Source: PHYS 212 Final Examination, University of Illinois at Urbana-Champaign
Tags: Gauss's law, conductors, electrostatic equilibrium, electric field, surface charge density, concentric spheres, conducting shell, induced charge, electric potential, Coulomb's law, point charge
Difficulty: Intermediate | Prerequisites: Basic vector calculus, Coulomb's law, concept of electric field and flux.
This topic sits at the heart of PHYS 212. Conductors in electrostatic equilibrium are the gateway to understanding how charge distributes itself on real objects and how we can use symmetry (via Gauss's law) to calculate electric fields without painful integration. If you missed the earlier weeks on Coulomb's law and the concept of flux, revisit those first. Everything here builds on the idea that charges in a conductor are free to move, and in equilibrium, they have finished moving, which constrains the field inside and the charge on each surface. This material connects directly to capacitors, shielding, and circuit behaviour later in the course.
Inside a conductor in electrostatic equilibrium the electric field is zero, all excess charge sits on surfaces, and the whole conductor is at one potential. Gauss's law lets you find the field outside symmetric charge distributions by relating enclosed charge to electric flux through a cleverly chosen surface.
Electrostatic equilibrium
The state a conductor reaches when all internal charge movement has stopped. The net electric field inside the conducting material is zero, and any excess charge resides entirely on the conductor's surface. In simple terms, the charges have settled and nothing is flowing.
Gauss's law
The total electric flux through any closed surface equals the net enclosed charge divided by the permittivity of free space: Φ_E = Q_enc / ε₀. Think of it as a bookkeeping rule: the number of field lines poking out of a closed surface tells you how much charge is inside.
Gaussian surface
An imaginary closed surface chosen to exploit symmetry so that the electric field is either constant across the surface or perpendicular to it. It is not a physical object. In simple terms, it is a mathematical trick that makes the flux integral easy.
Equipotential
A region (volume or surface) where every point is at the same electric potential. The entire bulk of a conductor in equilibrium is equipotential, meaning no work is done moving a test charge anywhere inside it.
Induced charge
Charge that appears on the surfaces of a neutral conductor when an external charge is brought nearby. The conductor remains neutral overall, so if charge -Q appears on one surface, charge +Q appears on another. Think of it as the conductor's free electrons redistributing to cancel the external field inside.
Surface charge density (σ)
Charge per unit area on a surface: σ = Q / A. For a sphere of radius r carrying charge Q, the surface charge density is Q / (4πr²).
Electric potential (V)
The potential energy per unit charge at a point, measured relative to infinity (where V = 0). For a point charge q at distance r: V = kq / r. This is a scalar, not a vector, which often makes calculations simpler than working with the field directly.
The electric field inside the conducting material is exactly zero. If it were not, free charges would move, contradicting "equilibrium."
Any net charge resides entirely on the conductor's outer surface (or surfaces, if there is a cavity).
The electric field just outside the surface is perpendicular to that surface. A tangential component would push surface charges along, again breaking equilibrium.
The entire conductor, surface and interior alike, is at one electric potential (equipotential volume).
The exam's recurring setup: a solid conducting sphere of radius a carrying charge +Q, surrounded by a neutral conducting shell with inner radius b and outer radius c (a < b < c).
Inside the inner sphere (r < a): E = 0. It is a conductor in equilibrium.
Between sphere and shell (a < r < b): This is vacuum (or air). Draw a spherical Gaussian surface of radius r here. The enclosed charge is +Q, so E = kQ / r², directed radially outward.
Inside the shell material (b < r < c): E = 0. The shell is a conductor in equilibrium.
Outside everything (r > c): Draw a Gaussian surface of radius r. The shell is neutral, so the total enclosed charge is still +Q, giving E = kQ / r².
Because E must be zero inside the shell material, Gauss's law applied to a surface just inside the shell (radius between b and c) requires that the enclosed charge is zero. The inner sphere contributes +Q, so the inner surface of the shell must carry -Q.
The shell is neutral overall, so its outer surface carries +Q.
Inner surface charge density: σ_inner = -Q / (4πb²)
Outer surface charge density: σ_outer = +Q / (4πc²)
If a point charge +q is placed at the centre of the solid sphere, the total charge enclosed by a Gaussian surface just outside the sphere's surface becomes Q + q.
Inner surface of the shell: -(Q + q), to cancel the field inside the shell material.
Outer surface of the shell: +(Q + q), because the shell is still neutral overall.
Choose a spherical Gaussian surface of radius r > c, concentric with the system.
By symmetry, E is radial and constant over this surface.
Flux: Φ_E = E · (4πr²).
Enclosed charge: Q_enc = Q (the shell is neutral and contributes zero net charge).
