Course: ZJUI Physics 212 – University Physics: Electricity & Magnetism
Source: Gauss' Law (A) Check 2 | September 2024 Tags: Gauss's law, electric flux, Gaussian surface, charge symmetry, electric field, Coulomb's law, spherical shell, conducting sphere, infinite plane, superposition
Difficulty: Intermediate Prerequisites: Coulomb's law, electric field concept, vector addition, basic integral calculus. You should be comfortable with the idea that electric fields point away from positive charges and toward negative charges before starting here.
Gauss's law is one of Maxwell's four equations and provides a powerful shortcut for calculating electric fields when the charge distribution has enough symmetry. Instead of integrating Coulomb's law over every bit of charge (which can be brutal), you choose a clever closed surface, relate the total flux through it to the enclosed charge, and read off the field. The catch: the shortcut only works when symmetry lets you pull E out of the flux integral. This set of notes covers the main symmetry types you will see in Physics 212, the reasoning behind choosing a Gaussian surface, and common traps students fall into on exams.
Gauss's law says the total electric flux through any closed surface equals the enclosed charge divided by ε₀. Choosing the right Gaussian surface (one that matches the symmetry of the charge distribution) is the entire game. If the charge distribution lacks the right symmetry, Gauss's law still holds but cannot be used to solve for E directly.
Electric flux (Φ_E)
The measure of how much electric field passes through a surface. Formally, Φ_E = ∮ E · dA. Think of it as counting how many field lines pierce through a closed bag.
Gaussian surface
An imaginary closed surface you choose to apply Gauss's law. It is not a physical object. You pick its shape so that E is either constant and perpendicular to the surface, or parallel to it (contributing zero flux), everywhere on that surface.
Gauss's law
∮ E · dA = Q_enc / ε₀. The net electric flux through any closed surface equals the total charge enclosed divided by the permittivity of free space. In simple terms, this means the number of field lines leaving a closed surface is proportional to the charge inside it.
Charge density (ρ)
Charge per unit volume, measured in C/m³. Used when charge is spread through a volume rather than sitting on a surface. For a uniform volume charge density, Q_enc = ρ × (enclosed volume).
Surface charge density (σ)
Charge per unit area, measured in C/m². Appears when charge sits on a thin sheet or the surface of a conductor.
Permittivity of free space (ε₀)
The constant 8.854 × 10⁻¹² C²/(N·m²). It sets the scale for how strong electric interactions are in vacuum.
Superposition principle
The electric field at any point is the vector sum of the fields produced by each source independently. You calculate each source's contribution separately, then add them as vectors.
Gauss's law is always true for any closed surface, but it is only useful for finding E when the surface exploits the symmetry of the charge distribution.
The goal: choose a surface where E · dA simplifies. That means E must be constant in magnitude over the surface and either perpendicular or parallel to it at every point.
Three symmetry types dominate Physics 212:
Spherical symmetry (point charges, spherical shells, solid spheres): use a concentric spherical Gaussian surface.
Cylindrical symmetry (infinite line charges, long cylinders): use a coaxial cylindrical Gaussian surface.
Planar symmetry (infinite charged planes): use a "pillbox" or rectangular Gaussian surface straddling the plane.
If the charge distribution does not have one of these symmetries, you cannot use Gauss's law to solve for E, even though the law itself still holds.
A cube of side length a carrying charge +Q does not have spherical, cylindrical, or planar symmetry.
A spherical Gaussian surface centred on the cube would enclose all the charge, so Gauss's law gives you the total flux. However, E is not constant over that sphere (the field from a cube is stronger near the corners, weaker near the face centres), so you cannot pull E out of the integral.
A cubical Gaussian surface of dimension R + a/2 matches the shape of the charge but still does not help, because E varies across each face.
A cylindrical surface has the same problem.
Correct conclusion: this field cannot be calculated using Gauss's law. Gauss's law still tells you the flux, but it does not let you isolate E.
Setup: a shell with inner radius a, outer radius b, and uniform volume charge density ρ throughout the shell material. The hollow centre (r < a) contains no charge.
For a Gaussian sphere of radius r < a (inside the hollow):
The Gaussian surface encloses zero charge.
By Gauss's law, Φ_E = 0, and by the spherical symmetry of the shell, E = 0 everywhere inside the cavity.
For a < r < b (inside the shell material):
Q_enc = ρ × (4/3)π(r³ − a³)
E × 4πr² = ρ(4/3)π(r³ − a³) / ε₀
E = ρ(r³ − a³) / (3ε₀r²)
For r > b (outside the shell):
Q_enc = ρ × (4/3)π(b³ − a³)
E = ρ(b³ − a³) / (3ε₀r²)
This looks like the field of a point charge Q_total at the centre (shell theorem).
Setup: a solid conducting sphere carrying charge +q, surrounded by a conducting spherical shell carrying charge −q (equal magnitude, opposite sign).
Inside a conductor in electrostatic equilibrium, the electric field is always zero. All excess charge resides on surfaces.
Between the spheres (r between the inner sphere's surface and the shell's inner surface):
A Gaussian sphere here encloses only the +q on the inner conductor.
E = q / (4πε₀r²), pointing radially outward.
The field points outward because the enclosed charge is positive.
On the inner surface of the outer shell:
The conductor's interior must have E = 0. A Gaussian surface just inside the shell's material encloses +q from the inner sphere, so the shell's inner surface must carry −q to make Q_enc = 0 inside the conductor.
Outside the outer shell:
Total enclosed charge = +q + (−q) = 0.
E = 0 everywhere outside.
The shell's outer surface carries zero net charge. The charges have cancelled.
An infinite plane with surface charge density +σ produces a uniform field E = σ/(2ε₀) pointing away from the plane on both sides.
