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The electric field is characterized by the equations

$$\nabla\cdot \mathbf{E} = \dfrac{\rho}{\epsilon_0}$$

$$\nabla \times \mathbf{E} = 0$$

Or equivalently, $\nabla^2 V = -\rho/\epsilon_0$ and then $\mathbf{E} = -\nabla V$. The boundary conditions which should be used are the discontinuity of the normal component of $\mathbf{E}$ when crossing a charged surface and the continiuty of the tangential component. That is:

$$\mathbf{n}\cdot (\mathbf{E}_2-\mathbf{E}_1)=\sigma/\epsilon_0,$$

$$\mathbf{n}\times(\mathbf{E}_2-\mathbf{E}_1)=0.$$

I'm trying to show mathematically, just with the equations, that the electric field inside a conductor is zero. I've seen many "conceptual" arguments that if there was a field the charges would move and produce a field canceling this one out.

That's fine, but still I wanted to see a more concrete proof of this. I believe that it is a matter of picking $\rho$ correctly and using the correct boundary conditions. In truth I believe it all boils down to: how do we model a conductor? The concept is simple, but I mean, how the equations take form for a conductor and how using them we can show that $\mathbf{E}=0$ inside a conductor?

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  • $\begingroup$ It requires no further proof because it was an implied part of the problem description that the conductor was under static conditions. Otherwise you are not constraining the charge distribution at all. $\endgroup$ – WhatIAm Jul 10 '16 at 16:28
  • $\begingroup$ @Qmechanic are you sure this should have the homework-and-exercises tag? It doesn't look like it to me. $\endgroup$ – David Z Jul 10 '16 at 17:13
  • $\begingroup$ @DavidZ: Nah, you're probably right. I removed it again. One could argue it is conceptional. (But that said, there are probably duplicates.) $\endgroup$ – Qmechanic Jul 10 '16 at 17:19
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The boundary conditions by themselves can't tell you anything about a conductor. The boundary conditions can't even tell which side of the surface has the conductor!

One way to model a conductor is as an Ohmic conductor where there is a constant $\sigma$ (different than the surface charge density listed in your boundary conditions) and then you assert the Ohmic condition:

$$\vec J=\sigma \vec E$$ and then you can take the divergence of both sides and get $$\sigma\frac{\rho}{\epsilon_0}=\sigma\vec \nabla \cdot \vec E = \vec \nabla \cdot \vec J$$

Where we used the Maxwell equation $\dfrac{\rho}{\epsilon_0}=\vec \nabla \cdot \vec E$ and we can also take the divergence of $$\vec \nabla \times \vec B=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial \vec E}{\partial t}$$ to get the continuity equation

$$\vec \nabla \cdot \vec J=-\epsilon_0\vec \nabla \cdot\frac{\partial \vec E}{\partial t}=-\frac{\partial \rho}{\partial t}.$$

This means we have $$\frac{\partial \rho}{\partial t}=-\vec \nabla \cdot \vec J=-\frac{\sigma}{\epsilon_0}\rho.$$

So you might have started with an initial charge density but every place inside the conductor it exponentially decays over time.

And this relates to the initial value formulation of Electrodynamics. You start with an actual physical electromagnetic field at a moment and then it evolves according to $$\frac{\partial \vec B}{\partial t}=-\vec\nabla \times \vec E, \text{ and}$$

$$\frac{\partial \vec E}{\partial t}=\frac{1}{\epsilon_0}\left(\frac{1}{\mu_0}\vec\nabla \times \vec B-\vec J\right)$$

So the fields at a later time are a consequence of the fields at an earlier time (and the current) and the evolution equations above.

So for an Ohmic material we know $\vec J$ so we can evolve the fields because the evolution equations are just Maxwell solved for the time rates of change.

So you had an initial charge distribution and an initial electric field. They might have been zero, they might have been nonzero.

