Source: ZJUI Physics 212, Electric Flux and Field Lines worksheet (Version A)
Tags: electric flux, Gauss's law, Gaussian surface, cylindrical surface, spherical shell, enclosed charge, flux through closed surface, ε₀, permittivity, Coulomb, charge density, lambda, PHYS 212, UIUC, ZJUI
Difficulty: Intermediate Prerequisites: Coulomb's law, electric field of a point charge, electric field of an infinite line charge, dot product, surface integrals (conceptual level).
Electric flux quantifies how much electric field "passes through" a given surface. It bridges the gap between knowing the field at a point and knowing the total effect of charges enclosed by a surface. Gauss's law then provides a powerful shortcut: instead of integrating Coulomb contributions from every charge, you wrap a closed surface around the region of interest and relate the total flux to the enclosed charge. This is one of the four Maxwell equations, and it underpins nearly every electrostatics result you will meet for the rest of the course.
If you are comfortable with electric fields from point charges and line charges but have not yet seen surface integrals, start there. The conceptual intuition (field lines piercing a surface) matters more at this stage than the calculus.
Electric flux through a closed surface depends only on the charge enclosed inside that surface, not on the surface's shape or size. Gauss's law expresses this: Φ = Q_enc / ε₀. Moving a charge around inside a closed surface changes the local flux distribution across patches of the surface, but the total flux through the entire surface stays the same.
Electric flux (Φ or Φ_E)
The surface integral of the electric field over a given area: Φ = ∮ E · dA. In plain terms, it measures the net number of field lines passing outward through a surface. Positive flux means a net outward flow of field lines; negative means net inward.
Gaussian surface
An imaginary closed surface you choose to exploit Gauss's law. It does not need to be a physical object. You pick its shape and size to make the integral easy, typically matching the symmetry of the charge distribution.
Gauss's law
Φ_E = Q_enc / ε₀. The total electric flux through any closed surface equals the enclosed charge divided by the permittivity of free space. Think of it as a charge-counting tool: no matter how you reshape the bag, the flux tells you exactly how much charge is inside.
Linear charge density (λ)
Charge per unit length along a line or rod, measured in C/m. For a uniformly charged infinite rod, the enclosed charge inside a cylindrical Gaussian surface of length L is simply λL.
Enclosed charge (Q_enc)
The total charge sitting inside the chosen Gaussian surface. Charges outside contribute zero net flux through that surface, even though they affect the field at every point on it.
Permittivity of free space (ε₀)
A fundamental constant, approximately 8.85 × 10⁻¹² C²/(N·m²). It appears in the denominator of Gauss's law and sets the scale for how much flux a given charge produces.
The source problem presents an infinitely long rod with uniform charge density λ passing through a cylindrical Gaussian surface. Two cases are compared:
Case 1: Cylinder of radius S and length L.
Case 2: Cylinder of radius 2S and length L/2.
The key reasoning:
By Gauss's law, Φ = Q_enc / ε₀.
The enclosed charge depends only on how much length of the rod sits inside the cylinder, because the charge is distributed uniformly along the rod.
Case 1 encloses a length L of rod, so Q_enc = λL.
Case 2 encloses a length L/2 of rod, so Q_enc = λ(L/2).
The radius of the cylinder is irrelevant to the total flux. A wider cylinder intercepts the same field lines at a greater distance (where the field is weaker), and the effects cancel exactly.
Result: Φ₁ = 2 Φ₂. (Answer: a)
This is the single most important conceptual point in the problem: the radius of the Gaussian surface does not affect the total flux; only the enclosed charge matters.
A positive charge +Q sits inside a spherical shell. Two positions are considered:
Position 1: Charge near the centre of the shell.
Position 2: Charge displaced toward the upper-right region of the shell (near surface element dA).
Local flux through surface elements dA and dB (Question 3):
When the charge moves from Position 1 to Position 2, it gets closer to dA and farther from dB. The field through dA becomes stronger and more aligned with the outward normal, so dΦ_A increases. The field through dB becomes weaker and less aligned, so dΦ_B decreases. (Answer: a)
The local flux through a small patch depends on both the field strength and the angle between the field and the surface normal at that patch. Moving the charge redistributes how the flux is shared across the surface.
Total flux through the entire shell (Question 5):
Gauss's law again: Φ_E = Q_enc / ε₀. The enclosed charge is still +Q regardless of where inside the shell it sits. The total flux does not change. (Answer: c)
This is a classic exam trap. Students see the local fluxes changing and conclude the total must change too. It does not. Every field line that gains strength through one patch loses it through another; the integral over the whole closed surface is fixed by the enclosed charge alone.
