In electrostatics, why the electric field inside a conductor is zero?

Question

In electromagnetism books, such as Griffiths or the like, when they talk about the properties of conductors in case of electrostatics they say that the electric field inside a conductor is zero.

I have 2 questions:

1. We know that conductors (metallic) have free electrons which randomly moves in all directions, so how come we can talk about electrostatics which by definition means stationary charges?

2. When the textbooks try to show why the electric field inside a conductor is zero they say let us put our conductor in an electric field. What happens then is that there will be an induced surface charge density which consequently induces an electric field within the conductor such that the total electric field within the conductor will be zero. That is perfectly understood, but my problem is the following: the original claim was that the electric field within a conductor is 0, not the electric field after putting the conductor in an external electric field it became zero. I do not understand the logic!

Answer 1

Conductors are defined by the freedom of some of the charges inside to move with little resistance.

So, if there were a non-zero field, what would happen? Answer: some of the free charges move until the field is again zero.

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You might be wondering if there are limits to this claim, but a introductory book of that sort is not worrying about extreme situations. In any case, try choosing a simple geometry, make an estimate of the fraction of charges that are free to move and calculate the saturation field.

Answer 2

Gauss's law states that the electric field flux through a closed surface is equal to the quotient of the load inside the surface divided by $ \epsilon_0$.

A driver is characterized by the charge carriers can move freely within it. If the charges in a conductor in equilibrium at rest, the electric field intensity in all interior points of the same must be zero, otherwise, would move the loads caused an electric current.

Within a conductor arbitrarily draw a closed surface $S$, and it follows that:

The electric field is zero, $E = 0$ on all points of said surface.

The flow through the closed surface $S$ is zero.

The net charge q on the inside of said surface is zero. As the closed surface S we can make it as small as we conclude that at any point P inside a conductor there is no excess burden, so this should be placed on the surface of the conductor. this should answer your question.

Answer 3

In electrostatics free charges in a good conductor reside only on the surface.

There are at least two ways to understand this

• So the free charge inside the conductor is zero. So the field in it is caused by charges on the surface. Since charges are of the same nature and distribution is UNIFORM, the electric fields cancel each other.

• Consider a Gaussian surface inside the conductor. Charge enclosed by it is zero (charge resides only on surface). Therefore electric flux =0 Furthermore, electric flux = electric field * area. Since area cannot be zero, electric field is zero

Answer 4

Since I'm not satisfied with the answers and it seems that people still stumble upon this question googling, I'll try to answer it.

First we need to understand what are some basic assumptions of the classical electrodynamics. As every other field in science it uses models to describe the nature. Some well known models are point mass, point charge, continuum etc.

Electrodynamics uses charge continuum and point charge models to describe charges in the real world. Charge continuum is given by one main quantity and that is charge density.

Charge density in a point $A$ is defined using averaging of all charges in a small volume of space $\Delta V$ around the point $A$. Also we average the charge density over some small time interval $\Delta t$. Now I will not go into details of what $\Delta V$ and $\Delta t$ actually are, but you can read about physically infinitesimal volumes and time intervals. The point is that $\rho(A)$ is not the "exact" charge density at that point, but rather the averaged value.

Now back to the question:

We know that conductors (metallic) have free electrons which randomly moves in all directions, so how come we can talk about electrostatics which by definition means stationary charges?

Yes, they do randomly move in all directions and that is the point. When you average out over small space and time intervals (given that electrons usually don't cross a long distance and don't have a great velocity) - you will get zero charge density. The key is the randomness of thermal motion which averages to zero. In jargon you would say that classical electrodynamics doesn't see the quantum and thermal effects because of its zoomed out scale.

That is perfectly understood, but my problem is the following: the original claim was that the electric field within a conductor is 0, not the electric field after putting the conductor in an external electric field it became zero. I do not understand the logic!