I've seen many "conceptual" arguments that if there was a field the charges would move and produce a field canceling this one out.

If you think of statics as the long time limit of dynamics then you don't have to go conceptual. An Ohmic material literally does have a nonzero current where there is a nonzero electric field. But that current causes the charge to go away, you can imagine places where the charge density is positive and negative initially and the initial electric field streamlines could connect some of those and/or it can connect those places to the surface. And since the current points the same way we can see that this exponentially decreasing charge density is becasue opposite charge densities are canceling each other as charge flows or the charge imbalance is moving to the surface, thus increasing the charge density of the surface over time.

The charge density on the surface can change in a different way than exponentially decreasing over time. Why? Because $\sigma$ (from the Ohmic condition) is not constant across the surface on the boundary of the conductor. In fact, the boundary of the conductor might have a vacuum with $\vec J=\vec 0$ on the other side.

Can we argue that the electric field is zero inside? Yes and no. On the one hand if we assert that part of statics is $\vec J=\vec 0$ then we have $\vec E=\vec 0$ right away. But that's more like assuming. But if you allow $\vec J\neq \vec 0$ then your conductor could have a zero charge density everywhere but have a steady current as long as the boundary of the conductor is supplied with the current it needs for that steady current.

It's totally a valid solution of Maxwell to have a cylindrical infinite wire pointing in the $\hat z$ direction with a uniform nonzero $\vec J$ pointing in the $\hat z$ direction inside the cylindrical infinite wire.

So whenever you have a counterexample you know you need to strengthening your hypothesis. That situation can have a static unchanging electric field, but it has a nonzero current.

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You need to use Ohms law: $J = \sigma E$ which has to be added to Maxwell's equations as a bulk observation, as explained by this answer.

You can then conclude that the electric field is zero in a conductor for:

  • perfect conductor where $\rho = 1/\sigma = 0$ and $J$ is finite
  • static case where $J = 0$ and $\sigma$ is finite
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  • $\begingroup$ What is the situation if a certain part of conductivity electrons is "stripped" off a bulk of iron? $\endgroup$ – Vladimir Kalitvianski Jul 10 '16 at 18:00
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In the electrostatic case, according to Poisson's equations, the electric field equation for an empty cavity space $\mathcal V$ with no electric charges $\rho (\vec r) = 0$ and electostatic potential $\Phi (\vec r)$ at the position $\vec r$ is: \begin{equation} \nabla^2 \Phi = \frac{\rho}{\epsilon} = 0. \label{poisson} \tag{1} \end{equation} If we take the integral of the square of the electric field over the volume of the cavity: \begin{equation} I = \int_{\mathcal V} d V \; | \nabla \Phi |^2 = \int_{\mathcal V} d V \; \left[ \vec \nabla . \left( \Phi \vec \nabla \Phi \right) - \Phi \nabla^2 \Phi \right]. \label{efint} \tag{2} \end{equation} According to (\ref{poisson}), the 2nd term in (\ref{efint}) can vanish, so we can write: $$\int_{\mathcal V} d V \; | \nabla \Phi |^2 = \int_{\mathcal V} d V \; \vec \nabla . \left( \Phi \vec \nabla \Phi \right).$$ Now, using Gauss' Divergence Theorem, this volume integral can be re-written into a surface boundary integral: \begin{equation} \int_{\mathcal V} d V \; | \nabla \Phi |^2 = \int_{\mathcal S} d \vec S . \left( \Phi \vec \nabla \Phi \right). \label{sint} \tag{3} \end{equation} Since we are talking about a cavity in a conductor, the electrostatic potential everywhere in the conducting material, including the boundary walls of the cavity is uniformly constant, ie. $\Phi (\vec r) = \Phi_s \ \forall \ \vec r \in \mathcal S = \partial \mathcal V$. Thus, the surface integral of the cavity can be re-written as: \begin{equation} \int_{\mathcal S} d \vec S . \left( \Phi \vec \nabla \Phi \right) = \Phi_s \int_{\mathcal S} d \vec S . \left( \vec \nabla \Phi \right) = \Phi_s \int_{\mathcal V} d V \; \nabla^2 \Phi. \label{nsint} \tag{4} \end{equation} Thus, applying Poisson's equation (\ref{poisson}) again, we have $$\int_{\mathcal V} d V \; | \nabla \Phi |^2 = \Phi_s \int_{\mathcal V} d V \; \nabla^2 \Phi = 0,$$ and since $| \nabla \Phi |^2 \geq 0$, we can say that the only way for the integral $I$ to vanish is \begin{equation} \boxed{\vec E = \vec \nabla \Phi (\vec r) = 0.} \label{nullef} \tag{5} \end{equation}

