Exercises — Applications — sphere, cylinder, infinite plane

Symmetry: spherical. Surface: a sphere of radius $r=0.15\text{m}$, which is outside $R=0.05\text{m}$. Why the flux collapses here: on this concentric sphere, spherical symmetry forces $\vec{E}$ to have the same magnitude everywhere and to point straight out — exactly along every tile's $d\vec{A}$. So each tile contributes $E\times (\text{tile area})$, and summing the tile areas gives the sphere's area $4\pi r^2$. That is why $\oint \vec{E}\cdot d\vec{A}=E(4\pi r^2)$ and the integral becomes plain multiplication. Since we are outside, all the charge is enclosed and it acts like a point charge: $E_{out}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}=\frac{(9\times 10^9)(3\times 10^{-6})}{(0.15{)}^2}$ Numerator $(9\times 10^9)(3\times 10^{-6})=2.7\times 10^4$; denominator $(0.15{)}^2=0.0225$. Dividing the top by the bottom (this is just $E=\text{numerator}/r^2$) gives: $E=\frac{2.7\times 10^4}{0.0225}=1.2\times 10^6\text{N/C}.$ Direction: radially outward (positive charge pushes a test charge away).

Symmetry: planar. Why the flux collapses here: on a pillbox piercing the sheet, $\vec{E}$ is perpendicular to the sheet and so it is parallel to $d\vec{A}$ on the two flat faces (each contributes $EA$) but perpendicular to $d\vec{A}$ on the side walls (those tiles skim sideways → zero). That is why only the two faces count and $\oint \vec{E}\cdot d\vec{A}=2EA$. The isolated-sheet result is distance-independent: $E=\frac{\sigma }{2{\epsilon }_0}=\frac{2\times 10^{-6}}{2(8.85\times 10^{-12})}=1.13\times 10^5\text{N/C}.$ Same value at $4\text{cm}$ and at $40\text{cm}$ — that is the whole point: an infinite sheet's field never weakens with distance. Direction: the sheet is positive, so the field points away from the sheet on both sides — pointing left on the left face and right on the right face, everywhere perpendicular to the sheet.

Symmetry: axial (cylindrical). Why the flux collapses here: on a coaxial cylinder, $\vec{E}$ points radially outward, so on the curved side it is parallel to $d\vec{A}$ and constant (contributing $E\times 2\pi sL$), while on the two flat end caps $\vec{E}$ is perpendicular to $d\vec{A}$ → zero. That is why $\oint \vec{E}\cdot d\vec{A}=E(2\pi sL)$. Line-charge result: $E=\frac{\lambda }{2\pi {\epsilon }_{0}s}=\frac{8\times 10^{-8}}{2\pi (8.85\times 10^{-12})(0.10)}=1.44\times 10^4\text{N/C}.$ Direction: radially away from the wire.

Symmetry: spherical. Surface: sphere of radius $r=0.04\text{m}<R$, so we are inside. Only the charge within radius $r$ counts.

Set up Gauss's law explicitly. On our sphere $E$ is constant and points radially, so at every tile $\vec{E}$ is parallel to $d\vec{A}$ (recall the definition box: $d\vec{A}$ is the outward arrow of each tile). Hence each tile gives $E\times (\text{tile area})$ and the sum over tiles is the sphere's area: $\oint \vec{E}\cdot d\vec{A}=E(4\pi r^2).$ Why: $E$ factors out (constant on the surface), and $\oint dA$ is just the sphere's area $4\pi r^2$.

What charge is enclosed? Not all of $Q$ — only the fraction inside radius $r$. With uniform density $\rho$, that is the density times the small ball's volume: $Q_{enc}=\rho \cdot \frac{4}{3}\pi r^3.$ Why: charge is spread evenly through the volume, so enclosed charge = density × enclosed volume.

