Mid Term 1 Prep

ENSC 316 Midterm 1 Prep

The ME1 topic list from the professor’s study guide, matched to Griffiths (4th ed.) and sorted by how likely each topic is to show up. A mock midterm built from the 2017 and 2018 papers follows the map.

Exam format: 1 advanced (35 min), 2 medium (20 min each), 3 beginner (10 min each). 105 minutes total. Bring the printed formula sheet, unmodified except your name and student number.

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Topic map

The percentages are the professor’s odds that a topic appears as a medium or advanced question. Start at the top.

Odds	Topic (prof’s sub-items)	Griffiths	Be able to
80%	Path & line integrals: differential line element, work done	1.3.1, 1.3.3, 1.4.1–1.4.2, 2.3.1	Parameterize a curve with one parameter $u$, write $d\vec{l}=\frac{d\vec{r}}{du}du$, substitute, evaluate. Check conservative with $\nabla \times \vec{F}$.
80%	Scalar potential: circulation, gradient	1.2.2–1.2.3, 1.2.5, 1.3.3, 1.3.5, 2.2.4, 2.3.1–2.3.2	$\vec{E}=-\nabla V$ in all three coordinate systems. Explain why $\oint \vec{E}\cdot d\vec{l}=0$ lets a potential exist.
80%	Surface integrals: differential area element, flux	1.3.1, 1.4.1–1.4.2, 2.2.1	Build $d\vec{a}$ on spheres, cylinders, planes and cones. On a parameterized surface, $d\vec{a}=({\partial }_u\vec{r}\times {\partial }_t\vec{r})dudt$.
80%	Divergence theorem: Gauss’ law, integral/differential laws	1.2.4, 1.3.4, 2.2.1–2.2.3	Swap a flux for $\int \nabla \cdot \vec{F}d\tau$. Gauss’s law with a written symmetry argument, surface drawn, flux and charge computed separately.
70%	Coulomb’s law: convolution integral	2.1.2–2.1.4, 2.3.4, (3.4.4)	Set up $V({\vec{r}}_f)=\frac{1}{4\pi {\epsilon }_0}\int \frac{\rho d{\tau }_s}{\Vert {\vec{r}}_f-{\vec{r}}_s\Vert }$ for lines, discs and shells. Law-of-cosines trick for a spherical shell.
45%	Coordinate systems: coordinate curves & surfaces, general fields	1.1.4, 1.4.1–1.4.2	Name the surface when one coordinate is fixed and the curve when two are. Convert unit vectors.
45%	Spatial integrals: regions of integration, classification of types	1.3.1, 1.4	Set limits on odd regions like the torus. Get $d\tau$ from the triple product of tangent vectors.
40%	Metals: equilibrium properties, boundary conditions, shielding	2.5.1–2.5.3, 2.3.5	$\vec{E}=0$ inside, metal is equipotential, $E_{\perp }=\sigma /{\epsilon }_0$. Cavity charge induces equal and opposite charge on the cavity wall. The outer surface can’t tell where it is.
30%	Dielectrics I: micro/macro fields, polarization	4.1.1–4.1.4, 4.2.1–4.2.3	Torque $\vec{N}=\vec{p}\times \vec{E}$. Bound charges ${\sigma }_b=\vec{P}\cdot \hat{n}$, ${\rho }_b=-\nabla \cdot \vec{P}$.
30%	Dielectrics II: susceptibility, permittivity, displacement	4.3.1, 4.3.3, 4.4.1	$\vec{D}={\epsilon }_0\vec{E}+\vec{P}=\epsilon \vec{E}$, $\oint \vec{D}\cdot d\vec{a}=Q_{f,enc}$. Boundary conditions on $D_{\perp }$ and $E_{\parallel }$.

