Exercises — Gauss's law — integral form, choosing Gaussian surfaces

The shape does not matter — Gauss's law only asks for the charge trapped inside. Using $\Phi \equiv {\oint }_S\vec{E}\cdot d\vec{A}$ (the leakage symbol defined above), $\Phi =\frac{Q_{\text{enc}}}{{\epsilon }_0}=\frac{3.0\times 10^{-9}}{8.85\times 10^{-12}}\approx 339{\text{N⋅m}}^2/\text{C}.$ A nanocoulomb is $10^{-9}\text{C}$. We divided the trapped charge by ${\epsilon }_0$; the cube's size and orientation are irrelevant.

Zero. With $d\vec{A}$ pointing outward everywhere (see the sign-convention box), a field line from an outside charge that enters the sphere gives $\vec{E}\cdot d\vec{A}<0$ ($-$), and where the same line leaves it gives $\vec{E}\cdot d\vec{A}>0$ ($+$). Each entry is matched by an exit (Step 3 of the parent derivation), so they cancel. No charge is enclosed, so $Q_{\text{enc}}=0$ and $\Phi =0/{\epsilon }_0=0$. Note: the field $\vec{E}$ on the surface is not zero — the outside charge is felt pointwise — but the net flux cancels.

Match the surface to the symmetry so that on every face, $\vec{E}$ is either parallel to $d\vec{A}$ (flux $=EA$) or perpendicular (flux $=0$).

• (a) point charge → concentric sphere (spherical symmetry).
• (b) infinite line → coaxial cylinder (cylindrical symmetry).
• (c) infinite plane → pillbox straddling the sheet (planar symmetry).

This is a genuine ambiguity — a mathematical point on a zero-thickness surface is neither in nor out, so $Q_{\text{enc}}$ is not well defined and Gauss's law gives no clean answer.

• The physical resolution: real charges have finite size, so nudge the surface an infinitesimal amount to one side. If the surface passes symmetrically through an ideal point charge, half its field lines poke out and half stay in, so you count half the charge: $Q_{\text{enc}}=q/2$, giving $\Phi =q/2{\epsilon }_0$.
• Practical advice: never draw a Gaussian surface through a charge. Choose it so every charge is unambiguously inside or outside; then all the bookkeeping is clean.

Surface: sphere of radius $r$. On it $E$ is constant and parallel to $d\vec{A}$, so $\Phi =E(4\pi r^2)$. $E=\frac{q}{4\pi {\epsilon }_0{r}^2}=\frac{4.0\times 10^{-9}}{4\pi (8.85\times 10^{-12})(0.20{)}^2}\approx 899\text{N/C}.$ Using $\frac{1}{4\pi {\epsilon }_0}\approx 8.99\times 10^9$: $E=8.99\times 10^9\cdot \frac{4.0\times 10^{-9}}{0.04}=899\text{N/C}$.

Read the figure: the field arrows cross the curved wall perpendicularly (parallel to $d\vec{A}$ → flux $=EA$) but run parallel to the flat end caps (never pierce them → flux $=0$).

Surface: coaxial cylinder, radius $r$, length $L$.

• Curved wall: $\vec{E}$ radial $\parallel d\vec{A}$, constant → flux $=E(2\pi rL)$.
• End caps: $\vec{E}\perp d\vec{A}$ → flux $=0$.
• $Q_{\text{enc}}=\lambda L$. The $L$ cancels: $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{2.5\times 10^{-8}}{2\pi (8.85\times 10^{-12})(0.05)}\approx 8990\text{N/C}.$

Surface: pillbox straddling the sheet. Field exits both flat faces, so $\Phi =2EA$; the side wall gives $0$. With $Q_{\text{enc}}=\sigma A$: $E=\frac{\sigma }{2{\epsilon }_0}=\frac{1.0\times 10^{-6}}{2(8.85\times 10^{-12})}\approx 5.65\times 10^4\text{N/C}.$ This is independent of distance — the field is uniform.

