A Gaussian sphere encloses a point charge of +Q. A separate point charge of −2Q sits outside the sphere. What is the total electric flux through the Gaussian sphere?
A+Q/ε₀ — only the enclosed charge determines the flux
B−2Q/ε₀ — the larger external charge dominates
C−Q/ε₀ — the net of all charges (+Q and −2Q) determines the flux
D+3Q/ε₀ — both charges contribute and the signs reinforce
Gauss's law states ∮E·dA = Q_enclosed/ε₀. The external charge −2Q contributes field lines that enter and exit the Gaussian surface in equal numbers — its net flux contribution is exactly zero. Only the enclosed +Q produces an unbalanced flux. This is a key test of whether students have grasped the 'enclosed charge only' principle versus confusing total flux with total charge in the vicinity.
Question 2 Multiple Choice
You want to use Gauss's law to find the electric field at a distance r from the center of a uniformly charged cube. You draw a cubic Gaussian surface concentric with the cube. Why does this fail to simplify the calculation, even though Gauss's law still applies?
AGauss's law only applies to spherical Gaussian surfaces
BThe enclosed charge for a cube cannot be calculated
CThe electric field is not constant in magnitude and not everywhere perpendicular to the surface, so the surface integral does not reduce to E × Area
DThe Gaussian cube must be larger than the charged cube to be valid
Gauss's law is always mathematically valid — the flux through any closed surface equals Q_enclosed/ε₀. The problem is calculational. To solve for E, you need the integral ∮E·dA to collapse to a simple product E × A. That only works when E is constant in magnitude and perpendicular to the surface everywhere. A cube lacks the spherical symmetry needed to guarantee this: the field near an edge of the cube is different in both magnitude and direction from the field near the center of a face.
Question 3 True / False
A charge placed outside a closed Gaussian surface contributes zero net electric flux through that surface.
TTrue
FFalse
Answer: True
Every field line from an external charge that enters the closed surface must also exit it — the surface is closed, so no field line can terminate inside without a charge source. The inward flux and outward flux from any external charge cancel exactly, producing zero net contribution. This cancellation is why only the enclosed charge appears on the right side of Gauss's law.
Question 4 True / False
Gauss's law can be used to directly calculate the electric field at any point near any charge distribution.
TTrue
FFalse
Answer: False
Gauss's law ∮E·dA = Q_enclosed/ε₀ is always true, but 'always true' does not mean 'always useful for finding E.' To extract E from the integral, you need to choose a Gaussian surface where E is constant in magnitude and either perpendicular or parallel to dA at every point. This requires the charge distribution to have spherical, cylindrical, or planar symmetry. For an irregular or non-symmetric distribution, the integral cannot be simplified and Gauss's law provides no computational advantage over Coulomb's law.
Question 5 Short Answer
Explain why Gauss's law is 'always true but not always useful' for calculating electric fields. What specific condition must be satisfied to make it a practical calculation tool?
Think about your answer, then reveal below.
Model answer: Gauss's law ∮E·dA = Q_enclosed/ε₀ holds for every closed surface, regardless of symmetry or charge distribution. However, knowing the total flux does not by itself tell you E at any particular point — the integral may be a complicated function of both magnitude and direction varying across the surface. The law becomes useful only when you can choose a Gaussian surface where E is constant in magnitude and everywhere perpendicular to the surface (or parallel, contributing zero). Then ∮E·dA = E × (surface area), and E can be solved algebraically. This requires the charge distribution to have high symmetry: spherical symmetry (use a concentric sphere), cylindrical symmetry (use a coaxial cylinder), or planar symmetry (use a pillbox).
The choice of Gaussian surface is the key skill in applying Gauss's law. The surface is mathematical — it does not need to coincide with any physical object. The art is recognizing which geometry exploits the symmetry of the charge distribution to simplify the integral.