Skip or skim for ME1

• 1.1.5 (how vectors transform), 1.3.6, 1.6 (Helmholtz)
• 1.5 Dirac delta: read it once, don’t drill it
• 2.4 energy and 2.5.4 capacitors. Those landed on Midterm 2 and the final instead.
• Ch. 3, except 3.4.4 (dipole field). You need that one for the 2017 needle question.
• 4.4.2–4.4.4

Where Griffiths falls short

The professor parameterizes curves and surfaces with tangent vectors ($d{\vec{T}}_t$, $d{\vec{T}}_{\alpha }$, …) and builds $d\vec{a}$ and $d\tau$ from cross and triple products. Griffiths §1.4 only hands you the standard elements for spherical and cylindrical coordinates. For general parameterized surfaces, use the parametric surfaces section of a multivariable calc book (Stewart has one), or Notaros Ch. 1.

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Notation translation

The exams don’t use Griffiths’ symbols. Keep this nearby while reading.

Meaning	Your exams	Griffiths
Cylindrical radial coordinate	$\gamma$, $\hat{\mathbf{\gamma }}$	$s$, $\hat{\mathbf{s}}$
Field point (where you measure)	${\vec{r}}_f$	$\mathbf{r}$
Source point (where the charge is)	${\vec{r}}_s$	${\mathbf{r}}^{\prime }$
Separation vector	${\vec{r}}_{sf}={\vec{r}}_f-{\vec{r}}_s$	script r, $\mathbf{r}-{\mathbf{r}}^{\prime }$
Free charge densities	${\rho }_{vg}$, ${\rho }_{sg}$, ${\rho }_{\ell g}$ (g = “glued”)	$\rho$, $\sigma$, $\lambda$
Volume element	$dv$ or $d\tau$	$d\tau$

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Study order

Seven sessions. Each one leans on the one before it. Read the sections, then try the matching problem before opening the solution.

1. Coordinates and elements. 1.1.4, 1.4.1, 1.4.2. Memorize $d\vec{l}$, $d\vec{a}$, $d\tau$ in all three systems. Drill X7.
2. Gradient, potential, circulation. 1.2.2–1.2.5, 1.3.3, 2.2.4, 2.3.1–2.3.2. Problems B1 and X6.
3. Line integrals on strange curves. 1.3.1 plus the parameterization note above. M2, then redo 2018 MT1 #4 without notes.
4. Flux and the divergence theorem. 1.2.4, 1.3.1 (surface part), 1.3.4. M1 and X2.
5. Gauss’s law and Coulomb convolution. 2.1, 2.2.1–2.2.3, 2.3.4. B3, X1, X3. Redo 2018 MT1 #5 and #6.
6. Metals and dipoles. 2.5.1–2.5.3, 3.4.4, 4.1.3. B2 and X4.
7. Dielectrics. 4.1–4.3, 4.4.1. X5, then A1 on a 35-minute timer. Finish with the whole mock midterm at 105 minutes.

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What the grader wants

The 2018 sample papers still have the professor’s marking notes in the margins. Every note says some version of “show the work”:

• math. A symmetry argument ends in an equation like $\vec{E}=E_{\gamma }(\gamma )\hat{\mathbf{\gamma }}$. A sentence in English isn’t enough.
• Write out the zeros. In a curl or divergence, show every component that vanishes.
• Name the identities you use, e.g. $\nabla \cdot (\nabla \times \vec{A})=0$.
• One operation per line in Gauss’s-law problems. Draw the Gaussian surface, define every symbol, and label which part is flux and which is enclosed charge.
• Derive before you quote. $C=\epsilon S/d$ only counts if you derived it for that problem.
• With dielectrics, find $\vec{D}$ first. Gauss’s law for $\vec{D}$ only needs free charge. Then $\vec{E}=\vec{D}/\epsilon$.

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Mock midterm

Same layout as the real exam. Each problem copies the shape of a 2017/2018 question with new numbers or geometry. Use ${\epsilon }_0=8.854\times 10^{-12}$ F/m and time yourself.

B1 · Electric field from a potential

Beginner · 10 min · like 2017/18 #1

a) $V({\vec{r}}_f)=x^{2}yz-2y^3+4x{z}^2$. Find $\vec{E}$ at ${\vec{r}}_f=\hat{x}+\hat{y}+2\hat{z}$.
b) $V={\gamma }^2\sin \phi +3z$ (cylindrical). Find $\vec{E}$ at $\gamma =2,\phi =\pi /6,z=0$.