Expression | Meaning |
|---|---|
Φ_E = ∮ E · dA | Definition of electric flux through a closed surface |
Φ_E = Q_enc / ε₀ | Gauss's law |
Q_enc = λL | Enclosed charge for a length L of a uniformly charged rod |
E (infinite line) = λ / (2πε₀r) r̂ | Electric field at distance r from an infinite line charge |
ε₀ ≈ 8.85 × 10⁻¹² C²/(N·m²) | Permittivity of free space |
Gauss's law is how engineers determine the electric field inside coaxial cables (the cylindrical symmetry of the inner conductor and outer shield is a direct application of the cylindrical Gaussian surface from this worksheet). It also governs the design of Faraday cages and electrostatic shielding, where the principle that enclosed charge dictates flux is used to block external fields.
"A bigger Gaussian surface means more flux." It does not. The total flux depends only on the enclosed charge, not on the surface's size or shape. A larger surface intercepts a weaker field over a larger area, and the two effects cancel.
"Moving a charge inside a closed surface changes the total flux." The total flux stays constant. Only the distribution of flux across different patches of the surface changes. This is a direct consequence of Gauss's law.
"Charges outside the Gaussian surface contribute to the flux." External charges create fields that enter and exit the surface, contributing zero net flux. They affect the field at each point, but their inward and outward contributions cancel exactly.
"Electric flux is always positive." Flux can be negative. If the net enclosed charge is negative, field lines point inward through the surface, giving a negative total flux.
⚠️ Comparing flux through Gaussian surfaces of different sizes is a standard exam question. Always go straight to Q_enc / ε₀ rather than trying to integrate the field.
⚠️ The "move the charge inside the shell" scenario tests whether you understand that Gauss's law fixes the total, even when local flux patches change. Expect this in conceptual multiple-choice sections.
⚠️ Know the distinction between "flux through the entire closed surface" and "flux through one patch." Gauss's law governs the former; the latter requires detailed knowledge of the field and geometry.
True or false: Doubling the radius of a spherical Gaussian surface around a point charge doubles the electric flux through it.
Fill in the blank: The total electric flux through a closed surface is determined by ________ alone.
True or false: If a charge is moved from the centre of a spherical shell to a point near the shell wall, the total flux through the shell changes.
Fill in the blank: For an infinite line charge with linear density λ, the charge enclosed by a cylindrical Gaussian surface of length L is ________.
True or false: A charge located outside a Gaussian surface contributes zero net flux through that surface.
Answers: 1. False (flux is unchanged). 2. The enclosed charge (Q_enc / ε₀). 3. False (total flux is unchanged). 4. λL. 5. True.
Q: An infinitely long rod with charge density λ passes through two cylindrical Gaussian surfaces. Surface A has radius R and length L. Surface B has radius 3R and length 2L. What is the ratio Φ_A / Φ_B?
A: Φ_A / Φ_B = λL / (λ · 2L) = 1/2. The radius is irrelevant; only the enclosed length of the rod matters.
Q: A point charge +Q is at the centre of a cube. What is the flux through one face of the cube?
A: Total flux = Q / ε₀. By symmetry the cube has six identical faces, so the flux through one face is Q / (6ε₀).
Q: A +3 μC charge and a −1 μC charge are both inside a closed surface. What is the total electric flux through the surface?
A: Φ = Q_enc / ε₀ = (3 × 10⁻⁶ − 1 × 10⁻⁶) / ε₀ = 2 × 10⁻⁶ / ε₀ ≈ 2.26 × 10⁵ N·m²/C.
Q: If you move the +Q charge inside a spherical shell from the centre to a point very close to the shell wall, what happens to the flux through a small patch of shell nearest the charge?
A: The flux through that small patch increases, because the field is stronger and more closely aligned with the outward normal. The total flux through the entire shell, however, stays the same.
Q: Why does the radius of a cylindrical Gaussian surface not affect the total flux through it when used with an infinite line charge?
A: Because changing the radius does not change the enclosed charge. The field strength decreases with distance (E ∝ 1/r), but the lateral surface area increases proportionally (A ∝ r), so the product E · A, and therefore the flux, remains the same.
This material connects directly to the calculation of electric fields for symmetric charge distributions (infinite planes, spherical shells, solid spheres), which rely on choosing an appropriate Gaussian surface and applying Gauss's law. It also lays the groundwork for understanding capacitance, where the field between capacitor plates is derived via Gauss's law, and for Faraday's law later in the course, which uses the same "flux through a surface" concept but for magnetic fields.
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