Why is the field within a solid conducting material zero?

[madafo3435]

Homework Statement

I have seen that in an introductory book on electricity and magnetism, they make the observation that in a solid conductive material, charges are redistributed by being able to move freely in the solid due to its conductivity, and they move until they reach an electrostatic configuration. , that is, after some time they stop. All of this is understandable to me. My problem is that, then the book mentions that the field inside the solid will be 0. This last one is not clear to me, because, if I had a conductive solid wait, and I put two equal positive charges in it, they will repel and they will reach an electrostatic configuration at some point, but the field in the entire sphere will not be 0. I will place the fragment of the book where it justifies my concern:

"In the interior of such a conductor, in the static case, we can state confidently that the electric field must be zero. If it weren’t, charges would have to move."

I do not understand this justification. Could someone clarify this for me please?

Relevant Equations

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[jbriggs444]

Normally when we consider a solid conductor we are ignoring the fact that charge is quantized. Instead, we treat it as a continuous fluid that permeates the conductor. A conducting sphere with only two charges on it does not fit this model. A conducting sphere with 6.02 x 10²³ charges on it fits the model tolerably well.

 

[etotheipi]

@jbriggs444 is bang on the money here (as always!). A common model of electrical conduction is the Drude Model, where the charge carriers are essentially treated using the kinetic theory of gases. Under the assumptions of the Drude model you can derive the equation $\vec{j} = \sigma \vec{E}$, where $\vec{j}$ is the current density [try to derive it!]. With that in place, let's first consider Maxwell IV,$$\begin{align*} \nabla \times \vec{B} &= \mu_0 \vec{j} + \mu_0 \varepsilon_0 \partial_t \vec{E} \\ \\

\nabla \cdot (\nabla \times \vec{B}) &= \nabla \cdot (\mu_0 \vec{j} + \mu_0 \varepsilon_0 \partial_t \vec{E})\end{align*}$$Now since $\nabla \cdot (\nabla \times \vec{v}) = 0$ for any vector field $\vec{v}$ , we have that$$\nabla \cdot \vec{j} = - \nabla \cdot (\varepsilon_0 \partial_t \vec{E})$$Now since $\nabla$ and $\partial_t$ are both linear differential operators, Schwarz' theorem allows us to interchange them, so$$\nabla \cdot \vec{j} = -\varepsilon_0 \partial_t (\nabla \cdot \vec{E}) = - \partial_t \rho$$where we've used that $\nabla \cdot \vec{E} = \rho/\varepsilon_0$ (Maxwell I). Also, since $\vec{j} = \sigma \vec{E}$ from the Drude model, $\nabla \cdot \vec{j} = \frac{\sigma \rho}{\varepsilon_0}$, and we have$$\partial _t \rho = - \frac{\sigma}{\varepsilon_0} \rho$$You might know the solution to this differential equation... it's the exponential decay equation! Specifically,$$\rho = \rho_0 e^{-\frac{\sigma}{\varepsilon_0} t}$$Interpreted physically, this means that if we start with some non-zero charge density $\rho$ in the bulk, it will decay toward zero and soon become negligible! This quantifies what your book is trying to say.

But of course, exactly like @jbriggs444 said, if you only have 2 charged particles, then $\vec{j} = \sigma \vec{E}$ will not apply, and by extension nor will this result!

 

[PeterDonis]

if I had a conductive solid wait, and I put two equal positive charges in it, they will repel and they will reach an electrostatic configuration at some point, but the field in the entire sphere will not be 0.

While @jbriggs444 is correct that the model being used by your textbook doesn't really apply to just two charges, your scenario doesn't just have two charges. The conductive solid itself contains charges that can move; that's why it's a conductor. So if you put two equal positive charges on the surface of the solid, the charges inside the solid will move so that the fields of those two positive charges are canceled inside the conducting solid. As long as there is any nonzero field inside the conducting solid, that field will provide an electromotive force that will cause charges inside the conducting solid to move, as your textbook says. So the only possible static configuration inside a conducting solid is that the field is zero.

 

[madafo3435]

thanks @jbriggs444, @etotheipi and @PeterDonis. Your considerations have helped me to understand my problem much better. But I still have problems seeing because the field inside the solid is going to be 0. I understand that initially the charge is distributed in such a way that some time later they produce an electrostatic configuration. What I can't see is because after this configuration is reached, in some region of the solid where there is no charge, the field has to be 0. I know that the interactions between the charges are going to be cancerous, but why Will they also cancel at any solid point? Doesn't this depend on the symmetry of the solid? @PeterDonis talks about an electromotive force, but I don't know about this interaction.

 

[etotheipi]

What I can't see is because after this configuration is reached, in some region of the solid where there is no charge, the field has to be 0

This is also a (very!) good question. We deduced that the charge density is zero inside the bulk of the conductor, but how do we deduce from this that the electric field is zero inside the conductor? After all, $\rho = 0$ definitely does not necessarily imply $\vec{E} = \vec{0}$...

I'm afraid it's also a bit involved. I'm going to need to take a little while to figure this out again, but you'll need to start with the Poisson equation$$\Delta \phi = \frac{\rho}{\varepsilon_0}$$We already deduced that $\rho = 0$ inside the bulk in the steady state, so this reduces to Laplace's equation, $\Delta \phi = 0$. We need some boundary conditions. The electric field at the surface of the conductor has no tangential component, so $\phi$ is constant, $\phi = k$, along the boundary. Now we need Green's first identity. Let's define $\phi' := \phi - k$ (so that $\phi' = 0$ on the surface of the conductor), and then since $\Delta \phi = \Delta \phi' = 0$, Green's first identity reduces to $$\int_{\Omega} (\nabla \phi') \cdot (\nabla \phi') dV = \int_{\partial \Omega} \vec{n} \cdot (\phi' \nabla \phi') dS$$The RHS is zero since $\phi' = 0$ everywhere on $\partial \Omega$, and thus we must have $\nabla \phi' = \nabla \phi = 0$, because $\nabla \phi' \cdot \nabla \phi' \geq 0$. Since $\vec{E} = -\nabla \phi$, we deduce that $\vec{E} = \vec{0}$ everywhere inside $\Omega$.

 

[PeroK]

What I can't see is because after this configuration is reached, in some region of the solid where there is no charge, the field has to be 0.

If the electric field is zero inside the conductor, then the conductor must be an equipotential. Any chargeless cavity in the conductor, therefore, is surrounded by an equipotential. Clearly one solution to Laplace's equation (for this boundary condition) is an equipotential within the cavity as well. And, with sufficient boundary conditions, Laplace's equation has a unique solution. Hence that is the only solution and any chargeless cavity within the conductor also has no electric field.

 

[etotheipi]

If the electric field is zero inside the conductor, then the conductor must be an equipotential. Any chargeless cavity in the conductor, therefore, is surrounded by an equipotential. Clearly one solution to Laplace's equation (for this boundary condition) is an equipotential within the cavity as well. And, with sufficient boundary conditions, Laplace's equation has a unique solution. Hence that is the only solution and any chargeless cavity within the conductor also has no electric field.

Wow, nice insight! Uniqueness theorem does make this very easy!

 

[PeterDonis]

in some region of the solid where there is no charge

If the solid is a conductor, there are charges that can move everywhere inside the solid. That's the definition of a conductor. If there is some region of the solid where there are no charges that can move, that region is not a conductor.

 

[madafo3435]

I am satisfied with your considerations . thanks for your help.

PeroK PeterDonis @etotheipi