Worked examples — Gauss's law — integral form, choosing Gaussian surfaces

This page is the exercise gym for the parent Gauss's law note. We don't teach new theory here — we grind through every kind of situation the law can hand you, so that when an exam invents a "new" problem, you recognise which cell of the matrix it lives in.

Everything below rests on the one boxed law: ${\oint }_S\vec{E}\cdot d\vec{A}=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

If any symbol here feels unfamiliar, pause and reread the parent — this page assumes you already met Electric flux (the "how many field lines poke out" quantity), Coulomb's law (the $1/r^2$ field of a point charge), and the symmetry-matching rule for picking surfaces.

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The scenario matrix

Before working examples, let's name every case class this topic can produce. Each cell is a distinct kind of trap. The examples that follow are labelled by the cell(s) they cover.

#	Case class	What's different about it	Covered by
A	Positive point charge (baseline sphere)	field points outward, flux $>0$	Ex 1
B	Negative charge / sign flip	field points inward, flux $<0$	Ex 2
C	Charge outside the surface (degenerate: $Q_{\text{enc}}=0$)	flux is exactly zero though $\vec{E}\ne 0$	Ex 2
D	Line symmetry, field far vs near	$E\propto 1/r$, caps give zero	Ex 3
E	Planar symmetry, factor-2 trap	$E$ constant, both faces leak	Ex 4
F	Solid sphere, two regions ($r<R$ and $r>R$)	limiting values, continuity at $r=R$	Ex 5
G	Spherical shell / cavity (degenerate: empty inside)	$E=0$ inside, jump outside	Ex 6
H	Conductor vs insulated sheet ($\sigma /{\epsilon }_0$ vs $\sigma /2{\epsilon }_0$)	which geometry gives which factor	Ex 7
I	Real-world word problem	translate SI numbers, get a real force	Ex 8
J	Exam twist — off-centre / non-uniform	law valid but not solvable; reasoning-only	Ex 9

We hit every row. Let's go.

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Example 1 — Positive point charge (cell A)

Forecast: Guess before reading — will the flux depend on the radius you pick? Will it be positive or negative?

Figure s01 — A positive charge $+q$ (coral dot) at the centre of a dashed Gaussian sphere of radius $r$; lavender arrows show $\vec{E}$ pointing radially outward, all equal in length on the circle.

1. Pick a concentric sphere of radius $r$. Why this step? The charge has spherical symmetry, so $E$ can only depend on $r$ and must point radially (look at the lavender arrows in the figure — all the same length on the dashed circle). This is the only surface on which $E$ is constant and parallel to $d\vec{A}$.
2. Write flux $=E\cdot (4\pi r^2)$. Why this step? Because $\vec{E}\parallel d\vec{A}$ everywhere ($\theta =0$, so $\cos \theta =1$) and $E$ is constant, $\oint EdA=E\oint dA=E\cdot A_{\text{sphere}}$, where $A_{\text{sphere}}=4\pi r^2$ is the whole area of $S$.
3. Set equal to $q/{\epsilon }_0$ and solve. $E\cdot 4\pi r^2=\frac{q}{{\epsilon }_0}\Rightarrow E=\frac{q}{4\pi {\epsilon }_0{r}^2}=\frac{(9.0\times 10^9)(2.0\times 10^{-9})}{(0.30{)}^2}=200\text{N/C}.$ Why this step? $1/(4\pi {\epsilon }_0)=9.0\times 10^9$ is the Coulomb constant $k$ — we just recovered Coulomb's law.
4. Total flux $=q/{\epsilon }_0=\frac{2.0\times 10^{-9}}{8.85\times 10^{-12}}=226{\text{N⋅m}}^2/\text{C}$. Why this step? The total flux is the right-hand side of Gauss's law read directly: the whole surface integral just equals $Q_{\text{enc}}/{\epsilon }_0$, so we don't even need $E$ — the flux depends only on the trapped charge.

Recall Verify Example 1

Units of $E$: $\frac{\text{C}}{({\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}){\text{m}}^2}=\text{N/C}$ ✓. The flux answer contains no $r$ — pick any radius, same flux. Forecast check: flux is $+$ (charge positive), independent of $r$. ✓

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Example 2 — Negative charge, and a charge outside (cells B & C)

Forecast: Which of these gives zero flux? Which gives negative flux?

Figure s02 — Left: negative charge $-q$ (lavender) with coral arrows pointing inward, so net flux is negative. Right: a positive charge $Q$ sits outside a dashed surface $S$; mint field lines pass straight through, entering and leaving, for zero net flux.

