Electrostatics with Conductors: Study Guide (Physics C: E&M 10.1)

Course note. This guide is for the calculus-based Physics C: Electricity and Magnetism course, Topic 10.1, the first topic of Unit 10 (Conductors and Capacitors). It builds directly on Gauss’s law (Topic 8.6) and electric potential (Unit 9).

Constants used throughout: ε₀ = 8.85 × 10⁻¹² C²/(N·m²), so 1/(4πε₀) = 8.99 × 10⁹ N·m²/C², and e = 1.602 × 10⁻¹⁹ C.

What makes a conductor different

In an ideal conductor, some electrons are free to move through the whole material. In a metal, these are the outer electrons of the atoms. The positive ions stay fixed in the lattice.

• A conductor with a negative net charge has extra electrons. They are the excess charge carriers.
• A conductor with a positive net charge has lost electrons. Only electrons move, but you may model the result as if positive carriers sat where the missing electrons would be.

If there is any field inside the material, the free electrons feel a force qE and move. They keep moving until their new positions produce a field that cancels the original field everywhere inside. The conductor is then in electrostatic equilibrium: no charge is moving. The time this takes is so short that the course treats it as instant. So whenever you analyse a conductor “at rest”, you may assume equilibrium has already been reached.

Four results for a conductor in equilibrium

1. E = 0 inside the material. If E were not zero at some point inside, free electrons there would accelerate. That contradicts “no charge is moving”. This is the starting point for everything else.

2. No net charge inside; all excess charge is on the surface. Draw a Gaussian surface of any shape that lies entirely inside the metal, just below the surface. E = 0 at every point on it, so ∮E·dA = 0, and Gauss’s law gives q_enc = 0. You can shrink or move this surface anywhere inside the metal, so no region of the interior can hold net charge. Any excess charge must sit on the surface. Physically, the excess carriers repel one another and spread out until they can go no further.

3. The conductor is an equipotential. Between any two points A and B in the conductor, take a path that stays inside the metal. Then V_B − V_A = −∫E·dl = 0, because E = 0 along the whole path. Every point in the conductor, including every point on its surface, has the same potential.

4. The field just outside is perpendicular to the surface. Suppose E just outside had a component along the surface. It would push surface charges sideways, so they would not be in equilibrium. Another way to see it: the surface is an equipotential, and field lines always cross equipotentials at 90°.

These four results apply to any shape of conductor in equilibrium. They do not apply to insulators, which can hold charge anywhere in their volume (Topic 8.3).

The field just outside: E = σ/ε₀

Let σ be the surface charge density at some point on the conductor. Use a small pillbox Gaussian surface with end area A, with one end just outside the surface and the other end just inside the metal.

• Inside end: E = 0, so no flux.
• Curved side: just outside, E is perpendicular to the surface, so it is parallel to the curved side. No flux. (The part of the side inside the metal has E = 0 anyway.)
• Outside end: E is perpendicular to it, so the flux is EA.

Gauss’s law: EA = σA/ε₀, so E = σ/ε₀, pointing away from the surface if σ > 0 and towards it if σ < 0.

Compare this with E = σ/(2ε₀) for a thin isolated sheet (Topic 8.6). There, flux leaves through both ends of the pillbox. Here, one end is inside the metal, so all the flux goes out through one end.

The result is local: if a field meter reads 3.0 × 10⁴ N/C just outside a flat part of a charged metal object, the charge density there is σ = ε₀E = (8.85 × 10⁻¹²)(3.0 × 10⁴) = 2.7 × 10⁻⁷ C/m².

Where the surface charge gathers: points and edges

Surface charge on a conductor is not spread evenly unless the shape is a sphere far from other charges. It is denser at sharp points and edges than on flat or gently curved parts. Because E = σ/ε₀, the field just outside is also strongest there.

