To find the electric field at distance r from a uniformly charged sphere, a physicist chooses a spherical Gaussian surface centered on the sphere. Why is this the right choice?
AThe divergence theorem only applies to spherical surfaces
BThe electric field is always strongest on a sphere, maximizing the flux integral
CBy symmetry, E is radial and uniform in magnitude on the sphere, so it factors out of the flux integral, reducing ∯ E·dA to E · 4πr²
DA sphere encloses the maximum possible charge for a given radius
The power of Gauss's law in integral form is that it lets you factor E out of the integral when symmetry guarantees E is constant in magnitude and perpendicular to the surface everywhere. A concentric sphere achieves this for a spherically symmetric charge distribution. Without this symmetry, the integral form is harder to use than the differential form — you cannot simply 'pull E outside the integral.'
Question 2 Multiple Choice
Why did Maxwell add the displacement current term (ε₀ ∂E/∂t) to Ampère's law?
ATo account for current flowing through the dielectric material of a capacitor
BTo correct for the magnetic permeability of free space
CSo that Ampère's law gives the same result regardless of which surface bounded by the same Amperian loop is chosen — without it, a charging capacitor produces contradictory answers
DTo include the contribution of magnetic monopoles to the circulation of B
When a capacitor charges, current flows in the wire but not through the gap. If you compute the Amperian circulation using a flat surface cutting through the wire, you get μ₀I. If you use a bulging surface that passes through the capacitor gap (no conduction current passes through it), you get zero — a contradiction for the same loop. Maxwell resolved this by adding ε₀ ∂E/∂t, which is nonzero in the gap during charging and restores consistency regardless of surface choice.
Question 3 True / False
Maxwell's equations in integral form contain additional physical laws beyond those expressed in the differential forms.
TTrue
FFalse
Answer: False
The integral and differential forms are mathematically equivalent — they express exactly the same physics. The integral forms are derived from the differential forms by applying the divergence theorem (to Gauss's laws) and Stokes' theorem (to Faraday's and Ampère-Maxwell's laws). No new physics enters; you are simply converting local point-by-point relationships into global statements about finite surfaces and loops.
Question 4 True / False
For Gauss's law in integral form to be a practical tool for calculating electric field strength, the charge distribution must have enough symmetry that E is constant in magnitude over a well-chosen Gaussian surface.
TTrue
FFalse
Answer: True
This is the key condition. When E is constant and perpendicular to the surface everywhere, ∯ E·dA simplifies to E · A (total surface area), and you can immediately solve for E. Without this symmetry, E varies over the surface and the integral cannot be evaluated without knowing E in advance — defeating the purpose. This is why Gauss's law in integral form is most useful for spherical, cylindrical, and planar symmetry.
Question 5 Short Answer
Why are the integral forms of Maxwell's equations more practically useful than the differential forms for high-symmetry problems, and which two theorems connect the two forms?
Think about your answer, then reveal below.
Model answer: The integral forms are more useful in symmetric problems because symmetry allows you to choose a Gaussian surface or Amperian loop where the field is constant in magnitude and either parallel or perpendicular to the surface/path element everywhere — letting you pull the field outside the integral and solve in one algebraic step. The two connecting theorems are the divergence theorem (converts the divergence equations ∇·E = ρ/ε₀ and ∇·B = 0 into surface flux integrals) and Stokes' theorem (converts the curl equations for E and B into line integrals around closed loops).
The differential forms are more general — they apply at every point in arbitrary geometries and are the starting point for deriving wave equations. But for textbook problems involving a point charge, an infinite wire, or a solenoid, integral forms with a smart surface choice turn a PDE problem into a single equation.