Part (a):

1. Flux $=q/{\epsilon }_0$. Why? Gauss's law doesn't care about sign — plug the signed charge straight in. $\Phi =\frac{-3.0\times 10^{-9}}{8.85\times 10^{-12}}=-339{\text{N⋅m}}^2/\text{C}.$
2. Interpret the minus. Why this step? Negative flux means field lines point into the surface — the coral inward arrows in the figure. A negative charge swallows field lines.

Part (b):

1. Identify $Q_{\text{enc}}$. Why? Gauss counts only charge inside the closed surface $S$. The $+5$ nC is outside, so $Q_{\text{enc}}=0$.
2. Conclude $\Phi =0$. $\Phi =\frac{0}{{\epsilon }_0}=0.$ Why this step? Look at the right panel: each field line that enters the surface on one side leaves on the other. Enter $=-$, leave $=+$, they cancel in pairs. The field $\vec{E}$ on the surface is not zero — but the net flux is.

Recall Verify Example 2

(a) Sign matches: negative charge → negative flux ✓. (b) $Q_{\text{enc}}=0\Rightarrow \Phi =0$ exactly, even though $E\ne 0$ pointwise — this is the classic degenerate case C. ✓

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Example 3 — Infinite line charge, near and far (cell D)

Forecast: Doubling the distance — does $E$ drop by a factor of 2, or 4?

Figure s03 — A coral line charge $+\lambda$ along the axis, wrapped by a dashed coaxial cylinder of radius $r$ and length $L$; mint arrows show $\vec{E}$ pointing radially outward through the curved wall, while the flat end caps face along the axis and catch no flux.

1. Coaxial cylinder, radius $r$, length $L$. Why? Line symmetry: $E$ depends only on distance from the wire and points straight out (mint arrows). A cylinder shares that symmetry.
2. Curved wall: flux $=E\cdot (2\pi rL)$. Why? On the wall $\vec{E}\parallel d\vec{A}$ and $E$ constant; here the wall area is $A=2\pi rL$.
3. End caps: flux $=0$. Why? On the flat caps $\vec{E}$ (radial) is perpendicular to $d\vec{A}$ (axial). $\cos 90°=0$.
4. Solve with $Q_{\text{enc}}=\lambda L$: $E(2\pi rL)=\frac{\lambda L}{{\epsilon }_0}\Rightarrow E=\frac{\lambda }{2\pi {\epsilon }_{0}r}.$ Why $L$ cancels? Both the trapped charge and the wall area grow with $L$.
5. Plug numbers: $E(0.020)=\frac{4.0\times 10^{-6}}{2\pi (8.85\times 10^{-12})(0.020)}=3.60\times 10^6\text{N/C},$ $E(0.040)=1.80\times 10^6\text{N/C}.$

Recall Verify Example 3

Ratio $E(0.020)/E(0.040)=2$. Because $E\propto 1/r$ (not $1/r^2$), doubling $r$ halves $E$ — forecast answer is factor 2. ✓

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Example 4 — Infinite charged sheet, the factor-2 trap (cell E)

Forecast: Will the field be weaker at 10 cm than just above?

Figure s04 — A coral charged sheet with a dashed pillbox straddling it; butter arrows show $\vec{E}$ leaving both flat caps (each of area $A$) perpendicular to the sheet, while the side wall is parallel to the field and catches no flux.

1. Pillbox straddling the sheet, cap area $A$. Why? Planar symmetry: field is perpendicular to the sheet, same on both sides. A symmetric box captures that. Here $A$ is the area of one flat cap of the box.
2. Two caps leak, side wall gives zero. Why? Field exits both faces (butter arrows point away on each side) → total cap flux $=2EA$. The wall is parallel to $\vec{E}$ → zero.
3. Solve with $Q_{\text{enc}}=\sigma A$: $2EA=\frac{\sigma A}{{\epsilon }_0}\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}=\frac{2.0\times 10^{-6}}{2(8.85\times 10^{-12})}=1.13\times 10^5\text{N/C}.$ Why does $A$ vanish? It multiplies both sides — the answer can't depend on the size of box we imagined.
4. At 10 cm: still $1.13\times 10^5$ N/C. Why unchanged? No $r$ in the answer — an infinite plane gives a uniform field.

Recall Verify Example 4

Answer contains no distance → same at 0 cm and 10 cm. Forecast answer: no, it stays the same ✓. Note the factor 2 from two faces.