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Ohm's law is \begin{equation}J=\sigma E \end{equation}

Substituting Ohm's law in $ \nabla . E=\frac{\rho}{\epsilon} $ gives $$ \nabla .J=\frac{\sigma \rho}{\epsilon} $$

The continuity equation (charge conservation) is $$\nabla . J=- \frac{\partial\rho}{\partial t}$$

Equating the above two equations gives

$$\frac{\partial\rho}{\partial t}=-\frac{\sigma \rho}{\epsilon}$$ The solution to this equation is simply

$$\rho = \rho (0)\exp(-\frac{\sigma t }{\epsilon})$$ where $\rho (0)$ is the charge density at time, t=0.

We have derived the equation which describes how the charge density inside a material varies with time. For a conductor, the conductivity, $\sigma$ is large and therefore, $\rho$ will exponentially (and quickly) tend to zero with time, INSIDE the conductor. Therefore, inside a conductor, we can write Maxwell's first equation as:

$$\nabla . E =\frac{\rho}{\epsilon}=0$$

Gauss' law tells us that $$\int E.dS=\int (\nabla . E) dV=0 (from above)$$ [see https://en.wikipedia.org/wiki/Gauss%27s_law]

Since Gauss law should hold for any closed surface inside the conductor, we conclude that E should be identically zero inside a conductor.

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    $\begingroup$ Your last statement does follow from the rest. For instance if you have vacuum between two plates of a charged parallel plate capacitor. Then any closed surface between the plates has zero total net electric flux, but the electric field between the plates is not zero. $\endgroup$ – Timaeus Jul 12 '16 at 19:46
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I'd personally use the principle of least action: the point at which the charge density remains static will be the point at which field has a minimum energy. We can work out the energy of an electric field with the following formula:

$$E=\frac{1}{2}\int \varepsilon|\mathbf{E}|^2dV$$

If we assume a homogeneous conductor, then we can say:

$$E=\frac{\varepsilon}{2}\int|\mathbf{E}|^2dV$$

Now we get to my favourite way of looking at this. Say we have two numbers: $n$ and $m$. We want to minimise the sum of the squares of these two, when $m+n=o$. Now, let's first hypothesise that our solution will come when the two are equal:

$$n=m=\frac{o}{2}$$

And thus we can say that:

$$m^2+n^2=2\frac{o^2}{4}=\frac{o^2}{2}$$

My hypothesis is that by adding an arbitrary number to one value, and subtracting it from the other, the sum of their squares will then increase. And so we are looking for:

$$\Delta = \left(m-\delta\right)^2+\left(n+\delta\right)^2-\left(m^2+n^2\right)$$

Expanding out the brackets, and equating $m$ and $n$, it can be shown taht the change is always positive.

This can then be made analogous to the formula for the electric field. If you reduce the contribution to the electric field of a single region by reducing its charge, you then need to add that charge on elsewhere, which generates a larger effect. (Sorry, this is done from memory, from a proof I did ages ago, so I apologise if I've mucked this up at all!)

Basically, the minimum energy for the medium is the energy where the electric field distribution is even.

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