A plane with −σ produces the same magnitude field but directed toward the plane.
Case A (single positive plane, point P to its left):
E at P = σ/(2ε₀), pointing left (away from the plane).
Case B (three planes in order: +σ, −σ, +σ, point P to the left of all three):
Each plane contributes σ/(2ε₀) at point P.
The left +σ plane pushes the field leftward at P: +σ/(2ε₀) to the left.
The −σ plane pulls the field rightward at P: σ/(2ε₀) to the right.
The right +σ plane pushes the field leftward at P: σ/(2ε₀) to the left.
Net field at P = σ/(2ε₀) − σ/(2ε₀) + σ/(2ε₀) = σ/(2ε₀) to the left.
Result: the magnitude at P is the same in both cases. Superposition of three planes, with the middle one carrying opposite charge, cancels one plane's worth of contribution in the region to the left.
Formula | Meaning |
|---|---|
∮ E · dA = Q_enc / ε₀ | Gauss's law (integral form) |
E = Q / (4πε₀r²) | Field of a point charge or outside a spherical distribution |
E = σ / (2ε₀) | Field from one side of an infinite plane of charge |
E = σ / ε₀ | Field between two parallel plates with ±σ (a capacitor) |
Q_enc = ρ × Volume | Enclosed charge for uniform volume charge density |
ε₀ = 8.854 × 10⁻¹² C²/(N·m²) | Permittivity of free space |
Gauss's law with spherical symmetry is the reason we can treat the Earth (or any planet) as a point mass/charge when we are outside it. GPS satellites, for instance, model Earth's gravitational field this way, and the same shell theorem applies to electric fields.
The parallel-plate result (E = σ/ε₀ between the plates) is the operating principle behind every parallel-plate capacitor, which appears in essentially all electronic circuits, from phone screens to defibrillators.
"Gauss's law can always be used to find E." It cannot. Gauss's law is always true, but you can only solve for E when the symmetry lets you pull E out of the integral. A charged cube is the classic counterexample.
"If there is no charge inside a Gaussian surface, the field everywhere on that surface is zero." Not necessarily. It means the net flux is zero. Field lines can enter one side and exit the other, giving zero net flux but a nonzero field at every point on the surface.
"A conductor has no electric field anywhere." The field is zero inside the bulk of a conductor in electrostatic equilibrium. There is typically a nonzero field just outside the surface and in the surrounding space.
"Superposition means fields always add up to something bigger." Fields are vectors. When sources have opposite signs or are arranged so their fields oppose each other, the net field can be smaller than any individual contribution, or even zero.
⚠️ Expect a question that tests whether Gauss's law can be applied, not just how. The cube scenario is a favourite: students jump to a spherical surface without checking that E is constant over it.
⚠️ For concentric conductors, examiners love asking for the field in every region and the charge on every surface. Draw a table: region, Q_enc, E.
⚠️ Superposition with planes is a fast calculation, but only if you keep careful track of directions. Set a positive direction first, then assign signs to each plane's contribution.
⚠️ Remember that for a hollow insulating shell, E = 0 inside the cavity only because there is no charge there. If a point charge were placed at the centre of the cavity, E would not be zero inside.
True or False: Gauss's law can be used to find the electric field at a point near a uniformly charged cube.
False. The cube lacks the required symmetry.
Fill in the blank: Inside the hollow centre of a uniformly charged spherical shell, the electric field is ________.
Zero.
True or False: If a Gaussian surface encloses zero net charge, the electric field at every point on that surface must be zero.
False. Zero net flux does not mean zero field everywhere on the surface.
Fill in the blank: The electric field of an infinite plane of charge with surface charge density σ is ________ and does not depend on ________.
σ/(2ε₀); distance from the plane.
True or False: Inside a conductor in electrostatic equilibrium, the electric field is always zero.
True.
Q: A charged spherical insulating shell has inner radius a, outer radius b, and uniform volume charge density ρ. What is the electric field at a distance r from the centre, where r < a?
A: E = 0. A Gaussian sphere of radius r < a encloses no charge, and the spherical symmetry of the shell means the field must be zero everywhere inside the cavity.
Q: A solid conducting sphere with charge +q is surrounded by a conducting shell with charge −q (same magnitude). Describe the electric field in the region between the two conductors.
A: The field points radially outward with magnitude E = q/(4πε₀r²). The Gaussian surface in this region encloses only +q from the inner conductor.
Q: For the same concentric-conductor setup, what is the electric field outside the outer shell?
A: E = 0. The total enclosed charge is +q + (−q) = 0.
Q: You need to find the electric field at a distance R from a charged cube of side a. Can you use Gauss's law? Why or why not?
A: You cannot use Gauss's law to solve for E. While Gauss's law is valid, no Gaussian surface exists on which E is both constant and perpendicular (or parallel) at every point, because the cube lacks spherical, cylindrical, or planar symmetry.
Q: Three infinite planes (charges +σ, −σ, +σ from left to right) are parallel. A point P is to the left of all three. What is the magnitude of E at P?
A: E = σ/(2ε₀). By superposition, the three contributions partially cancel, leaving a net field equal to that of a single plane.
Q: In the three-plane setup above, is the field at P larger than, smaller than, or the same as the field at P due to a single +σ plane alone?
A: The same. The −σ plane's contribution at P cancels one of the two +σ planes' contributions, leaving σ/(2ε₀).
This material connects directly to electric potential, which you will study next. Once you know E from Gauss's law, you integrate to find V. The spherical-shell results here will reappear when you calculate the potential of capacitors and charged conductors.
The conducting-sphere problem is the foundation for understanding capacitance of spherical capacitors, covered later in the course.
Superposition of plane fields leads straight into parallel-plate capacitors and eventually dielectrics, both major exam topics in Physics 212.
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