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Example 5 — Solid sphere: inside and outside (cell F, limiting values)

Forecast: Does $E$ grow or shrink as you move outward from the centre, inside the ball?

Figure s05 — A lavender solid charged ball of radius $R$, with a coral dashed Gaussian sphere inside ($r<R$) enclosing only part of the charge, and a mint dashed sphere outside ($r>R$) enclosing all of it; mint arrows show the external field.

1. Inside ($r<R$): $Q_{\text{enc}}=Q\frac{r^3}{R^3}$. Why? Uniform volume density $\rho$ → enclosed charge scales with enclosed volume $\propto r^3$, so the fraction inside radius $r$ is $r^3/R^3$. $E\cdot 4\pi r^2=\frac{Q{r}^3/R^3}{{\epsilon }_0}\Rightarrow E_{\text{in}}=\frac{Qr}{4\pi {\epsilon }_0{R}^3}.$ At $r=0.025$: $E_{\text{in}}=\frac{(9.0\times 10^9)(8.0\times 10^{-9})(0.025)}{(0.050{)}^3}=1.44\times 10^4\text{N/C}.$
2. Outside ($r>R$): $Q_{\text{enc}}=Q$ (all of it). $E_{\text{out}}=\frac{Q}{4\pi {\epsilon }_0{r}^2}.$ At $r=0.10$: $E_{\text{out}}=\frac{(9.0\times 10^9)(8.0\times 10^{-9})}{(0.10{)}^2}=7.2\times 10^3\text{N/C}.$
3. Check continuity at $r=R$. Why? Because the charge is purely volumetric — there is no thin shell of surface charge sitting on the boundary — no extra $Q_{\text{enc}}$ appears exactly at $r=R$, and the field cannot jump. (A surface charge would cause a jump; here there is none.) Evaluate both formulas at $r=R$ and compare the actual numbers: $E_{\text{in}}(R)=\frac{QR}{4\pi {\epsilon }_0{R}^3}=\frac{Q}{4\pi {\epsilon }_0{R}^2}=\frac{(9.0\times 10^9)(8.0\times 10^{-9})}{(0.050{)}^2}=2.88\times 10^4\text{N/C},$ $E_{\text{out}}(R)=\frac{Q}{4\pi {\epsilon }_0{R}^2}=\frac{(9.0\times 10^9)(8.0\times 10^{-9})}{(0.050{)}^2}=2.88\times 10^4\text{N/C}.$ The two give the same $2.88\times 10^4$ N/C, so the field is continuous at the surface.

Recall Verify Example 5

Inside $E\propto r$ → grows linearly from $0$ at centre (forecast: grows). $E_{\text{in}}(R)=E_{\text{out}}(R)=2.88\times 10^4$ N/C — no jump. ✓

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Example 6 — Charged spherical shell / cavity (cell G, degenerate empty inside)

Forecast: What is the field inside a hollow charged shell?

Figure s06 — A hollow coral shell of radius $R$ carrying all the charge; the lavender cavity inside is labelled $E=0$, an inner dashed Gaussian sphere encloses no charge, and mint arrows show the external field beyond the shell.

1. Inside ($r<R$): $Q_{\text{enc}}=0$. Why? All charge lives on the shell at radius $R$; a Gaussian sphere of smaller radius encloses nothing. $E\cdot 4\pi r^2=\frac{0}{{\epsilon }_0}=0\Rightarrow E_{\text{in}}=0.$ Why exactly zero everywhere inside? Symmetry forces $E$ radial and constant on the inner sphere; zero flux + constant $E$ ⇒ $E=0$. The cavity is field-free (lavender region in the figure).
2. Outside ($r>R$): $Q_{\text{enc}}=Q$. $E_{\text{out}}=\frac{Q}{4\pi {\epsilon }_0{r}^2}=\frac{(9.0\times 10^9)(6.0\times 10^{-9})}{(0.20{)}^2}=1.35\times 10^3\text{N/C}.$ Why the same form as a point charge? From outside, a shell looks identical to a point charge at its centre.

Recall Verify Example 6

Inside: $E=0$ (forecast answer: zero — this is the shielding of a cavity, foundational to Electric field of conductors). Outside behaves like a point $Q$. ✓

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Example 7 — Conductor surface vs insulated sheet (cell H)

Forecast: Same $\sigma$ — same field? Or does one double the other?

