Difficulty: Intermediate | Prerequisites: Coulomb's law, electric field concept, integration basics
Gauss's law lets you calculate the electric field around highly symmetric charge distributions by choosing a clever closed surface (a "Gaussian surface") and relating the total flux through it to the enclosed charge. For infinite cylinders and spheres, this collapses into clean expressions where the field depends only on the enclosed charge and the distance from the axis or centre. The key to exam problems is identifying which charge is enclosed by your chosen surface, especially when conductors redistribute charge on their inner and outer surfaces.
Gauss's law
The total electric flux through any closed surface equals the net enclosed charge divided by ε0. In simple terms, this means the number of field lines poking out of a closed surface tells you exactly how much charge is inside.
Gaussian surface
An imaginary closed surface you choose to exploit the symmetry of a problem. Think of it as the mathematical bag you draw around the charge to make the integral easy.
Linear charge density (λ)
Charge per unit length along a line or cylinder, typically in nC/cm or C/m. In simple terms, it tells you how much charge sits on each centimetre of a long wire or cylinder.
Surface charge density (σ)
Charge per unit area on a surface, typically in nC/cm² or µC/m². Think of it as how densely charge is smeared across a flat or curved surface.
Cylindrical symmetry
A charge distribution that looks the same at every angle around a central axis and at every point along that axis. The electric field depends only on the radial distance from the axis.
Spherical symmetry
A charge distribution that looks the same in every direction from a central point. The electric field depends only on the distance from the centre.
Conductor (metal)
A material in which charges are free to move. In electrostatic equilibrium, the electric field inside is zero and all excess charge sits on the surfaces.
Insulator (dielectric)
A material in which charges are fixed in place. Charge can exist throughout the volume, not just on surfaces.
Grounding
Connecting a conductor to a reservoir of charge (the Earth) so its potential is fixed at zero. Charge flows to or from ground until the conductor reaches V = 0.
Setting up the Gaussian surface
For an infinite cylinder, choose a coaxial cylindrical Gaussian surface of radius r and length L.
The flux through the end caps is zero (field is radial, caps are perpendicular to the axis).
The entire flux passes through the curved surface: Φ = E(2πrL).
Gauss's law then gives E(2πrL) = Q_enclosed / ε0, so E = λ_enclosed / (2πε0 r), where λ_enclosed is the linear charge density enclosed.
Charge distribution on conductors in coaxial geometry
A solid metal inner cylinder of charge λ_inner places all its charge on its outer surface (field inside a conductor is zero).
A surrounding conducting shell must have charge –λ_inner on its inner surface (to make E = 0 inside the shell's metal).
The shell's outer surface then carries whatever is left: λ_shell,outer = λ_shell + λ_inner (total shell charge minus what went to the inner surface).
Example: λ_inner = −5 nC/cm and λ_shell = +2 nC/cm. Inner surface of shell: +5 nC/cm. Outer surface of shell: +2 + (−5) = −3 nC/cm.
Electric field in each region
r < a (inside inner conductor): E = 0.
a < r < b (between conductors): E = |λ_inner| / (2πε0 r). Only the inner cylinder's charge is enclosed.
b < r < c (inside shell metal): E = 0.
r > c (outside everything): E = |λ_inner + λ_shell| / (2πε0 r). The total enclosed charge is the sum of both.
Direction and force on a test charge
Outside the system, the net enclosed linear charge density determines the field direction.
If λ_inner + λ_shell is negative, the field points inward (toward the axis), and a positive test charge moves toward the origin.
If λ_inner + λ_shell is positive, the field points outward and a positive test charge moves away.
Potential difference between surfaces
The potential difference between the inner cylinder surface (r = a) and the outer shell surface (r = c) requires integrating E from a to c.
Between b and c (inside the shell), E = 0, so no contribution to the potential difference from that region.
Only the gap a < r < b contributes: V_a – V_c = –∫(a to b) E dr, using E from the inner cylinder alone.
Surface charge density vs linear charge density
For a cylinder of radius a and length L: total charge = λ L, surface area = 2πaL.
Surface charge density: σ = λ / (2πa).
Effect of grounding the shell
Grounding sets V_shell = 0. Charge flows to or from ground.
The inner surface of the shell is unaffected: it must still carry –λ_inner to shield the interior (Gauss's law still requires E = 0 in the conductor).
The outer surface charge changes because ground supplies or removes charge to bring V to zero.
Electric field outside (r > R)
The sphere looks like a point charge. Use a spherical Gaussian surface of radius r.
E = Q / (4πε0 r²), identical to a point charge at the centre.
Electric field inside (r < R)
Only the charge within radius r is enclosed: Q_enc = Q(r/R)³ (since the charge is uniform throughout the volume).
E = Qr / (4πε0 R³). The field grows linearly with r inside.
Potential with a shifted reference point
If V = 0 at the surface (r = R) instead of at infinity, integrate E from the surface outward.
At r > R: V(r) = –∫(R to r) E dr = (Q / 4πε0)(1/r – 1/R). This is negative outside because you are moving in the direction the field pushes a positive charge.
Shape of V(r) for a uniformly charged sphere (V = 0 at surface)
Inside (r < R): V rises parabolically toward the centre (the field is weaker near the centre, so the potential climb is gentle).
At r = R: V = 0 (by definition).
Outside (r > R): V drops as 1/r, going more negative with increasing r.
The correct graph is the one that peaks at r = 0, crosses zero at r = R, and falls off to the right.