Equate and solve. $E(4\pi r^2)=\frac{\rho \frac{4}{3}\pi r^3}{{\epsilon }_0}.$ Cancel $4\pi$ from both sides and one power of $r^2$ from the $r^3$ on the right (this is why the field ends up linear — area grows like $r^2$ but enclosed charge grows like $r^3$, leaving one power of $r$): $E_{in}=\frac{\rho r}{3{\epsilon }_0}.$

Numbers. Volume density: $\rho =\frac{Q}{\frac{4}{3}\pi R^3}=\frac{6\times 10^{-6}}{\frac{4}{3}\pi (0.10{)}^3}=1.43\times 10^{-3}{\text{C/m}}^3.$ $E_{in}=\frac{\rho r}{3{\epsilon }_0}=\frac{(1.43\times 10^{-3})(0.04)}{3(8.85\times 10^{-12})}=2.16\times 10^6\text{N/C}.$ Equivalent form (avoids computing $\rho$; substitute $\rho =Q/\frac{4}{3}\pi R^3$ into $E_{in}$): $E_{in}=\frac{Qr}{4\pi {\epsilon }_0{R}^3}=\frac{(9\times 10^9)(6\times 10^{-6})(0.04)}{(0.10{)}^3}=2.16\times 10^6\text{N/C}.$ ✓

Conductor surface, not an isolated sheet, so use the $\sigma /{\epsilon }_0$ version (all the flux goes to the outside; field inside the metal is zero — see Conductors in Electrostatics): $E=\frac{\sigma }{{\epsilon }_0}=\frac{5\times 10^{-6}}{8.85\times 10^{-12}}=5.65\times 10^5\text{N/C}.$

Superpose two isolated sheets (Parallel Plate Capacitor). Each gives $\sigma /2{\epsilon }_0$. Between: the two fields point the same way, so they add: $E_{between}=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}=\frac{1.0\times 10^{-6}}{8.85\times 10^{-12}}=1.13\times 10^5\text{N/C}.$ Outside: the two fields point opposite ways and cancel: $E_{outside}=0$.

(i) Inner wall. Inside the metal of the shell, $E=0$. Draw a Gaussian sphere inside the metal ($a<r<b$). Since $E=0$ there, the enclosed charge must be zero. The centre charge is $+4\mu \text{C}$, so the inner wall must carry $-4\mu \text{C}$ to cancel it: $Q_{inner}=-4\mu \text{C}$. (ii) Outer wall. The shell is neutral overall, so its two walls sum to zero: $Q_{inner}+Q_{outer}=0\Rightarrow Q_{outer}=+4\mu \text{C}$. (iii) Field at $r=0.20\text{m}$ (outside everything). A Gaussian sphere at $r=0.20$ encloses $q$ plus both walls: $+4-4+4=+4\mu \text{C}$ net. $E=\frac{(9\times 10^9)(4\times 10^{-6})}{(0.20{)}^2}=9.0\times 10^5\text{N/C},\text{ radially outward.}$

Inside ($r\le R$): $E_{in}=\frac{Qr}{4\pi {\epsilon }_0{R}^3}$ — rises linearly, biggest at $r=R$. Outside ($r\ge R$): $E_{out}=\frac{Q}{4\pi {\epsilon }_0{r}^2}$ — falls, biggest at $r=R$. Both branches meet and peak exactly at the surface, $r=R$: $E_{max}=\frac{Q}{4\pi {\epsilon }_0{R}^2}.$ The field is continuous there (a good sanity check): plugging $r=R$ into either formula gives the same thing.

Potential is the work per unit charge to bring a test charge in from infinity: $V(R)=-{\int }_{\infty }^R{E}_{out}dr={\int }_R^{\infty }\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}dr=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}.$ Why an integral here? $E$ tells you force per charge at each point; potential is the accumulated push over the whole path, which is exactly what an integral sums. Numerically: $V(R)=\frac{(9\times 10^9)(5\times 10^{-6})}{0.10}=4.5\times 10^5\text{V}.$

At the surface the inside and outside formulas agree; use $E(R)=\frac{\rho R}{3{\epsilon }_0}$ and solve for $\rho$: $\rho =\frac{3{\epsilon }_{0}E(R)}{R}=\frac{3(8.85\times 10^{-12})(3.0\times 10^5)}{0.10}=7.97\times 10^{-5}{\text{C/m}}^3.$

Between ($s=0.02$): Gaussian cylinder encloses only the inner wire's $\lambda$: $E=\frac{\lambda }{2\pi {\epsilon }_{0}s}=\frac{3\times 10^{-8}}{2\pi (8.85\times 10^{-12})(0.02)}=2.70\times 10^4\text{N/C, outward.}$ Outside ($s=0.5$): enclosed charge per length is $+\lambda -\lambda =0$, so $E=0$. The cable is externally "silent" — the reason coax shields.

