It’s been said that the electric field of a conductor is always perpendicular to its surface. However, I’m having difficulty understanding this. There are only a limited number of excess charges on its surface, and they will be scattered around to minimize repulsions. Electric field lines are drawn from the excess charges, and are not drawn from the parts of the metal where excess charges aren’t present. I know that an electric field exists even where there are no field lines, but if you do place a test charge in between field lines, the electric field would point… toward the part of the metal that has no excess charge? My comprehension is that the excess charges don’t have any field other than that which points perpendicular to the surface. How then can there be a field that isn’t directly across from an excess charge?


First of all field lines are very "crude" way of visualizing electromagnetic phenomena. They are very useful qualitative idea but one shouldn't give them too much importance specially quantitatively.

Yes, excess charges are responsible for "effective electric fields". If you take some "uniform surface" (plane sheet, sphere etc) they will spread around the whole surface uniformly. Thus field lines will be perpendicular to the surface and uniformly dense. Here by "uniform surface" I mean a surface with same curvature everywhere. But if the surface has different curvature at different points (ellipsoid, squashed sphere, etc) then the excess charges will accumulate more around the high curvature region resulting more dense field lines there.

Edit : These are all classical arguments and therefore one shouldn't worry about atomic length scales. As far as classical physics is concerned charging of a conductor involves huge amount of elementary charges and therefore one should think them as a continuum of charges (like a water drop) which spreads around the conductor. It is in the same spirit that we do not need to consider a water drop as collection of tiny atoms and most of it is void. We cannot "see" the void because we are not looking at it in that length scale (viz, the atomic length scale).

  • $\begingroup$ But what if a test charge is directly across from a point on the surface of the conductor where the excess charge does not reside? Why is there still a field? $\endgroup$ – lightweaver Oct 31 '15 at 13:03
  • 1
    $\begingroup$ I thing these are all classical arguments. Charging a conductor involves a huge number of elementary charges. So for all practical purposes you can think of them as a continuum like adding a drop of liquid over the hole surface. You know that liquids are also made up of tiny atoms but for all practical purposes in our length scale we think of them as some continuum (without any "holes" or "gaps".) $\endgroup$ – Physics Moron Oct 31 '15 at 13:09

You can imagine replacing the perfect conductor with an ohmic material with $\vec E =\sigma \vec J,$ with a really huge conductivity, $\sigma$.

While not accurate this kind of an object rapidly approaches the state that a perfect conductor has. And when you use the full time dependent Maxwell Equation you find that any non zero charge density on the inside decreases exponentially in time (with a time factor inversely proportional to $\sigma$) and that current flows from surface charge to surface charge with fields lines always going through the body of the conductor.

Sometimes starting and stopping on the interior region of nonzero charge density and sometimes connecting places on the surface that will develop a surface charge over time.

To be more clear, the electric fields lines inside get smaller and smaller as the charge densities inside get exponentially smaller. And the field line components on the surface pointing normal approach a steady limiting value while any other part gets smaller over time.

To see this take a charge on the inside, $\vec \nabla \cdot \vec E=\rho/\epsilon_0$ and take it's time derivative to and get:

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

Thus $\mu_0\frac{\partial \rho}{\partial t}=-\mu_0\vec \nabla \cdot \vec J=-\mu_0\vec \nabla \cdot \left(\sigma\vec E\right).$

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

So $\rho(\vec r,t)=\rho(\vec r,0)e^{-\sigma t/\epsilon_0}.$ So if you imagine the fields due to to the charges originally inside they just decay over time. But the fields of charges on the surface have a $\vec J$ that isn't proportional to $\vec E$ anymore. The part of $\vec E$ parallel to the surface can have a current proportional to it and can decay away but any part orthogonal to the surface fails to have a current since current is not free to floe outside of the conductor.

So you can imagine two electric fields. One that is inside the conductor or parallel to the surface and it can make a current and over time that current makes the charge get less and the charge makes there be less current.

And you can a imagine a second charge distribution, one entirely on the surface and that makes an electric field entirely orthogonal to the surface on the surface and entirely zero inside. And any charge like that can grow in time and get make more of itself until it has no more charge from elsewhere to steal.

Since it itself has no electric fields inside or along the surface it can't more charge around at all. It is at the mercy of the other charge the one that just just get exponentially weaker in time, and it is at the mercy of those fields to have current to supply the charge.

So over time such a material, one with $\vec E=\sigma\vec J$ starts to change the charge distribution into one where it is all at the surface and has all the fields point orthogonal to the surface. Which is what it approaches.

How then can there be a field that isn’t directly across from an excess charge?

To be clear. The field on the inside just on one side of the surface is the field from the charge right there next to it, plus the field from all he other surface charge across the whole conductor. And on the inside the two cancel out 100% perfectly.


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