A quick model shows why. For an isolated conducting sphere of radius R, the potential is V = Q/(4πε₀R) and σ = Q/(4πR²), so σ = ε₀V/R. Two parts of the same conductor must be at the same V. A sharply curved region behaves like a sphere of small R, so it needs a larger σ to reach the same potential. Topic 10.2 makes this exact for two spheres joined by a wire.

Polarization in an external field

Put a neutral conductor in an external field E₀. Free electrons move against E₀ until the induced charges produce a field that cancels E₀ everywhere inside the metal. The side facing “upstream” (where E₀ comes from) gains negative charge and the far side is left positive. The net charge stays zero. This is polarization. It happens because the conductor must stay an equipotential: without the shift, points along E₀ would be at different potentials.

Outside, the field lines bend so that they meet the surface at 90°. They end on the negative induced charge and start again from the positive induced charge, as Figure 1 shows.

Cavities and electrostatic shielding

Now make the conductor hollow: a closed metal shell with an empty space (a cavity) inside.

Empty cavity. Take a Gaussian surface inside the metal, surrounding the cavity. E = 0 on it, so the net charge on the cavity wall is zero. Could the wall hold + charge in one place and − charge in another? Then a field line would run from the + patch, across the cavity, to the − patch. Along that line ∫E·dl ≠ 0, so the two patches would be at different potentials. But the whole conductor is one equipotential. So the cavity wall has no charge at all, and E = 0 everywhere in the empty cavity, whatever happens outside.

This is electrostatic shielding. Surround a region with a closed conducting shell, and external fields cannot reach it. Outside charges induce charges on the outer surface, and those induced charges cancel the external field everywhere inside the outer surface, including in the cavity. A closed metal mesh with small gaps shields in nearly the same way; the outside field leaks in only close to the gaps. This is why sensitive instruments are often housed in metal cases.

A charge inside the cavity. Place a charge q in the cavity, not touching the wall. A Gaussian surface inside the metal still has E = 0 on it, so its enclosed charge is zero. Therefore the cavity wall must carry −q. If the conductor has net charge Q, the rest, Q + q, sits on the outer surface.

• Moving q around inside the cavity changes how −q is spread over the wall, but not its total.
• On a spherical outer surface far from other charges, Q + q spreads uniformly, wherever q is in the cavity. The metal “hides” the position of q from the outside.
• Shielding works one way only. Outside fields cannot get in, but the charge Q + q on the outer surface still produces a field outside. Grounding the shell can remove it (Topic 10.2).

Worked example 1: a sphere inside a conducting shell

Question. A solid metal sphere of radius a = 0.040 m carries q₁ = +3.0 nC. It sits at the centre of a thick, concentric metal shell with inner radius b = 0.10 m and outer radius c = 0.12 m. The shell’s net charge is −5.0 nC. Find (a) the charge on each surface, (b) E at r = 0.070 m, 0.11 m and 0.20 m, and (c) the charge density on the shell’s outer surface, then check it with E = σ/ε₀.

(a) Surface charges.

1. The solid sphere is a conductor, so its +3.0 nC is on its surface, r = a.
2. Take a Gaussian sphere of radius 0.11 m, inside the shell’s metal. E = 0 there, so q_enc = 0. It encloses +3.0 nC on the sphere, so the inner surface (r = b) must carry −3.0 nC.
3. The shell’s net charge is −5.0 nC, so the outer surface (r = c) carries −5.0 − (−3.0) = −2.0 nC.

(b) Fields. Every charge layer is spherically symmetric, so use concentric Gaussian spheres.

• r = 0.070 m (in the gap): q_enc = +3.0 nC, so E = (8.99 × 10⁹)(3.0 × 10⁻⁹) ÷ (0.070)² = 5.5 × 10³ N/C, outward.
• r = 0.11 m (in the shell’s metal): E = 0.
• r = 0.20 m (outside): q_enc = +3.0 − 5.0 = −2.0 nC, so E = (8.99 × 10⁹)(2.0 × 10⁻⁹) ÷ (0.20)² = 4.5 × 10² N/C, inward.