First the insulator's total charge: $Q_1=\rho \cdot \frac{4}{3}\pi R^3=(2\times 10^{-5})\frac{4}{3}\pi (0.10{)}^3=8.38\times 10^{-8}\text{C}$. (a) $r=0.05\text{m}$ (inside the insulator): only the charge within $r$ counts. $E=\frac{\rho r}{3{\epsilon }_0}=\frac{(2\times 10^{-5})(0.05)}{3(8.85\times 10^{-12})}=3.77\times 10^4\text{N/C.}$ (b) $r=0.15\text{m}$ (between insulator and shell): the Gaussian sphere at this radius encloses all of $Q_1$; the conducting shell sits farther out at $R_2=0.20\text{m}$, so it contributes nothing to what is enclosed. Treat $Q_1$ as a point charge at the centre: $E=\frac{(9\times 10^9)(8.38\times 10^{-8})}{(0.15{)}^2}=3.35\times 10^4\text{N/C.}$ (c) $r=0.30\text{m}$ (outside everything): now the Gaussian sphere encloses both the insulator's $Q_1$ and the shell's $Q_2$: $Q_{enc}=Q_1+Q_2=8.38\times 10^{-8}+1\times 10^{-6}=1.084\times 10^{-6}\text{C}.$ $E=\frac{(9\times 10^9)(1.084\times 10^{-6})}{(0.30{)}^2}=1.08\times 10^5\text{N/C.}$

Each isolated sheet makes $E=\sigma /2{\epsilon }_0$ pointing away from itself on both sides. Let rightward be positive. In field units, ${\sigma }_A/2{\epsilon }_0=2.26\times 10^5$ and ${\sigma }_B/2{\epsilon }_0=5.65\times 10^4\text{N/C}$. Left of both: both point left (away, toward −). $E=-(2.26\times 10^5+5.65\times 10^4)=-2.82\times 10^5\text{N/C}$ → magnitude $2.82\times 10^5$, pointing left. Between (A on left, B on right): A pushes right (+), B pushes left (−). $E=2.26\times 10^5-5.65\times 10^4=1.69\times 10^5\text{N/C}$ → magnitude $1.69\times 10^5$, pointing right. Right of both: both point right (+). $E=2.26\times 10^5+5.65\times 10^4=2.82\times 10^5\text{N/C}$ → magnitude $2.82\times 10^5$, pointing right.

Gauss's law gives $E=\lambda /2\pi {\epsilon }_{0}s$ only when the field is purely radial and constant along the axis — which requires the source to look the same as you slide along it (translational/axial symmetry), so that the end caps of the Gaussian cylinder carry zero flux.

Close up ($s\ll L$). When your distance $s$ is tiny compared with the rod's length $L$, the two ends are far away and effectively invisible. Locally the rod "looks" infinite: field is essentially radial, the end-cap flux is negligible, and the derivation's assumption holds. So the $1/s$ law is an excellent approximation here.

Far away ($s\gg L$). Now the whole rod subtends a tiny angle — it looks like a single dot of total charge $Q=\lambda L$. A point charge's field falls off as $1/r^2$, so the field must cross over from $1/s$ (near) to $1/r^2$ (far). Between these two regimes there is a smooth transition.

Which assumption fails? The axial / translational symmetry (invariance along the axis). A finite rod is not the same as you slide along it — near the ends the field tilts and is no longer purely radial, so the Gaussian cylinder's end caps no longer carry zero flux. Once end-cap flux matters, the clean $E(2\pi sL)=\lambda L/{\epsilon }_0$ step is invalid. Takeaway: the tidy $1/s$ formula is exact only for the idealised infinite line; for a finite rod it is a near-field approximation that decays into the point-charge $1/r^2$ far away.

Recognition ::: name the symmetry and pick $1/r^2$, $1/s$, or constant Application ::: decide inside vs outside, then plug once Analysis ::: use $E=0$ in metal + conservation to place induced charge Synthesis ::: link $E$ to $V$ by integrating, or combine two geometries Mastery ::: superpose vectors with correct signs and enclosed-charge bookkeeping