not find a proper explanation of this fact in all the book.

    Anyway he tries to justify the fact saying "For that same cavity could have been curved out of a huge spherical conductor with radius of 27 light years or whatever. In that case the density $\sigma_{A,ext}$ is simply too far away to produce a significant field and the two other fields ($\bf{E}_{B}$ and $\bf{E}_{\sigma_{A,int}}$) would have to accomplish the cancellation by themselves".

    This does make sense to me, but I'm looking for a more rigourous explanation (or at least where I can find one).

  2. I would guess that, for the conservativity of electrostatic fields: $$\oint_{\gamma} \bf{E}_{\sigma_{A,int}} \cdot \bf{ds}=\int_{\gamma_1}\bf{E}_{\sigma_{A,int}}\cdot \bf{ds}+\int_{\gamma_2}\bf{E}_{\sigma_{A,int}}\cdot \bf{ds}=0\,\,\,\, \forall \gamma_1,\gamma_2\tag{*}$$

    Where $\gamma_1$ is any curve like the red one in picture (connecting any two points in the cavity $C$ and $D$ passing through the cavity) and $\gamma_2$ is any curve like the green one in picture (hence passing inside the conductor). enter image description here Since $\gamma_2$ passes in the conductor, surely $$\int_{\gamma_2}\bf{E}_{\sigma_{A,int}}\cdot \bf{ds}=0\,\,\,\,\,\forall \gamma_2$$

    Therefore, from $(*)$

$$\int_{\gamma_1}\bf{E}_{\sigma_{A,int}}\cdot \bf{ds}=0 \,\,\,\,\forall \gamma_1$$

Nevertheless I'm not totally sure of the following implication

$$\int_{\gamma_1}\bf{E}_{\sigma_{A,int}}\cdot \bf{ds}=0 \,\,\,\,\forall \gamma_1\implies \bf{E}_{\sigma_{A,int}}=0 \,\,\,\,\, \mathrm{inside} \,\,\, \mathrm{the} \,\,\, \mathrm{cavity}$$

Can I come to this conclusion with this reasoning? (The same reasoning would lead to the conclusion that also $\bf{E}_{\sigma_{A,ext}}=0$ inside the cavity).

  • $\begingroup$ Why is this not a standard Farady Cage problem that everyone knows well? (Have you got the quote correctly applied to the right issue? Or the circular path integral proof was too well hidden in the book) $\endgroup$ – JMLCarter Dec 25 '16 at 0:45
  • $\begingroup$ arxiv.org/abs/1609.04248, this is an article published in the European Journal of Physics, where I tried to explain two different proofs to this. T'he second proof there is essentially the first answer here, explained more completely. $\endgroup$ – Aritro Pathak Jul 26 '17 at 21:30

Let $V$ be the total potential that is zero at $\infty$. We know that V is constant in both conductors. Let $V_B$ be its value inside B and $V_A$ its value inside A. That's all we need to reconstruct the potential and answer the question in detail.

Let $V_1$ be the potential such that \begin{align} &V_1 = V_B-V_A \quad \text{inside B,}\\ &V_1 = 0 \quad \text{inside A and on the exterior outside of A,}\\ &\nabla^2 V_1 = 0 \quad \text{in the free space between A and B.} \end{align} Let $V_2$ be the potential such that \begin{align} &V_2 = V_A \quad \text{inside B and inside A as well as in the free space between A and B,}\\ &\nabla^2 V_2 = 0 \quad \text{on the exterior outside of A,}\\ &V_2 = 0 \quad \text{at} \quad \infty . \end{align}

By construction, we have $$V = V_1+V_2.$$

Let $\bf{E}$ be the total electrostatic field defined in the question: $${\bf{E} = \bf{E}_{B} + \bf{E}_{\sigma_{A,int}} + \bf{E}_{\sigma_{A,ext}}}.$$ We have $${\bf{E}} = -\nabla V.$$

We need now to identify each component.

By construction, the charges associated to $V_1$ are located on the surface of B and the inner surface of A, and the charges associated to $V_2$ are located on the outer surface of A. Hence \begin{align} &{\bf{E}_{B} + \bf{E}_{\sigma_{A,int}}} = -\nabla V_1,\\ &{\bf{E}_{\sigma_{A,ext}}} = -\nabla V_2. \end{align}

By the definition of $V_1$, the first equation proves point 1 of the question:
$${\bf{E}_{B} + \bf{E}_{\sigma_{A,int}}} = -\nabla 0 = 0\quad \text{inside A and on the exterior outside of A.}$$

By the definition of $V_2$, the second equation proves half of point 2 of the question: $${\bf{E}_{\sigma_{A,ext}}} = -\nabla V_A = 0 \quad \text{inside B and A as well as in the free space between A and B}.$$

Finally, regarding ${\bf{E}_{\sigma_{A,int}}}$, it won't be zero everywhere inside the cavity except for the special case where the associated charges generate a constant potential inside that cavity (for example in the case of a spherical symmetry).