(c) Outer surface density. σ = q/(4πc²) = (−2.0 × 10⁻⁹) ÷ (4π × 0.12²) = −1.1 × 10⁻⁸ C/m². Check: |σ|/ε₀ = 1.25 × 10³ N/C. Gauss’s law just outside, kQ/c² = (8.99 × 10⁹)(2.0 × 10⁻⁹) ÷ 0.12² = 1.25 × 10³ N/C. They agree, and the field points inward because σ < 0.

Interpretation. The field outside depends only on the net charge, −2.0 nC, as if it were a point charge at the centre. The shell’s metal is field-free even though there are charges on both sides of it.

Worked example 2: a neutral slab in a uniform field

Question. A large, neutral metal slab, 0.020 m thick with face area 0.50 m², is placed with its faces perpendicular to a uniform field E₀ = 2.0 × 10⁴ N/C. Edge effects can be ignored. (a) Find the induced surface charge density and the induced charge on each face. (b) Use superposition to show that the field inside the slab is zero while the field outside is unchanged. (c) Compare the potential difference across the slab’s 0.020 m with the potential difference across the same 0.020 m of empty space.

(a) The field just outside each face has size E₀ and is perpendicular to the face, so σ = ε₀E₀ = (8.85 × 10⁻¹²)(2.0 × 10⁴) = 1.77 × 10⁻⁷ C/m². The face where E₀ enters (the upstream face) is negative and the face where E₀ leaves is positive. Each face holds a charge of magnitude σA = (1.77 × 10⁻⁷)(0.50) = 8.9 × 10⁻⁸ C. That is about 5.5 × 10¹¹ electrons moved from one face to the other. The net charge is still zero.

(b) Model each face as a sheet. Each gives a field of size σ/(2ε₀) = 1.0 × 10⁴ N/C, pointing away from the + face and towards the − face.

• Inside the slab, both sheet fields point against E₀: total induced field σ/ε₀ = 2.0 × 10⁴ N/C, so the net field is E₀ − E₀ = 0.
• Outside, on either side, the two sheet fields point in opposite directions and cancel. The net field is just E₀.

(c) In empty space, ΔV = E₀d = (2.0 × 10⁴)(0.020) = 400 V. Across the slab, ΔV = 0: the two faces are at the same potential, as they must be for one conductor.

Interpretation. The induced charges arrange themselves exactly so that the conductor stays an equipotential. That is the meaning of polarization for a conductor.

Common misconceptions

• “A charged conductor has charge spread through its volume.” That is true for some insulators. For a conductor in equilibrium, all excess charge is on the surface.
• “E = 0 inside, so V = 0 inside.” E = 0 means V is constant, not zero. A charged metal sphere can be at thousands of volts throughout.
• “The charge density is the same all over a conductor.” Only for an isolated sphere. It is larger at points and edges.
• Using σ/(2ε₀) just outside a conductor. The field just outside a conductor is σ/ε₀.
• “A neutral conductor in a field has no charges on it.” It has equal and opposite induced charges. The net charge is zero; the local charge is not.
• “Shielding blocks the field of a charge inside the shell.” It does not. The charge induces Q + q on the outer surface, which produces a field outside.
• Putting the cavity charge’s partner on the outer surface. The induced −q is on the inner wall; Gauss’s law inside the metal forces this.
• “Equilibrium takes noticeable time.” For the electrostatic situations in this course, it is reached effectively at once.

Where this leads

Next, in Topic 10.2, you will connect conductors to each other and to ground and see how charge moves until their potentials are equal: Redistribution of Charge Between Conductors. The results here, E = 0 inside and E = σ/ε₀ just outside, are also the basis of capacitors (Topic 10.3). Practise now with the practice questions, then use the revision notes and the topic checklist.