Solution

(a) $\vec{E}=-\nabla V$ with
$\nabla V=(2xyz+4z^2)\hat{x}+(x^{2}z-6y^2)\hat{y}+(x^{2}y+8xz)\hat{z}$
At $(1,1,2)$: $\nabla V=20\hat{x}-4\hat{y}+17\hat{z}$, so $\boxed{\vec{E}=-20\hat{x}+4\hat{y}-17\hat{z}\text{V/m}}$

(b) Watch the $1/\gamma$ on the $\phi$ term:
$\nabla V=\frac{\partial V}{\partial \gamma }\hat{\mathbf{\gamma }}+\frac{1}{\gamma }\frac{\partial V}{\partial \phi }\hat{\mathbf{\phi }}+\frac{\partial V}{\partial z}\hat{z}=2\gamma \sin \phi \hat{\mathbf{\gamma }}+\gamma \cos \phi \hat{\mathbf{\phi }}+3\hat{z}$
At the point: $\boxed{\vec{E}=-2\hat{\mathbf{\gamma }}-\sqrt{3}\hat{\mathbf{\phi }}-3\hat{z}\text{V/m}}$

B2 · Electrostatic shielding

Beginner · 10 min · like 2017/18 #2

A metal sphere of radius 30 cm carries net charge +2 nC. Inside it is an irregular cavity holding $Q_1=+5$ nC, $Q_2=-1$ nC and $Q_3=-2$ nC at unknown positions. Take $V(\infty )=0$. Find:
a) the total charge on the cavity wall
b) the surface charge density on the outer surface
c) the potential of the metal

Solution

(a) In equilibrium $\vec{E}=0$ inside the metal. A Gaussian surface inside the metal that wraps the cavity has zero flux, so $Q_{enc}=0$:
$q_{wall}=-(5-1-2)\text{nC}=-2\text{nC}$
(b) The metal’s charge is conserved: $q_{outer}=+2-(-2)=+4$ nC. Cavity charges plus wall charge give zero field in the metal, so the outer surface has no idea where they sit. It spreads uniformly over the sphere:
${\rho }_s=\frac{4\times 10^{-9}}{4\pi (0.3{)}^2}=3.54{\text{nC/m}}^2$
(c) Outside, it looks like a 4 nC point charge at the centre:
$V=\frac{q_{outer}}{4\pi {\epsilon }_{0}R}=\frac{(8.99\times 10^9)(4\times 10^{-9})}{0.3}\approx 120\text{V}$
Griffiths 2.5.1–2.5.2 (the cavity discussion)

B3 · Two charged sheets

Beginner · 10 min · like 2017 #3

Plane 1 at $z=-2$ cm has ${\rho }_{sg1}=+8$ nC/m². Plane 2 at $z=5$ cm has ${\rho }_{sg2}=+4$ nC/m². Find $\vec{E}$ in all three regions.

Solution

A single infinite sheet gives $|\vec{E}|={\rho }_s/(2{\epsilon }_0)$, pointing away from positive charge on both sides (Gauss pillbox, 2.2.3). Superpose:
$\frac{{\rho }_{s1}}{2{\epsilon }_0}=451.8\text{V/m},\frac{{\rho }_{s2}}{2{\epsilon }_0}=225.9\text{V/m}$

• $z>5$ cm: both point up, $\vec{E}=+677.7\hat{z}$ V/m
• $-2<z<5$ cm: sheet 1 up, sheet 2 down, $\vec{E}=+225.9\hat{z}$ V/m
• $z<-2$ cm: both point down, $\vec{E}=-677.7\hat{z}$ V/m

M1 · Divergence theorem

Medium · 20 min · like 2018 #3

$\vec{F}({\vec{r}}_f)=4\gamma \hat{\mathbf{\gamma }}+7\gamma z\hat{\mathbf{\phi }}-3z\hat{z}$
a) Show the net flux out of a cube of side $L$ doesn’t depend on where the cube sits.
b) Check the divergence theorem directly on a cylinder of radius $a$, $0\le z\le h$, centred on the $z$-axis.