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Paraphrasing Griffith's "Second Uniqueness Theorem" (section 3.1.6, 4th ed.):

The electric field in a volume bounded by conductors is uniquely determined if the total charge on each conductor is given.

(1) Since the charge is only placed on B, this tells you that the field in the cavity is uniquely solved by the shape of the cavity and $Q$. In other words, the outside surface cannot contribute to the cavity: $\vec{E}_{A,ext}=0$. However, by the same reasoning I think it is wrong to say $\vec{E}_{A,int}=0$ in the cavity because the inner surface (and therefore $\sigma_{A,int}$) must contribute to uniquely solving the field.

(2) The outer surface will have $Q$ distributed on it regardless of the cavity's shape. So even though the outer conductor technically has no charge, just think of the volume outside A as bounded by A's exterior (with $Q$ on it) and infinity. Then apply the same uniqueness theorem to show the irrelevance of the cavity (in other words, $\vec{E}_{A,int}+\vec{E}_B=0$ outside).

Addressing your attempts:

(1) The rigorous explanation lies in the uniqueness theorems presented in Chapter 3, and their associated proofs.

(2) The final step is wrong: A vanishing integral does not imply a vanishing integrand in general. Besides, that's not a conclusion you want to come to. Otherwise it is good reasoning, and very much in the spirit of $\Phi$ as a conservative field (replacing $\vec{E}$ with $-\nabla \Phi$ makes the integrals very easy to compute moving between the surfaces of a single conductor).

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Why is this not a standard Farady Cage problem that everyone knows well?

(Have you got the quote correctly applied to the right issue? Or the circular path integral proof was too well hidden in the book)

For reasoning consider a symmetric perfectly circular ring of charge. Your conductor allows charges to move, so they will distribute evenly in your irregular conductor, matching this arrangement.

For each point inside the conductor the force due to the nearest points of the ring will be exactly compensated for by the force from those further away (although the force is weaker, there are more of them), rendering no net force on a charge at that point. (A mathematical integral proof has been done, but that's not what you asked for).

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Ok, we have conductor A, conductor B, we have charged conductor B and now wander what is the distribution of charges in our system.

Let's try to guess. Let's take a conductor like A, but without cavity. Let's charge it with total charge $q$. This charge would distribute over the surface of conductor. Distribution would be such that the total electric field anywhere inside the conductor is zero: there are no more electric current. Let's remember how the charge is distributed: $\sigma_{A, ext}$.

Now let's take an infinite conductor with a cavity and another charged conductor B inside the cavity. Charges would get distributed on the inner surface of cavity in a way, that electric field anywhere outside the cavity is zero. Let's remember how the charges are distributed now: $\sigma_{A, int}$ and $\sigma_{B}$.

Now get back to original problem. Let's manually distribute charges exactly like them were distributed: $\sigma_{A, int}, \sigma_{A, ext}, \sigma{B}$. And now let them free! What would happen?

Actually nothing. Charges distributed over B and inner surface were in equilibrium, we bring some charges distributed over outer surface, but these outer charges are distributed such in a way that they do not produce any electrical field in the area of inner surface. And inner charges does not affect charges on outer surface. We still have an equilibrium.

So, we have a solution to our problem! We know now, that there is a solution such that "inner" charges do not produce any electrical field anywhere outside the cavity. (Item 1. of your question). This equilibrium distribution is such that $\sigma_{A, ext}$ produce no field inside the cavity. Please note, that $\sigma_{A, int}$ does produce some electric field inside the cavity, so your question is incorrect.

May be actual distribution would be different from what we have constructed? No, there are no other solutions.

Suppose there is some other distribution. Let's change the sign of all the charges and combine both pictures. We would get a new distribution, still equilibrium one, total charges of all the conductors would be zero, but still distribution is not: there would be some positively and negatively charged areas. That means there is some electric field in our system. Electric field has potential energy. The position can not be in equilibrium because charges can redistribute, eliminate all the electrical field and the potential energy would decrease.

The proof is not extremely rigorous, I know :)

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This is a worthy question to ask, and the answer is not trivial or totally immediate.

It's not clear why an effort is not usually made to address this question properly: I gave two complete arguments for why the combined field due to 'fixed' and induced charges in any one 'domain' of space doesn't penetrate into another domain, where any two domains are completely separated from each other by conducting material (and thus there's exactly one domain that extends to infinity and all other domains have finite volume). It's not a difficult argument, starting from either of the two standard Electrostatic existence and uniqueness theorems.

The most important thing to come to a consensus on, is whether there is a need for logically complete arguments in the first place!


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