Gauss's law (integral form)
\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}Electric field outside an infinite line charge / cylinder (r > outermost radius)
E = \frac{\lambda_{\text{enc}}}{2\pi \varepsilon_0 \, r}Electric field outside a sphere (r > R)
E = \frac{Q}{4\pi \varepsilon_0 \, r^2}Electric field inside a uniformly charged sphere (r < R)
E = \frac{Q \, r}{4\pi \varepsilon_0 \, R^3}Relating linear and surface charge density for a cylinder of radius a
\sigma = \frac{\lambda}{2\pi a}Potential difference (cylindrical, integrating from a to b)
V_a - V_b = -\int_a^b E \, dr = \frac{\lambda_{\text{enc}}}{2\pi \varepsilon_0} \ln\!\left(\frac{b}{a}\right)Note: the sign depends on the sign of λ_enc. When λ_enc is negative, V_a – V_b comes out negative (the inner conductor is at lower potential).
Potential outside a sphere with V = 0 at the surface
V(r) = \frac{Q}{4\pi \varepsilon_0}\left(\frac{1}{r} - \frac{1}{R}\right)Students often think the total charge on a conductor determines the field between the conductors. It does not. Only the enclosed charge matters for Gauss's law, and in a coaxial geometry the field between the cylinders depends solely on the inner cylinder's charge.
Students often confuse linear charge density (λ, charge per length) with surface charge density (σ, charge per area). They are related by the cylinder's circumference: σ = λ / (2πr). Keep track of which one the question asks for.
Students often assume grounding a conductor removes all its charge. It does not. Grounding changes the outer surface charge to bring the potential to zero, but the inner surface charge is dictated by Gauss's law and the requirement of E = 0 inside the metal. The inner surface is shielded from ground.
Students often think "V = 0 at the surface" is the same as "V = 0 at infinity." When the reference point shifts, the sign and value of V everywhere change accordingly. Read the problem's reference convention carefully.
⚠️ Determining which charge sits on which surface of a conducting shell is tested repeatedly. Walk through the logic: inner surface cancels the enclosed charge, outer surface gets whatever remains.
⚠️ Calculating E in the gap between coaxial conductors: only the inner charge matters, and the formula is λ / (2πε0 r).
⚠️ Potential difference integrals: know that E = 0 inside a conductor contributes nothing, so you only integrate across the gap.
⚠️ Grounding questions test whether you understand that the inner surface charge is locked by Gauss's law while the outer surface adjusts.
⚠️ For spherical insulators, be ready to compute E both inside and outside, and to sketch or identify V(r) for a non-standard reference point.
True or false: The electric field inside a conducting shell is always zero, regardless of charges outside it. (True, provided the shell is complete and in electrostatic equilibrium.)
Fill in the blank: For a coaxial cable, the charge on the inner surface of the outer shell equals ______ . (–λ_inner, i.e. equal in magnitude and opposite in sign to the inner conductor's charge.)
True or false: Grounding a conducting shell removes the charge on its inner surface. (False. The inner surface charge is dictated by Gauss's law and stays the same.)
Fill in the blank: Inside a uniformly charged insulating sphere, the electric field is proportional to ______ . (r, the distance from the centre. It grows linearly.)
True or false: If V = 0 at the surface of a positively charged sphere, V is positive at infinity. (False. V becomes more negative as you move away from the surface.)
Q: A coaxial cable has an inner conductor with λ_inner = −5 nC/cm and an outer shell with λ_shell = +2 nC/cm. What is the net linear charge density on the outer surface of the shell?
A: The inner surface of the shell carries +5 nC/cm (to cancel the inner conductor). The shell's total is +2 nC/cm, so its outer surface carries 2 – 5 = −3 nC/cm.
Q: For the same cable, what direction does the electric field point at a radius well outside the shell?
A: The net enclosed charge is −5 + 2 = −3 nC/cm (negative), so the field points radially inward, toward the axis.
Q: A positive test charge is released from rest outside this cable. Which way does it move?
A: Toward the axis (toward the origin), because the net enclosed charge is negative and the field points inward.
Q: A solid insulating sphere of radius R carries total charge Q uniformly distributed. If V = 0 at r = R, what is V at r = 2R?
A: V(2R) = (Q / 4πε0)(1/(2R) – 1/R) = –Q / (8πε0 R). The potential is negative outside the surface under this convention.
Q: For a uniformly charged insulating sphere, how does E depend on r inside the sphere?
A: E grows linearly with r: E = Qr / (4πε0 R³). At the centre, E = 0.
Q: If a conducting shell surrounding a charged inner cylinder is connected to ground, what happens to the charge on the shell's inner surface?
A: It remains unchanged. Gauss's law still requires the inner surface to carry –λ_inner regardless of what happens at the outer surface.
Gauss's law connects directly to capacitance: the coaxial cylinder geometry is the basis of the cylindrical capacitor formula, and the parallel-plate capacitor is a limiting case of Gauss's law for infinite planes. Understanding how charge distributes on conductors feeds into every capacitor and circuit problem in the course.
The relationship between E and V (the field is the negative gradient of the potential) shows up everywhere from here on, in both the capacitor unit and later in circuits with EMF sources.
Gauss's law, electric flux, cylindrical Gaussian surface, spherical Gaussian surface, coaxial cable, coaxial cylinder, linear charge density, surface charge density, λ, σ, enclosed charge, conductor electrostatics, insulator charge distribution, grounding a conductor, shielding, E = 0 inside a conductor, potential difference integration, V(r) graph, PHYS 212 Exam I, University of Illinois, electricity and magnetism