Solution

(a) Every term written out:
$\nabla \cdot \vec{F}=\frac{1}{\gamma }\frac{\partial }{\partial \gamma }(\gamma \cdot 4\gamma )+\frac{1}{\gamma }\frac{\partial }{\partial \phi }(7\gamma z)+\frac{\partial }{\partial z}(-3z)=8+0-3=5$
$\vec{F}$ is continuously differentiable everywhere, so the divergence theorem holds for any closed volume:
${\oint }_S\vec{F}\cdot d\vec{a}={\int }_V\nabla \cdot \vec{F}d\tau =5{\int }_{V}d\tau =5L^3$
The divergence is a constant. Only the volume matters, not the position.

(b) $F_{\phi }$ is tangent to every face and contributes nothing.

• Side $\gamma =a$, $d\vec{a}=ad\phi dz\hat{\mathbf{\gamma }}$: ${\int }_0^h{\int }_0^{2\pi }4a\cdot ad\phi dz=8\pi a^{2}h$
• Top $z=h$, $d\vec{a}=\gamma d\gamma d\phi \hat{z}$: $(-3h)(\pi a^2)=-3\pi a^{2}h$
• Bottom $z=0$, $d\vec{a}=-\gamma d\gamma d\phi \hat{z}$: $F_z=0$ there, so $0$

Total $5\pi a^{2}h=5\times$ volume. It matches.

Griffiths 1.2.4, 1.3.4, 1.4.2

M2 · Line integral on a tilted semicircle

Medium · 20 min · like 2018 #4

Sphere $S:x^2+y^2+z^2=A^2$. Plane $P:x-z=0$ (the $xy$-plane rotated 45° about the $y$-axis). Contour $C=S\cap P$ with $x>0$, from ${\vec{r}}_{start}=A\hat{y}$ to ${\vec{r}}_{end}=-A\hat{y}$.

a) Define a parameter $u$ and write the Cartesian coordinates of $C$.
b) Write $d\vec{l}$.
c) Evaluate ${\int }_C{\vec{F}}_1\cdot d\vec{l}$ for the uniform field ${\vec{F}}_1=F_0\hat{z}$.
d) Evaluate ${\int }_C{\vec{F}}_2\cdot d\vec{l}$ for ${\vec{F}}_2=\frac{F_0}{A}(y\hat{x}-x\hat{y})$.
e) Which field is conservative? Prove it, then confirm with a second path for ${\vec{F}}_2$.

Solution

(a) On $P$, $z=x$. Put that into $S$: $2x^2+y^2=A^2$. With $u\in [0,\pi ]$:
$x=\frac{A}{\sqrt{2}}\sin u,y=A\cos u,z=\frac{A}{\sqrt{2}}\sin u$
Check: $u=0$ gives $(0,A,0)$, $u=\pi$ gives $(0,-A,0)$, and $x\ge 0$ the whole way.

(b)
$d\vec{l}=\frac{d\vec{r}}{du}du=A\left(\frac{1}{\sqrt{2}}\cos u\hat{x}-\sin u\hat{y}+\frac{1}{\sqrt{2}}\cos u\hat{z}\right)du$
(c) ${\vec{F}}_1\cdot d\vec{l}=\frac{F_{0}A}{\sqrt{2}}\cos udu$, and ${\int }_0^{\pi }\cos udu=0$. The answer is 0, which makes sense: ${\vec{F}}_1=\nabla (F_{0}z)$ and $z=0$ at both ends.

