Electrostatics in Conductors: Surface Charge and Potential

This time we’ll study electrostatics in conductors. Nothing to it! But first—what is a conductor???

In my original 2014 notes, I gave this bit of general background: “If the energy gap between the CB (conduction band) and VB (valence band) is huuuge, we call it an insulator; if there is almost no gap, a conductor; and if it’s somewhere in between, a semiconductor.”

Scientific clarification — 1 October 2026. This is a rough band-theory sketch, not a general classification rule. In the usual band description, metals have a partially filled band or overlapping bands that leave available states near the Fermi level. Semiconductors and band insulators have a gap, but conductivity also depends on carrier density and mobility, temperature, and doping; no single gap-size cutoff classifies every material. MIT’s semiconductor lecture discusses band filling and ways to create mobile carriers.

A conductor contains mobile charge carriers. In a metal, conduction electrons can redistribute when an electric field acts on them. That freedom makes electrostatic problems much simpler: first find the charge distribution consistent with equilibrium, then use it to calculate the field and potential.

In the original notes I skipped the detailed band-structure discussion here, since it wasn’t needed for this part on conductor electrostatics, hehe.

The original notes then say: “So~ think of an ideal conductor here—that is, a conductor with infinite electrical conductivity~.”

Scientific clarification — 1 October 2026. Infinite conductivity is a classical modeling limit; ordinary good conductors have finite conductivity. Here the charges have settled into electrostatic equilibrium, with no sustained current or imposed electromotive force, so the field in the metal vanishes. Infinite conductivity alone does not imply the superconducting Meissner effect. See Ideal Conductors and MIT’s superconductivity lab guide.

Here we use the classical, macroscopic model of an ideal conductor in electrostatic equilibrium, after charge redistribution has finished. There is no sustained current or imposed electromotive force. A real good conductor approximates this behavior after its initial transients have died away.

Why the field vanishes in the metal

If a nonzero electric field remained in the conducting material, it would drive its mobile charges. The distribution would still be changing. Equilibrium therefore requires

$$ \mathbf E=\mathbf 0 \qquad\text{in the metal}. $$

Imagine giving a solid, isolated conductor a net charge $q$. The first sketch represents an initial charge placement; a localized excess charge in the bulk cannot remain there at equilibrium. The red plus signs illustrate the case $q>0$. In a metal, positive excess charge means an electron deficit, not positive lattice ions migrating outward.

The mobile electrons rearrange until the field in the metal vanishes. In the continuum model, the excess charge then resides on the external surface. Its surface density need not be uniform for an arbitrary shape or in an external field.

The marked point $r$ in the next sketch is in the conducting material, so the equilibrium field there is zero.

Gauss’s law now tells us something about the charge in the bulk. Choose a closed Gaussian surface lying wholly in the metal, as sketched in blue.

With $d\mathbf A$ directed outward, the electric flux is

$$ \oint_S\mathbf E\cdot d\mathbf A =\frac{Q_{\mathrm{enc}}}{\epsilon_0}=0. $$

Thus every small volume wholly within the metal has zero net macroscopic charge. Equivalently, the bulk charge density is $\rho=0$. Electrons and positive lattice ions are still present; their charges cancel in this macroscopic description.

In a way, this feels even easier than the earlier case with charge spread continuously through a region.

The order of the argument matters. Equilibrium gives a zero field, and Gauss’s law then gives zero net bulk charge. Zero enclosed charge alone would give only zero net flux. For example, a uniform nonzero field also has zero flux through a closed sphere.

A sphere of radius $k$ has area $4\pi k^2$. When spherical symmetry makes the field radial and its radial component constant on that sphere, the flux simplifies to

$$ \oint_S\mathbf E\cdot d\mathbf A =4\pi k^2 E_r(k). $$

That simplification needs the stated symmetry; merely choosing a spherical Gaussian surface does not establish it. The equilibrium and bulk-charge arguments are also explained in Richard Fitzpatrick’s Ideal Conductors.

Induction and cavities

A nearby charge can redistribute the electrons in an initially neutral conductor, producing regions of positive and negative surface charge. This is electrostatic induction. An isolated conductor remains neutral overall unless charge is supplied or removed. Connecting it to ground would permit charge exchange and change the problem.

Zero field in the metal does not mean zero field everywhere within its outer boundary. A charge in a cavity can produce a field in the cavity and induce charge on its internal wall, while the surrounding metal still has zero field. A completely enclosed, charge-free cavity has zero electrostatic field; the gap surrounding a charged inner conductor in the example below is a different situation. OpenStax discusses these distinctions in Conductors in Electrostatic Equilibrium.

A sphere inside a neutral conducting shell

Keep that property in mind, and the problems should be easier than before!

Let’s work through a problem. A solid conducting sphere of radius $R$ carries a fixed signed net charge $q$. A concentric conducting shell occupies $a\lt r\lt b$, with

$$ 0\lt R\lt a\lt b. $$

The shell is initially neutral, isolated, and ungrounded. The gap and exterior are vacuum, there is no external field, and we choose $V(\infty)=0$. The coordinate $r$ now denotes distance from the common center. We want the potential at that center.

The next four sketches build up the charge distribution in stages. They are bookkeeping diagrams, not four separate equilibrium states; only the last shows the complete final distribution. The central charge label in the first sketch indicates the inner sphere’s net charge, not a deposit that remains at its center.