Figure s07 — Left: an insulated sheet with butter arrows leaving both faces (field on both sides → $\sigma /2{\epsilon }_0$). Right: a conductor (lavender block, $E=0$ inside) with butter arrows on the outer side only (one face leaks → $\sigma /{\epsilon }_0$).

1. Insulated sheet: pillbox leaks from both faces. Why? Field exists on both sides, so $2EA=\sigma A/{\epsilon }_0$ (with $A$ = area of one cap): $E_{\text{sheet}}=\frac{\sigma }{2{\epsilon }_0}=\frac{4.0\times 10^{-6}}{2(8.85\times 10^{-12})}=2.26\times 10^5\text{N/C}.$
2. Conductor surface: pillbox leaks from one face only. Why? Inside a conductor $\vec{E}=0$ (charges rearrange until it is), so the inner cap catches no flux. Only the outer cap leaks: $EA=\sigma A/{\epsilon }_0$: $E_{\text{cond}}=\frac{\sigma }{{\epsilon }_0}=\frac{4.0\times 10^{-6}}{8.85\times 10^{-12}}=4.52\times 10^5\text{N/C}.$
3. Compare. The conductor field is exactly twice the sheet field for the same $\sigma$.

Recall Verify Example 7

$E_{\text{cond}}/E_{\text{sheet}}=2$ exactly — the factor comes purely from how many faces leak (one vs two), not from any new physics. Forecast answer: conductor doubles it. ✓

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Example 8 — Real-world word problem (cell I)

Forecast: A negative speck near a positive wire — pulled in or pushed out?

1. Field of the line at that radius. Why start here? Force needs the field the speck sits in; use the line-charge result from Example 3. $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{1.5\times 10^{-8}}{2\pi (8.85\times 10^{-12})(1.0\times 10^{-3})}=2.70\times 10^5\text{N/C}.$
2. Force magnitude $F=|q|E$. Why the absolute value $|q|$? We first compute only the size of the force; $|q|$ (the magnitude of the charge, ignoring its sign) times $E$ gives a positive number of newtons. We handle the direction separately in the next step, so we deliberately strip the sign here to avoid double-counting it. $F=|q|E=(2.0\times 10^{-15})(2.70\times 10^5)=5.39\times 10^{-10}\text{N}.$
3. Direction. Why? $\vec{E}$ points away from the positive wire, but $q<0$, so $\vec{F}=q\vec{E}$ (with the sign restored) points toward the wire — the speck is pulled in. That's exactly how toner is attracted to the drum.

Recall Verify Example 8

Units: $(\text{C})(\text{N/C})=\text{N}$ ✓. Sign of $q$ flips the force inward. Forecast answer: pulled in. ✓

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Example 9 — Exam twist: valid but not solvable (cell J)

Forecast: Does the off-centre position change the flux? Does it change your ability to solve for $E$?

1. Net flux (part a). Why unchanged? Gauss counts only enclosed charge, not its position. As long as $q$ is inside the closed surface $S$: $\Phi =\frac{q}{{\epsilon }_0}\text{(same as if it were centred).}$
2. Finding $\vec{E}$ (part b). Why not solvable? The charge being off-centre breaks spherical symmetry: $E$ now varies from point to point on the surface (stronger on the near side, weaker on the far side). Because $E$ is no longer constant over $S$, you cannot pull it out of the integral $\oint \vec{E}\cdot d\vec{A}$. The equation $\oint \vec{E}\cdot d\vec{A}=q/{\epsilon }_0$ stays true, but it is one number balancing infinitely many unknown values of $E$ across the surface — so Gauss's law alone cannot give you $\vec{E}$ here.
3. What to do instead. Why switch tools? Since symmetry is gone, use Coulomb's law / superposition directly for $\vec{E}$; keep Gauss only for the total flux. Conclusion: flux is fixed at $q/{\epsilon }_0$, but the field is not obtainable by Gauss's law without symmetry.

Recall Verify Example 9

Flux $=q/{\epsilon }_0$ regardless of position (validity). Symmetry broken ⇒ not solvable for $E$ (usability). Forecast: flux unchanged, but $E$ not obtainable by Gauss. ✓

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Recap of the matrix coverage

Recall Which example hit which cell?

A→Ex1, B & C→Ex2, D→Ex3, E→Ex4, F→Ex5, G→Ex6, H→Ex7, I→Ex8, J→Ex9. Every cell filled.

Related: Divergence theorem converts this integral law to the local form in Maxwell's equations; the shielding in Ex 6 underlies Electric field of conductors.