(d) On $C$, ${\vec{F}}_2=F_0\left(\cos u\hat{x}-\frac{1}{\sqrt{2}}\sin u\hat{y}\right)$:
${\vec{F}}_2\cdot d\vec{l}=F_{0}A\left(\frac{1}{\sqrt{2}}{\cos }^{2}u+\frac{1}{\sqrt{2}}{\sin }^{2}u\right)du=\frac{F_{0}A}{\sqrt{2}}du\Rightarrow {\int }_C=\frac{\pi F_{0}A}{\sqrt{2}}$
(e) $\nabla \times {\vec{F}}_1=0$ since its components are constant. For ${\vec{F}}_2$:
$\nabla \times {\vec{F}}_2=({\partial }_y{F}_z-{\partial }_z{F}_y)\hat{x}+({\partial }_z{F}_x-{\partial }_x{F}_z)\hat{y}+({\partial }_x{F}_y-{\partial }_y{F}_x)\hat{z}=0\hat{x}+0\hat{y}-\frac{2F_0}{A}\hat{z}$
Not zero, so ${\vec{F}}_2$ isn’t conservative. Along the straight path on the $y$-axis ($x=0$), ${\vec{F}}_2=\frac{F_0}{A}y\hat{x}$ and $d\vec{l}=dy\hat{y}$. That integral is 0. Two paths, two answers.

Griffiths 1.3.1, 1.3.3, 1.2.5

A1 · Inhomogeneous dielectric sphere

Advanced · 35 min · like 2017 #4

A sphere of radius $A$ sits in vacuum. It’s made of a dielectric with ${\epsilon }_r(r)=B\frac{A}{r}$ for $0<r<A$, $B\ge 1$. A free point charge $Q$ is at the centre. Use Gauss’s law and state every symmetry argument.

a) Find $\vec{D}$, $\vec{E}$, $V$ for $r>A$.
b) Find $\vec{D}$, $\vec{E}$, $V$ for $0<r<A$.
c) Find $\vec{P}$, the bound surface charge density at $r=A$, and the bound volume charge density.
d) Show the total bound charge is zero.

Solution

Symmetry. The free charge is a point at the origin and $\epsilon$ depends only on $r$. Rotating about the origin changes nothing, so no component can depend on $\theta$ or $\phi$, and no tangential direction is preferred:
$\vec{D}=D_r(r)\hat{r}$
Gauss for $\vec{D}$ on a sphere of radius $r$. It works in both regions because it only needs free charge:
$\underset{\text{flux}}{\underbrace{\oint \vec{D}\cdot d\vec{a}=D_r(r)4\pi r^2}}=\underset{Q_{f,enc}}{\underbrace{Q}}\Rightarrow \vec{D}=\frac{Q}{4\pi r^2}\hat{r}\text{for all }r$
(a) $r>A$: $\vec{E}=\frac{Q}{4\pi {\epsilon }_0{r}^2}\hat{r}$, $V=\frac{Q}{4\pi {\epsilon }_{0}r}$

(b) $0<r<A$: $\epsilon ={\epsilon }_{0}BA/r$, so
$\vec{E}=\frac{\vec{D}}{\epsilon }=\frac{Q}{4\pi {\epsilon }_{0}BAr}\hat{r}$
$V(r)=V(A)+{\int }_r^{A}Ed{r}^{\prime }=\frac{Q}{4\pi {\epsilon }_{0}A}\left[1+\frac{1}{B}\ln \frac{A}{r}\right]$
(c)
$\vec{P}=\vec{D}-{\epsilon }_0\vec{E}=\frac{Q}{4\pi r^2}\left(1-\frac{r}{BA}\right)\hat{r}$
${\rho }_{sb}(A)=\vec{P}\cdot \hat{r}{|}_{r=A}=\frac{Q}{4\pi A^2}\left(1-\frac{1}{B}\right),{\rho }_{vb}=-\frac{1}{r^2}\frac{d}{dr}\left(r^2{P}_r\right)=\frac{Q}{4\pi BA{r}^2}$
(d) Bound volume charge: ${\int }_0^A\frac{Q}{4\pi BA{r}^2}4\pi r^{2}dr=Q/B$. Bound surface charge: $Q(1-1/B)$. Near $r\to 0$, ${\epsilon }_r\to \infty$, and the polarization packed around $Q$ carries $-\oint \vec{P}\cdot d\vec{a}\to -Q$. Sum: $-Q+Q/B+Q-Q/B=0$.