(a) Inside ($r<R$): the Gaussian sphere of radius $r$ traps only the charge inside its own volume. With $V_{\text{enc}}=\frac{4}{3}\pi r^3$ and $V_{\text{total}}=\frac{4}{3}\pi R^3$ (defined above), $\frac{V_{\text{enc}}}{V_{\text{total}}}=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}=\frac{r^3}{R^3}\Rightarrow Q_{\text{enc}}=Q\frac{r^3}{R^3}.$ The $\frac{4}{3}\pi$ cancels, leaving the clean $r^3/R^3$. Then $E(4\pi r^2)=\frac{Q{r}^3/R^3}{{\epsilon }_0}\Rightarrow E=\frac{Qr}{4\pi {\epsilon }_0{R}^3}.$ $E=8.99\times 10^9\cdot \frac{6.0\times 10^{-9}(0.05)}{(0.10{)}^3}=2697\text{N/C}.$ (b) Outside ($r>R$): the whole $Q$ is enclosed; it acts like a point charge: $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}=8.99\times 10^9\cdot \frac{6.0\times 10^{-9}}{(0.20{)}^2}=1349\text{N/C}.$ Degenerate check: at $r=0$ (centre), $E=0$ — by symmetry no direction is preferred, and indeed $E\propto r\to 0$ (the origin dot in the figure).

(a) Inside ($r<R$): the Gaussian sphere encloses no charge (all charge is on the outer shell), so $Q_{\text{enc}}=0$ and $E=0.$ The field inside a uniform shell is exactly zero everywhere — the pulls from opposite sides cancel. (b) Outside ($r>R$): all $Q$ enclosed, point-charge field: $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}=8.99\times 10^9\cdot \frac{6.0\times 10^{-9}}{(0.20{)}^2}=1349\text{N/C}.$ Contrast with L3.1: outside, solid and shell agree (both look like a point charge). Inside, the solid sphere has $E\ne 0$ but the shell has $E=0$ — the enclosed charge differs.

The enclosed charge is the signed sum: $Q_{\text{enc}}=+3+(-8)=-5\text{nC}$. (a) Flux: $\Phi =\frac{Q_{\text{enc}}}{{\epsilon }_0}=\frac{-5.0\times 10^{-9}}{8.85\times 10^{-12}}\approx -565{\text{N⋅m}}^2/\text{C}.$ The flux is negative — with $d\vec{A}$ pointing outward, a negative $\Phi$ means the field points, on net, inward. Field lines are terminating on the net negative charge inside. (b) Magnitude and direction: $|E|=\frac{|Q_{\text{enc}}|}{4\pi {\epsilon }_0{r}^2}=8.99\times 10^9\cdot \frac{5.0\times 10^{-9}}{(0.20{)}^2}=1124\text{N/C},$ pointing radially inward (toward the net negative charge).

Each sheet alone makes a uniform field of magnitude $\frac{\sigma }{2{\epsilon }_0}$ pointing away from a $+$ sheet and toward a $-$ sheet. Superpose (add the vectors):

• Between (red arrows region): both fields point the same way (from $+$ toward $-$), so they add: $E=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}=\frac{2.0\times 10^{-6}}{8.85\times 10^{-12}}\approx 2.26\times 10^5\text{N/C}.$
• Outside: the two fields point in opposite directions and cancel: $E=0.$ This is why a parallel-plate capacitor confines its field to the gap.

Why we may add the flux contributions (superposition): The real field is the vector sum of the field made by the wire and the field made by the point charge, $\vec{E}={\vec{E}}_{\text{wire}}+{\vec{E}}_{\text{point}}$ (this is Coulomb's law plus superposition of forces/fields). The flux integral is linear in $\vec{E}$: $\oint ({\vec{E}}_{\text{wire}}+{\vec{E}}_{\text{point}})\cdot d\vec{A}=\oint {\vec{E}}_{\text{wire}}\cdot d\vec{A}+\oint {\vec{E}}_{\text{point}}\cdot d\vec{A}.$ A sum inside an integral splits into a sum of integrals, so the total flux is just the flux each charge would produce on its own — that is why fluxes superpose. Each piece equals its own enclosed charge over ${\epsilon }_0$, so the totals combine into one $Q_{\text{enc}}$.

