I am trying to understand the idea (or the fact) that most books introduce which is about the electric field inside a charged solid conductor.

Books tell that the field has to be zero everywhere inside solid conductor, otherwise charges will move around. Using this idea and Gauss's law, the charges inside the solid conductor is zero.

Now let us take for example four extra positive charges (each =1.2x10^-10 coulomb) inside a solid conductor of radius 1.

According to the idea of charge at the surface and due to the symmetry, the charges will distribute as follows:

Enter image description here

I have plotted the electric potential (V=Q/(4πε0r)) and electric field (E=-∇V) using principle of superposition and the plot is:

Enter image description here

Clearly the electric potential inside the conductor is not constant and the electric field is not zero.

How can this issue be explained?


2 Answers 2


You have ignored the mobile charges in the conductor. In your plot the field lines are not perpendicular to the surface, particularly near the charges. That will cause the conduction electrons to move. The positive charges will attract electrons until the field inside the conductor is zero. This means that the whole conductor, including the inner surface, is an equipotential. After that, Gauss' law says the field inside is zero.

You figure is a fine one if the four charges are in empty space.

  • $\begingroup$ Yes that is correct, E is not perpendicular to the surface. So if want to plot E that represent the case where E=0 inside the conductor, what should I do? $\endgroup$ Commented May 18, 2015 at 3:37
  • 2
    $\begingroup$ If the charges are embedded in a conductor, the charges don't matter any more. The conductor is an equipotential with that Q. If it is a spherical shell, the field inside is zero and the field outside is the same as if it were a point charge. If it has a more complicated shape, the field inside is still zero, but you need to solve Poisson's equation outside. Far away, the field will be similar to a point charge. The shape will introduce dipole, quadrupole, etc. moments, but those will die away with higher factors of $r$ than the square. $\endgroup$ Commented May 18, 2015 at 3:51
  • $\begingroup$ @RossMillikan: It ignores any quantum effects though, doesn't it? If there are just four elementary charges, they shouldn't be able to redistribute themselves (but at that scale treating the conductor as “solid” is no longer substantiated either). $\endgroup$
    – Jan Hudec
    Commented May 18, 2015 at 7:05
  • $\begingroup$ @JanHudec: This whole calculation ignores both quantum effects and the fact that electrons are indivisible. When we call something a "perfect conductor" we do that. Charges have to be able to move to eliminate the fields inside the conductor. We also don't worry about the thickness of the screening layer. These are all simplifying assumptions that are close enough for many practical purposes. $\endgroup$ Commented May 18, 2015 at 14:05
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    $\begingroup$ @Algohi: you are approaching this question at the proper level. I would suggest ignoring quantum mechanics and the unit electron charge for this purpose. The point is that the charges in the conductor will distribute themselves so that the internal field is zero. You will get a minus charge next to each positive charge so there is no field going into the conductor. Your are correct that the conductor is neutral, so the positive charge must appear somewhere. It will be spread around the exterior surface in such a way that the field lines hit the conductor perpendicularly. $\endgroup$ Commented May 19, 2015 at 0:37

We assume that the the electric field is uniform for a charged solid sphere. It follows that the electric charge of the sphere is equal to

$$ Q = \rho V$$

Where $\rho$ is the charge density and $V$ is the volume. Therefore,

$$ Q = \rho V = \frac{4}{3}\rho R^3$$

We create a Gaussian surface in the form of a sphere of radius $r <R$. Thus, using Gauss's Law,

$$ E(4\pi r^2) = \frac{Q}{\epsilon_0} \implies E = \frac{1}{\epsilon_0}\frac{4\pi}{3}R^3\rho \implies E(r) = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^3}r$$

From the center of the sphere, the electric field is $0$. As you move away from the center of the sphere to $R$, the electric field increases in linear fashion that is proportional to $r$.

  • $\begingroup$ why electric field is 0 inside the conductor? the equation of electric field of charges (using principle of superposition) results in non-zero electric field inside the sphere. E =0 only at the center of the sphere $\endgroup$ Commented May 18, 2015 at 2:50
  • $\begingroup$ @Algohi Yes, at the center $E = 0$ as I said. However, the electric field increases linearly in proportion to $r$ as you move away from the center. $\endgroup$
    – user43617
    Commented May 18, 2015 at 2:52
  • $\begingroup$ Ok. Now the question is why books say E=0 everywhere inside the conductor? $\endgroup$ Commented May 18, 2015 at 3:01
  • $\begingroup$ @algohi Perhaps you are confused about a solid sphere vs a hollow sphere. See this link, physics.stackexchange.com/questions/44233/… $\endgroup$
    – user43617
    Commented May 18, 2015 at 3:10

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