This second question is essentially already answered above. The authors usually assume trivial the question about field inside the conductor with external field $E_{ext}=0$, so they jump right away to $E_{ext}\not=0$

Answer 5

Question 1: There are two space scales at play: Microscopic scale: One considers the electrons individually. Their motion and the electromagnetic field they generate widely varies in both space and time. Macroscopic scale: Both the motion of individual electrons and the electromagnetic fields are not measurable with standard laboratories apparatus. At our scale one can only observe space time average. In plasma kinetic theory, one derives a method to calculate these average and how they vary in both space and time.

Electrostatics is only concerned with macroscopic fields. It does not exclude microscopic electron motion but assume the average motion to be null.

The proof for your second question is not difficult. First let's prove that any free charge diffuse towards the surface in a short time.

Combining the charge conservation, Ohm's law and Maxwell's second equation, one gets:

$$\begin{cases} \frac{\partial \rho }{\partial t} + \overrightarrow{ \nabla }. \overrightarrow{j} =0 \\\overrightarrow{j}= \sigma \overrightarrow{E} \\\overrightarrow{ \nabla }.\overrightarrow{E} = \frac{ \rho }{ \varepsilon _{0}} \end{cases} ~~\Rightarrow ~~ \frac{\partial \rho }{\partial t}+\frac{ \sigma \rho }{ \varepsilon _{0}}=0~~ \Rightarrow ~~\rho(t)=\rho(0)e^{-\frac{ \sigma }{ \varepsilon _{0}}t }$$

Wikipedia gives for copper:$$\sigma=16.8×10^{-9}~~Ω.m~~at~~20~~°C.$$ $$\varepsilon _{0}= 8.85×10^{-12}~F⋅m^{-1}$$

So: $\frac{ \sigma }{ \varepsilon _{0}} \approx 1900$

The time $\triangle t$ for 99% of $ \rho _{0}$ to diffuse to the surface is: $$ \triangle t =- \frac{ln(0.01)}{1900} \approx 2.10^{-3} s$$

So for any physics problem involving time scale greater than the milli-second, one can consider there is no volume charges in conductors.

In the second step, apply Gauss's law to any volume inside the conductor: $$ \int_ \Sigma \overrightarrow{E}. \overrightarrow{d \Sigma } = \frac{Q_{en}}{ \varepsilon _{0}} =0 $$ Since it is true for any $\Sigma$, one must have: $\overrightarrow{E}=\overrightarrow{0}$

Answer 6

In electrostatics we only look onto static charges, which is why the statement of zero electric field beholds, as if it was not zero then it would cause charges to move. which is the concern of current electricity

[as sir feynman said it to be "the only electrostatic solution"] https://www.feynmanlectures.caltech.edu/II_05.html#Ch5-S9

Answer 7

I will attempt a hopefully more rigorous derivation than is typically given. I assume a continuous charge distribution within the conductor.

First, Gauss's law states: $$\nabla \cdot E = \rho/\epsilon_0$$

For a system in static equilibrium the net force, and therefore the electric field, acting on any charge must be zero. If there is a region of non-zero density, $\rho(\vec x)$ then, due to the Gauss's law, the gradients of the electric field must not all be zero. Correspondingly, the electric field will be non-zero in a small-enough ball around $\vec x$. Due to the continuity of the charge distribution, $\rho$ will also be non-zero in a small-enough ball around $\vec x$. Therefore any continuous charge distribution (in 3d) cannot be in static equilibrium. This implies that if any charge is present it must be present at the free surface.

To derive that the electric field is zero within the conductor we cannot invoke Gauss's law for an arbitrary surface ($\nabla \cdot E = 0$ for a point charge and therefore a lack of local charge density says nothing about the local field).

Instead, one can note that the electric field tangent to the surface of the conductor must be zero at equilibrium. Since $E=-\nabla V=0$, the electric potential on the surface of the conductor will be a constant. Then one can note that Laplace's equation ($\nabla^2V=0)$ must hold within the conductor with a constant potential boundary condition. A constant potential within the conductor satifies both Laplace's equation and the boundary conditions (since solutions of Laplace's equation with Dirichlet boundary condition are unique this is the solution).

Lastly, since $E=-\nabla V$, we have $E=0$ within the conductor.