Apply Gauss's law: E · (4πr²) = Q / ε₀, so E = Q / (4πε₀r²) = kQ / r².
The correct expressions for the potential V at distance r from a point charge q, relative to infinity:
V = U / q (potential energy per unit test charge)
V = -∫ E · dl (line integral of the field from infinity to the point)
V = kq / r
Note: V = kq / r² is the electric field magnitude, not the potential. Confusing these two is a common exam error.
Gauss's law: Φ_E = ∮ E · dA = Q_enc / ε₀
Coulomb constant form: E = kQ / r² (outside a spherical charge distribution)
Surface charge density: σ = Q / (4πr²) for a sphere of radius r
Electric potential of a point charge: V = kq / r
Relationship: E = -dV/dr (for spherically symmetric systems)
Electrostatic shielding is why sensitive electronics are housed in metal enclosures (Faraday cages): the field inside a conducting shell is zero regardless of what is happening outside. This same principle protects passengers inside a car struck by lightning.
Students often assume the field inside the shell material depends on the charge on the inner sphere. It does not; the field inside any conductor in equilibrium is zero, full stop.
Mixing up V = kq / r (potential) with E = kq / r² (field). The potential falls off as 1/r, the field as 1/r². These are different quantities with different units.
Thinking the neutral shell "blocks" the field outside. It does not. Outside the shell, the field is exactly the same as if the shell were not there at all, because the shell's net charge is zero.
Believing that Gauss's law only works for spheres. It works for any closed surface and any charge distribution; spherical symmetry just makes the integral trivial.
⚠️ Questions 1, 5, 10, 13, and 15 of the final all draw from this material. Expect to be tested on the field in each region of the concentric sphere/shell setup and on induced surface charges.
⚠️ "E = 0 inside a conductor" is the single most-tested fact in this topic cluster. Know why it is true (free charges would move otherwise) as well as the result itself.
⚠️ The short-answer section asks you to derive E for r > c using Gauss's law. Practise writing the derivation in five clear lines: choose surface, state symmetry, write flux, state enclosed charge, solve.
⚠️ The multi-select question on conductor properties (Q10) expects you to select all four statements (equipotential, charge on surface, field perpendicular outside, field zero inside). Do not second-guess yourself into dropping one.
True or False: The electric field inside a solid conducting sphere carrying net charge is zero.
Fill in the blank: The induced charge on the inner surface of a neutral conducting shell surrounding a charge +Q is ______.
True or False: V = kq / r² is the correct expression for the electric potential of a point charge.
Fill in the blank: A conductor in electrostatic equilibrium is an _______ volume.
True or False: Outside a neutral conducting shell surrounding a point charge, the electric field is zero.
Answers: 1. True. 2. -Q. 3. False (that is the field; potential is kq / r). 4. Equipotential. 5. False (the field is kQ / r², as if the shell were absent).
Q: A solid conducting sphere of radius a carries charge +Q and is surrounded by a concentric neutral shell (inner radius b, outer radius c). What is E in the region b < r < c?
A: E = 0. The region b < r < c is inside the conducting material of the shell, and the field inside a conductor in electrostatic equilibrium is always zero.
Q: In the same setup, what is the surface charge density on the inner surface of the shell at r = b?
A: σ = -Q / (4πb²). A total charge of -Q is induced on the inner surface to ensure E = 0 inside the shell material. Distributing -Q over the area 4πb² gives the density.
Q: A point charge +q is now placed at the centre of the inner sphere. What is the total charge on the outer surface of the shell (at r = c)?
A: Q + q. The inner surface of the shell must carry -(Q + q) to cancel the enclosed charge. Since the shell is neutral, the outer surface carries +(Q + q).
Q: Using Gauss's law, derive E at a distance r > c from the centre.
A: Choose a concentric spherical Gaussian surface of radius r. By symmetry E is radial and constant over this surface, so the flux is E · 4πr². The enclosed charge is Q (shell is neutral). Gauss's law gives E · 4πr² = Q / ε₀, hence E = kQ / r².
Q: Which of V = U/q, V = -∫E · dl, V = kq/r, and V = kq/r² are correct expressions for the potential of a point charge?
A: V = U/q, V = -∫E · dl, and V = kq/r are all correct. V = kq/r² has the wrong dimensions and is the electric field expression, not the potential.
This material connects directly to capacitors: a parallel-plate capacitor is just two conductors in equilibrium (when not connected to a circuit), and the field between the plates is found by the same Gauss's law reasoning. It also underpins how current flows in circuits, because current only flows when equilibrium is broken (by an EMF source, for example). The concept of equipotential surfaces reappears in every discussion of voltage and potential energy throughout the course.
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