Cross-check: net charge inside radius $r$ is $Q-Q+Qr/(BA)=Qr/(BA)$. Dividing by $4\pi {\epsilon }_0{r}^2$ gives back the $E$ from (b).

Griffiths 4.2.1, 4.3.1, 4.4.1 (the metal sphere in a dielectric shell example in 4.4.1 is close)

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Extra drills

One per leftover topic, so every box on page 1 of the study guide gets a rep.

X1 · Gauss’s law, infinite charged cylinder

Medium · like 2018 #5

${\rho }_{vg}={\rho }_0(\gamma /A{)}^2$ for $0<\gamma \le A$, zero outside, infinite along $z$.
a) Argue the coordinate dependence. b) Argue which components survive. c) Find $\vec{E}$ inside. d) Find $\vec{E}$ outside.

Solution

(a) Translating in $z$ or rotating in $\phi$ leaves the charge unchanged. Components depend on $\gamma$ only: $E_i=E_i(\gamma )$.

(b) The reflection $z\to -z$ leaves the charge alone but would flip $E_z$, so $E_z=0$. Same with $\phi \to -\phi$ for $E_{\phi }$. Result: $\vec{E}=E_{\gamma }(\gamma )\hat{\mathbf{\gamma }}$.

(c) Gaussian cylinder of radius $\gamma$, length $L$. End caps contribute nothing since $\hat{\mathbf{\gamma }}\cdot \hat{z}=0$.
$\text{Flux: }{E}_{\gamma }2\pi \gamma L,\text{Charge: }{\int }_0^L{\int }_0^{2\pi }{\int }_0^{\gamma }{\rho }_0\frac{{\gamma }^{\prime 2}}{A^2}{\gamma }^{\prime }d{\gamma }^{\prime }d\phi dz=\frac{\pi {\rho }_{0}L{\gamma }^4}{2A^2}$
$\vec{E}=\frac{{\rho }_0{\gamma }^3}{4{\epsilon }_0{A}^2}\hat{\mathbf{\gamma }}(\gamma <A)$
(d) Enclosed charge stops growing at $\gamma =A$:
$\vec{E}=\frac{{\rho }_0{A}^2}{4{\epsilon }_0\gamma }\hat{\mathbf{\gamma }}(\gamma >A)$
Both give ${\rho }_{0}A/(4{\epsilon }_0)$ at $\gamma =A$. Continuous, as it should be with no surface charge.

X2 · Total charge on a torus

Medium · like 2017 #6

Same torus as the 2017 exam: central ring radius $B$, tube radius $A$, parameters $t\in (0,A)$, $\alpha ,\beta \in [0,2\pi ]$. This time ${\rho }_{vg}={\rho }_0(t/A{)}^2(1+\cos \alpha )$.
Find a) source coordinates, b) tangent vectors, c) $d\tau$, d) total charge.

Solution

(a) ${\gamma }_s=B+t\cos \alpha ,{\phi }_s=\beta ,z_s=t\sin \alpha$, so ${\vec{r}}_s=(B+t\cos \alpha )\hat{\mathbf{\gamma }}(\beta )+t\sin \alpha \hat{z}$

(b) Using $\partial \hat{\mathbf{\gamma }}/\partial \phi =\hat{\mathbf{\phi }}$:
$d{\vec{T}}_t=(\cos \alpha \hat{\mathbf{\gamma }}+\sin \alpha \hat{z})dt$
$d{\vec{T}}_{\alpha }=t(-\sin \alpha \hat{\mathbf{\gamma }}+\cos \alpha \hat{z})d\alpha$
$d{\vec{T}}_{\beta }=(B+t\cos \alpha )\hat{\mathbf{\phi }}d\beta$
(c) $d\tau =|d{\vec{T}}_t\cdot (d{\vec{T}}_{\alpha }\times d{\vec{T}}_{\beta })|=t(B+t\cos \alpha )dtd\alpha d\beta$