At equilibrium, the inner sphere’s charge lies on its surface at $r=R$. The sketches show $q>0$; reversing the sign of $q$ reverses the induced signs and electric-field direction.

That charge spreads right out~~~

Let $q_a$ and $q_b$ denote the charges on the shell’s inner and outer surfaces. A Gaussian sphere with $a\lt r\lt b$ lies in metal, where the field is zero. It encloses the inner sphere and the shell’s inner surface, so Gauss’s law gives

$$ q+q_a=0, \qquad q_a=-q. $$

Like this!!

The isolated shell has zero net charge. Charge conservation therefore supplies the second constraint:

$$ q_a+q_b=0, \qquad q_b=q. $$

Then onto the outer surface of the shell—spread right out, like this~~~

Concentric spherical symmetry and the absence of an external field make the charge density uniform on each surface. Dividing each charge by its corresponding area gives the three densities shown here.

Thus $\sigma_R=q/(4\pi R^2)$, $\sigma_a=-q/(4\pi a^2)$, and $\sigma_b=q/(4\pi b^2)$. These are local densities here because of symmetry; charge divided by total area gives only an average in a general geometry.

Oh~~ it looks as though the electric field differs between the regions. Shall we see?

The electric field in four regions

Spherical symmetry lets us write the field away from the origin as

$$ \mathbf E(r)=E_r(r)\,\hat{\mathbf r}, $$

where $\hat{\mathbf r}$ points outward. The signed radial component is $E_r$: it is positive for an outward field and negative for an inward one. The scalar $E$ in the two retained equation images below has this radial meaning.

(Luckily, so many of these problems are symmetric… it’s almost as if the problem itself is saying, “Use Gauss’s law^^”.)

Outside the shell, a Gaussian sphere encloses the total charge $q+q_a+q_b=q$. Its flux is $4\pi r^2E_r(r)$, giving

$$ E_r(r)=\frac{q}{4\pi\epsilon_0 r^2}, \qquad r>b. $$

Within the shell’s metal, equilibrium gives

$$ E_r(r)=0, \qquad a\lt r\lt b. $$

In the vacuum gap, a Gaussian sphere encloses only the inner sphere’s charge. Hence

$$ E_r(r)=\frac{q}{4\pi\epsilon_0 r^2}, \qquad R\lt r\lt a. $$

Finally, inside the solid inner conductor,

$$ E_r(r)=0, \qquad 0\lt r\lt R. $$

At the origin itself, the field is the zero vector; no radial direction needs to be assigned there. We use strict inequalities at the charged surfaces because the electric field has distinct one-sided limits. The potential remains continuous across each surface.

Integrating the potential to the center

Let’s pull the path all the way from infinity to the origin—in other words, hit it with an integral!

Starting from the reference $V(\infty)=0$, integrate inward along a radial path. Use $s$ as the integration variable. The path crosses the exterior, the shell’s metal, the vacuum gap, and the solid inner conductor:

$$ \begin{aligned} V(0)={}&-\int_{\infty}^{b}E_r(s)\,ds\\ &-\int_b^a 0\,ds\\ &-\int_a^R E_r(s)\,ds\\ &-\int_R^0 0\,ds. \end{aligned} $$

The two metal intervals contribute zero potential difference. Substituting the field in the other two intervals and keeping the leading minus sign gives

$$ \begin{aligned} V(0)&=-\frac{q}{4\pi\epsilon_0}\\ &\quad\times\left\{ \left[-\frac1s\right]_{\infty}^{b} +\left[-\frac1s\right]_{a}^{R} \right\}. \end{aligned} $$

The first bracket is $-1/b$ and the second is $-1/R+1/a$. Therefore

$$ V(0)=\frac{q}{4\pi\epsilon_0} \left(\frac1R-\frac1a+\frac1b\right). $$

One more comment: what about the potential at the inner sphere’s surface?

Since the field vanishes throughout the inner metal, its entire volume and surface share this value: $V(R)=V(0)$. The shell is also an equipotential, with

$$ V(a)=V(b)=\frac{q}{4\pi\epsilon_0 b}. $$

The two disconnected conductors can have different constant potentials. In fact, their potential difference is

$$ V(R)-V(a)=\frac{q}{4\pi\epsilon_0} \left(\frac1R-\frac1a\right). $$

Ahahahahah—so that’s it~~~!

This is the useful distinction: a zero electric field makes the potential constant within a conductor. It does not require that constant to be zero. An absolute value follows only after we choose a reference, as we did at infinity.

In my original 2014 notes, I ended this chapter by saying: “That finishes Chapter 2, electrostatics… This chapter is almost the same as high-school physics, so it may look a little rough… sob. I’ll try harder next time!”

Editorial note — 9 September 2026

This English edition restores the intended image sequence from the original Korean post and translates the four conductor labels. It also corrects the original flux argument: equilibrium establishes the zero field, and a spherical surface contributes area $4\pi k^2$, not volume. A missing intermediate minus sign in the potential derivation has been corrected; the original final expression for $V(0)$ was already correct. The two faulty displayed derivations have been replaced with accessible TeX, while the historical image files are preserved. The assumptions and the role of the successive charge sketches are made explicit above.

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