Question bank — Maxwell's equations — integral form, all four

False — its field lines enter and exit the surface, so the net flux they contribute is exactly zero; only enclosed charge $Q_{enc}$ counts in Gauss's Law.

False — zero net flux only means $Q_{enc}=0$; equal amounts of $+$ and $-$ charge inside, or an external field, can give strong $\vec{E}$ on the surface with lines going in and back out.

False — it is always zero because there are no magnetic monopoles; $\vec{B}$ lines are closed loops regardless of any current.

False — a steady $\vec{B}$ creates nothing; only a changing flux $d{\Phi }_B/dt$ drives a circulating $\vec{E}$ around the loop $C$.

False — the minus sign is the direction rule (Lenz's law), and it only makes sense once $d\vec{\ell }$ and $d\vec{A}$ are paired by the right-hand rule; dropping it lets induced currents reinforce their own cause, violating energy conservation.

False — no charge crosses the gap; $I_d={\epsilon }_{0}d{\Phi }_E/dt$ is a changing electric flux that produces a magnetic field like a current would.

False — Gauss uses a closed surface (a bag) trapping $Q_{enc}$; Ampère–Maxwell uses an open surface bounded by the loop $C$.

True — $I_d={\epsilon }_{0}d{\Phi }_E/dt=dQ/dt=I$, so the "current" is continuous through the gap even though no charge moves there. See Capacitors.

True — $Q_{enc}=0$, so ${\oint }_S\vec{E}\cdot d\vec{A}=0$; the field is not zero, but every line from the $+$ ends on the $-$ inside the bag.

The cube destroys the symmetry — $\vec{E}$ is neither constant nor perpendicular to $d\vec{A}$ on its faces, so ${\oint }_S\vec{E}\cdot d\vec{A}$ becomes unsolvable. Choose a sphere so $E$ is constant and radial (parallel to $d\vec{A}$) on the surface.

A sphere doesn't match the cylindrical symmetry, so $\vec{E}\cdot d\vec{A}$ varies unpredictably over it; use a coaxial cylinder where flux is uniform on the curved side and zero on the flat ends.

EMF is $-d{\Phi }_B/dt$, the rate of change of flux, not the flux itself. A steady field through a still loop gives $d{\Phi }_B/dt=0$, hence zero EMF.

It fails for a charging capacitor: a flat surface through the wire gives $I_{enc}=I$, but a bulged surface (bounded by the same loop $C$) passing between the plates catches no current — contradiction, fixed only by adding ${\mu }_0{\epsilon }_{0}d{\Phi }_E/dt$.

The integral is over a closed surface and only says lines in = lines out; the field itself is very much nonzero — flux cancels, field does not.

Once launched, the changing $\vec{E}$ feeds a circulating $\vec{B}$ and vice versa (Faraday + Ampère–Maxwell), so an EM wave sustains itself in empty space with no charges present.

Field strength falls as $1/r^2$ while sphere area grows as $r^2$, so their product (the flux ${\Phi }_E$) is $r$-independent — this is why Gauss's law holds for any radius.

They set circulation ${\oint }_C$ equal to flux through the surface that the loop $C$ bounds, with the right-hand rule pairing $d\vec{\ell }$ to $d\vec{A}$; a closed surface has no bounding loop, so there is no $C$ and no consistent normal to pair — the equation would have nothing on its left side.

Real current stops at the capacitor plate, but the growing $\vec{E}$ supplies exactly ${\epsilon }_{0}d{\Phi }_E/dt=dQ/dt$, so the total current threading any surface bounded by the loop $C$ is the same.

A charge's field points outward and does zero net work around a loop; the induced field circulates (${\oint }_C\vec{E}\cdot d\vec{\ell }\ne 0$), doing real work per charge — that's the EMF.

Each law's changing field drives the other; combining them forces a wave equation whose speed is fixed by the constants ${\mu }_0$ and ${\epsilon }_0$, giving $c=1/\sqrt{{\mu }_0{\epsilon }_0}$.

No isolated magnetic poles have ever been found, so there is no magnetic charge to enclose; the right side is structurally zero.

${\Phi }_B=B\times \text{area}=0$ for all time, so $d{\Phi }_B/dt=0$ and the EMF is zero — no area, no flux to change.

No — $\vec{E}$ is constant, so $d{\Phi }_E/dt=0$ and $I_d=0$; the displacement current vanishes and no $\vec{B}$ circulates.

Zero — a uniform field has as many lines entering as leaving, so ${\Phi }_E=0$, which by Gauss's law means no enclosed charge.

The surface integral is ambiguous at a mathematical point, so the honest fix is to nudge the surface infinitesimally: if you deform it to just enclose the charge you get $q/{\epsilon }_0$, and to just exclude it you get $0$. For a smooth surface the symmetric answer is half, $q/(2{\epsilon }_0)$ — but always resolve the ambiguity by deciding which side the charge is on, never leave it on the surface.

No — that would be a magnetic monopole; Gauss's law for magnetism forbids it, so the line must close on itself.

Yes — a changing $\vec{B}$ produces a real circulating $\vec{E}$ in space itself; the wire only reveals it by carrying current, it doesn't create it.

Yes — the integral law is exact for whatever charge is enclosed at that instant. The subtlety is that the field pattern on the surface is not the instantaneous Coulomb field: changes propagate at speed $c$ (retarded fields), so the local $\vec{E}$ lags the charge — yet those retardation effects rearrange the flux so the total still equals $Q_{enc}/{\epsilon }_0$ exactly.

Yes — the wire current is replaced by the displacement current from the changing $\vec{E}$ between the plates, giving the identical answer; that consistency was Maxwell's whole point.