Question bank — Applications — sphere, cylinder, infinite plane

False. Gauss's law $\oint \vec{E}\cdot d\vec{A}=Q_{enc}/{\epsilon }_0$ is always true; symmetry is only what lets us solve for $E$ by pulling it out of the integral. Without symmetry the law still holds — you just can't isolate $E$.

False. Zero net flux means as many arrows enter as leave (so $Q_{enc}=0$), but the field can still be strong and non-zero at every point — think of a surface sitting in the uniform field between capacitor plates.

False in general, true only when symmetry makes $E$ uniform on the surface. Doubling $Q_{enc}$ doubles the total flux, but if the field is uneven over the surface the individual point values need not double proportionally — only the integral is pinned down.

True. For $r>R$, both give $E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$. From outside, a spherically symmetric ball is indistinguishable from all its charge concentrated at the center.

False. Inside the insulator only the charge within radius $r$ counts, giving $E=\frac{\rho r}{3{\epsilon }_0}$, which grows linearly from zero at the center. It is zero only at the exact center, not throughout.

False. It falls off as $1/s$: $E=\frac{\lambda }{2\pi {\epsilon }_{0}s}$. The enclosed charge grows with the box length $L$ while area also grows with $L$, so one power of distance-dependence is "used up" differently than for a point.

False. It is constant, $E=\frac{\sigma }{2{\epsilon }_0}$, independent of distance. As you step back you see more of the sheet at wide angles, which exactly cancels the inverse-square weakening of each patch.

False — a classic trap. Outside charges contribute zero to the net flux (their arrows go in one side and out the other), but they absolutely contribute to $\vec{E}$ at each point. Only the integral is blind to them, not the field itself.

False. It is $\sigma /{\epsilon }_0$. A conductor keeps its interior field at zero, so all the flux from the surface charge is forced out one side, doubling the isolated-sheet value.

False. $\sigma /(2{\epsilon }_0)$ is the isolated sheet result (field on both sides). A conducting surface, or the region between two capacitor plates, gives $\sigma /{\epsilon }_0$. The material and the geometry decide which applies.

Error: inside a conducting shell $Q_{enc}=0$, so $E=0$. The point-charge formula only holds outside ($r>R$), where the enclosed charge is the full $Q$.

Error: the end caps carry zero flux because $\vec{E}$ there is parallel to the caps ($\perp$ to their outward normal). Only the curved side, area $2\pi sL$, contributes.

Error: $L$ cancels — it multiplies both $Q_{enc}=\lambda L$ and the curved area $2\pi sL$. Its cancellation is the proof that the field depends only on $s$, not on your arbitrary box length.

Error: on a cube $E$ is not constant over the faces (corners are farther than face centers) and $\vec{E}$ is not parallel to $d\vec{A}$ everywhere. You cannot pull $E$ out of the integral — a sphere is required to exploit the symmetry.

Error: the pillbox has two faces, each passing flux $EA$, so the left side is $2EA$, giving $E=\sigma /(2{\epsilon }_0)$. Forgetting the second face is the most common plane mistake.

Error: $Q_{enc}=(+q)+(-q)=0$, so net flux is zero. The field is far from zero — the dipole makes a rich pattern — but every arrow that leaves also returns.

Error: inside, only the fraction within radius $r$ is enclosed: $Q_{enc}=Q(r/R{)}^3=\rho \cdot \frac{4}{3}\pi r^3$. Using the full $Q$ inside overestimates the field.

Error: that formula assumes an infinite (or effectively infinite) line, valid only when $s$ is much smaller than the wire's length. At $s=2$ m from a 10 cm rod the symmetry is broken — the end caps no longer give zero flux and Gauss's law can't isolate $E$.

Because the Gaussian surface must match the symmetry of the source so that $E$ is constant and either parallel or perpendicular to $d\vec{A}$ on each piece. A sphere achieves this only for radial (point/ball) fields; a line's field is uniform on a coaxial cylinder, and a plane's on a pillbox.

On the curved side $\vec{E}$ points radially, straight along $d\vec{A}$, so $\vec{E}\cdot d\vec{A}=EdA$ (full contribution). On the flat caps $\vec{E}$ is perpendicular to $d\vec{A}$, so $\vec{E}\cdot d\vec{A}=0$.

The source geometry sets the power law. A point sends its flux outward through spheres whose area grows as $r^2$, diluting $E$ as $1/r^2$. An infinite plane's flux only ever crosses the two fixed pillbox faces of area $A$, which don't grow with distance — so nothing dilutes.

An isolated sheet radiates equally to both sides, splitting its flux ($\frac{\sigma }{2{\epsilon }_0}$ each way). A conductor holds its interior at $E=0$, so all the flux from the surface charge is forced out one side — the same total flux through half the area doubles $E$.

Because $Q_{enc}=\lambda L$ and the curved area $=2\pi sL$ both scale with $L$. Its cancellation confirms the answer is independent of an arbitrary choice we made — a self-consistency check that our surface was legitimate.

By symmetry (Symmetry in Physics): there is no special sideways direction on a spherically symmetric charge, so any non-radial component would have no reason to point one way rather than another — it must be zero. This argument, not the algebra, is what makes the problem solvable.

Between the plates both single-sheet fields ($\frac{\sigma }{2{\epsilon }_0}$ each) point the same way and add to $\frac{\sigma }{{\epsilon }_0}$. Outside, the $+\sigma$ sheet's field and the $-\sigma$ sheet's field point oppositely and cancel exactly.

Zero. A Gaussian sphere of radius $r\to 0$ encloses vanishing charge ($\propto r^3$) while its area $\propto r^2$, so $E=\frac{\rho r}{3{\epsilon }_0}\to 0$. Every direction is equivalent, so the field cancels by symmetry too.

For an insulating ball the inside formula $\frac{\rho R}{3{\epsilon }_0}$ and outside formula $\frac{Q}{4\pi {\epsilon }_0{R}^2}$ meet at the same value — the field is continuous. For a conducting shell it jumps from $0$ (inside) to $\frac{\sigma }{{\epsilon }_0}$ (just outside) across the charged surface.

No real line is infinitely thin, so a genuine wire has a finite radius where the formula stops applying (you're now inside the conductor, $E=0$). The divergence is an artifact of the zero-thickness idealization.

$Q_{enc}=0$, so net flux is zero, even though the atom produces a real (dipole-like) field. Flux counts net charge only.

Just inside, $Q_{enc}=0\Rightarrow E=0$; just outside, $Q_{enc}=Q\Rightarrow E=\frac{Q}{4\pi {\epsilon }_0{R}^2}$. The shell creates a sharp discontinuity of size $\frac{\sigma }{{\epsilon }_0}$ across itself.

Yes — the net flux is still $Q/{\epsilon }_0$ because $Q_{enc}=Q$. But you can no longer solve for $E$: the field is not constant over an off-center sphere, so the integral won't collapse to $E\times \text{area}$.

Yes — flux depends only on total $Q_{enc}=2q$, so it doubles. But the field pattern is not simply "twice one charge's" everywhere; only the integrated flux obeys the clean doubling.

No. An external charge contributes zero net flux (its arrows enter and leave). It does perturb $\vec{E}$ at points on the box, but $\oint \vec{E}\cdot d\vec{A}$ still equals $\sigma A/{\epsilon }_0$ from the enclosed sheet charge alone.