1.8.6 · D5Electromagnetism
Question bank — Gauss's law — integral form, choosing Gaussian surfaces
Prerequisites worth re-reading if a line stings: Electric flux, Coulomb's law, Symmetry in physics, Electric field of conductors, Divergence theorem.
True or false — justify
The inside is only the field made by the enclosed charge.
False — is the total field from all charges, inside and outside. Outside charges change at each point on the surface; they just add up to zero net flux.
If the net flux through a closed surface is zero, then everywhere on that surface.
False — zero net flux only means as many lines leave as enter. A surface around a dipole (or in a uniform external field) can have strong everywhere yet zero net flux.
Gauss's law is valid for any closed surface, in any situation.
True — the law always holds; it follows from Coulomb + superposition. Validity is unconditional — only solvability for needs symmetry.
Because only appears on the right, you can always solve for .
False — validity is not solvability. Without symmetry you cannot pull out of the integral, so the equation, though true, gives no value of (e.g. a finite rod).
Moving a charge around inside a fixed closed surface changes the net flux.
False — net flux depends only on total , not on where it sits inside. Re-arranging trapped charge changes pointwise but not the total leakage.
For an infinite sheet, the field magnitude is independent of distance from the sheet.
True — an infinite plane looks the same from any distance (constant solid-angle coverage), so field lines never spread out; stays uniform.
The field of a conductor's surface is , twice that of an isolated charged sheet.
True — a conductor has zero field inside the metal, so all flux escapes through the one outer face (). The isolated sheet leaks through both faces, splitting the field in half.
Doubling the length of the coaxial cylinder around a line charge doubles the computed .
False — both and the curved area scale with , so cancels. The answer is length-independent.
Spot the error
"A field line from an external charge that enters the surface must eventually leave, so external charge contributes flux then , totalling ."
The two contributions are then of the same magnitude, so they cancel to zero, not add. External charge contributes exactly zero net flux.
"For the infinite line charge I'll use a coaxial cylinder, and the flat end caps give flux each."
The end caps face along the axis, while points radially outward — perpendicular to the caps' normal. Their flux is , not .
"For an infinite plane I put a pillbox with only its top face outside; flux is , so ."
A symmetric pillbox straddling the sheet leaks through both faces: total flux , giving . The one-face pillbox is only valid for a conductor, where the inner face sits in field-free metal.
"The point charge is off-centre inside my spherical surface, so varies over the sphere — but I'll still write ."
With an off-centre charge is not constant on the sphere, so you cannot factor out of the integral. The flux is still (Gauss holds), but this surface can't give you .
"For a finite charged rod I use a coaxial cylinder just like the infinite case and get ."
A finite rod lacks translational symmetry — the field is not purely radial and varies along the length, so can't leave the integral. Gauss is true but useless here; you'd need direct integration of Coulomb's law.
"Inside a uniformly charged solid sphere, more volume is enclosed as grows, and , so blows up near the centre."
grows faster than : , so shrinks linearly to zero at the centre, not blows up.
"The net flux told me , so there is no charge anywhere inside."
Zero net enclosed charge means equal positive and negative charge inside (e.g. a dipole), not necessarily no charge. Only the algebraic sum is zero.
Why questions
Why does the sphere's radius disappear from the point-charge flux?
Coulomb's shrinks the field exactly as the sphere's area grows, so their product is -independent — this cancellation is the heart of Gauss's law.
Why can external charges affect on the surface but never the net flux?
Every field line from an outside charge that enters must exit (it terminates only on charge, and there's none inside to catch it), so its inward and outward flux cancel — while pointwise it still bends .
Why do we match the surface to the symmetry rather than pick any convenient shape?
Only a surface aligned with the field's symmetry keeps either constant-and-parallel () or perpendicular () on each face, which is the sole way to pull out of the integral. See Symmetry in physics.
Why does the solid-angle argument prove the surface shape doesn't matter?
Tilting a patch multiplies its area by but the flux-catching component by , and greater distance dilutes the field by exactly the compensating — all shape effects cancel, leaving only the total solid angle .
Why is the factor 2 present for the plane but absent for the line-charge caps?
The plane's field escapes through both pillbox faces (two leaking surfaces → ); the cylinder's caps are perpendicular to the radial field and leak nothing, so only the curved wall counts.
Why does Gauss's law reduce to the Divergence theorem statement about ?
The divergence theorem turns into a volume integral of ; equating it to for every volume forces — the differential form and one of Maxwell's equations.
Why is the field zero inside a hollow charged conducting shell?
A Gaussian sphere inside the cavity encloses no charge (), and spherical symmetry forces the constant radial to satisfy , so . See Electric field of conductors.
Edge cases
What is the net flux through a closed surface with no charge inside but a strong nearby external charge?
Exactly zero — so , even though everywhere on the surface due to the external charge.
What happens to the point-charge field formula as ?
— the idealized point charge gives a genuine singularity; real charges have finite size, so the divergence is a modelling artefact, not physics.
What is exactly at the centre of a uniformly charged solid sphere?
Zero — a Gaussian sphere of radius encloses vanishing charge, and by symmetry the field must point nowhere, so .
What is the flux through a surface that slices through a point charge (charge sitting exactly on it)?
Ill-defined / degenerate — "enclosed" is ambiguous when charge lies on the boundary; the idealization breaks. In practice split the charge or perturb the surface so the charge is cleanly in or out.
For a charge exactly on the boundary between two touching Gaussian surfaces, how is its flux shared?
By symmetry a point charge on a flat boundary sends half its lines each way, so each surface catches — but this only works for symmetric placement and is best avoided by shifting the surface.
What net flux comes from a perfectly neutral atom (dipole) fully enclosed?
Zero net flux, since ; the field is highly non-uniform on the surface but every outgoing line from returns to inside.
As the radius of a Gaussian sphere around a fixed point charge , what happens to the flux?
It stays exactly — flux never depends on . The field weakens as but the sphere's area grows as , keeping the product fixed.
Recall One-line survival summary
Gauss is always true but only solves for under symmetry; the right side counts only enclosed charge, the left side uses the total field. Reveal ::: Net flux = ; solvability needs a symmetry-matched surface where each face gives or .