The enclosed charge. The cylinder traps a length $L$ of the wire plus the whole point charge: $Q_{\text{enc}}=\lambda L+q=(1.0\times 10^{-8})(0.30)+2.0\times 10^{-9}=3.0\times 10^{-9}+2.0\times 10^{-9}=5.0\times 10^{-9}\text{C}.$

The total flux. $\Phi =\frac{Q_{\text{enc}}}{{\epsilon }_0}=\frac{5.0\times 10^{-9}}{8.85\times 10^{-12}}\approx 565{\text{N⋅m}}^2/\text{C}.$ Notice we found the flux even though the combined field is not symmetric enough to pull $E$ out of the integral — flux only ever needs the enclosed charge, never the detailed field.

The enclosed charge is the integral of $\rho$ over the ball of radius $r$. Using shells of thickness $d{r}^{\prime }$ (volume $4\pi r^{\prime 2}d{r}^{\prime }$): $Q_{\text{enc}}={\int }_0^r\alpha r^{\prime }(4\pi r^{\prime 2})d{r}^{\prime }=4\pi \alpha {\int }_0^r{r}^{\prime 3}d{r}^{\prime }=4\pi \alpha \frac{r^4}{4}=\pi \alpha r^4.$ Then by Gauss on a sphere of radius $r$: $E(4\pi r^2)=\frac{\pi \alpha r^4}{{\epsilon }_0}\Rightarrow E=\frac{\alpha r^2}{4{\epsilon }_0}.$ Plug in: $E=\frac{(3.0\times 10^{-5})(0.05{)}^2}{4(8.85\times 10^{-12})}=\frac{7.5\times 10^{-8}}{3.54\times 10^{-11}}\approx 2119\text{N/C}.$ Sanity: here $E\propto r^2$ (steeper than the uniform-density $E\propto r$) because charge piles up outward.

For a conductor, the field inside the metal is zero (see Electric field of conductors). Draw a pillbox with one face inside the metal (flux $=0$ there) and one face just outside (flux $=EA$). Only one face leaks: $EA=\frac{\sigma A}{{\epsilon }_0}\Rightarrow E=\frac{\sigma }{{\epsilon }_0}=\frac{1.0\times 10^{-6}}{8.85\times 10^{-12}}\approx 1.13\times 10^5\text{N/C}.$ The isolated sheet leaks through both faces → $2EA=\sigma A/{\epsilon }_0\Rightarrow E=\sigma /2{\epsilon }_0$. Same law, different number of leaking faces — that is the whole difference.

• As $r\to 0^{+}$: $E\to \infty$. The idealised line has all its charge crammed onto a zero-thickness filament, so hugging it gives an unbounded field. (Real wires have finite radius, capping this.)
• As $r\to \infty$: $E\to 0$, but only like $1/r$ — slowly.
• Contrast: a point charge falls off like $1/r^2$ because its lines spread over a sphere ($\propto r^2$). A line's lines spread over a cylinder ($\propto r$), one dimension less, so they thin out more gently — $1/r$. The geometry of how field lines dilute is the exponent.

1. Identify the symmetry → pick sphere / cylinder / pillbox, with $d\vec{A}$ pointing outward.
2. Split the surface into faces; mark each flux as $EA$, $0$, or $2EA$.
3. Count only the enclosed charge, adding charges with their signs (integrate $\rho$ if non-uniform); never let a Gaussian surface pass through a charge.
4. Solve for the field: set $E\times (\text{leaking area})=Q_{\text{enc}}/{\epsilon }_0$ and divide by the leaking area to get $E$.
5. Check limits and signs — a net negative $Q_{\text{enc}}$ gives negative flux and an inward field.