(d) Expand $(1+\cos \alpha )(B+t\cos \alpha )$ and integrate over $\alpha$. Only the constant and the ${\cos }^2\alpha$ term survive, leaving $2\pi B+\pi t$:
$Q=2\pi \frac{{\rho }_0}{A^2}{\int }_0^A{t}^3(2\pi B+\pi t)dt={\pi }^2{\rho }_0{A}^2\left(B+\frac{2A}{5}\right)$
The real 2017 density had $\cos (3\beta )$ in it. Since ${\int }_0^{2\pi }\cos 3\beta d\beta =0$, the answer there is 0. That’s the “not as scary as it looks” joke.

X3 · Coulomb convolution, disc and shell

Medium · like 2018 #6

a) A disc of radius $R$ in the $xy$-plane carries uniform ${\rho }_{s0}$. Write the convolution integral for $V$ on the $z$-axis and evaluate it.
b) Find $E_z$ for $z>0$. Check the limits $R\to \infty$ and $z\gg R$.
c) A spherical shell of radius $A$ carries uniform ${\rho }_{s0}$. Use the law of cosines to find $V$ on the $z$-axis inside the shell.

Solution

(a) ${\vec{r}}_s={\gamma }_s\hat{\mathbf{\gamma }}$, ${\vec{r}}_f=z\hat{z}$, $r_{sf}=\sqrt{{\gamma }_s^2+z^2}$, $d{a}_s={\gamma }_{s}d{\gamma }_{s}d{\phi }_s$
$V(z)=\frac{1}{4\pi {\epsilon }_0}{\int }_0^{2\pi }{\int }_0^R\frac{{\rho }_{s0}{\gamma }_{s}d{\gamma }_{s}d{\phi }_s}{\sqrt{{\gamma }_s^2+z^2}}=\frac{{\rho }_{s0}}{2{\epsilon }_0}\left(\sqrt{z^2+R^2}-|z|\right)$
(b)
$E_z=-\frac{dV}{dz}=\frac{{\rho }_{s0}}{2{\epsilon }_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)$
As $R\to \infty$ this becomes ${\rho }_{s0}/(2{\epsilon }_0)$, the infinite sheet. For $z\gg R$, $\sqrt{z^2+R^2}\approx z+R^2/(2z)$, so $V\approx \frac{\pi R^2{\rho }_{s0}}{4\pi {\epsilon }_{0}z}$. A point charge.

(c) $r_{sf}^2=z^2+A^2-2zA\cos {\theta }_s$. Substitute $w=\cos {\theta }_s$:
${\int }_0^{\pi }\frac{\sin \theta d\theta }{\sqrt{z^2+A^2-2zA\cos \theta }}=\frac{1}{zA}[(z+A)-|z-A|]$
Inside ($z<A$) the bracket is $2z$, which gives $2/A$:
$V=\frac{{\rho }_{s0}{A}^2\cdot 2\pi }{4\pi {\epsilon }_0}\cdot \frac{2}{A}=\frac{{\rho }_{s0}A}{{\epsilon }_0}$
Constant, so no field inside. Outside the bracket is $2A$, and you get the 2018 answer ${\rho }_{s0}{A}^2/({\epsilon }_{0}z)$.

Griffiths 2.1.4, 2.3.4. The spherical shell example there is this exact integral.

X4 · Needle in a dipole field

Medium · like 2017 #5

$+Q$ at $(d,0,0)$, $-Q$ at $(-d,0,0)$. A tiny polarized needle at ${\vec{r}}_f$ with $\Vert {\vec{r}}_f\Vert \gg d$ rotates freely until it settles. Find the angle $C$ between the needle and the $x$-axis. Evaluate it at ${\vec{r}}_f\propto (3,0,4)$.

Solution

The needle acts as a dipole ${\vec{p}}_n$. Torque $\vec{N}={\vec{p}}_n\times \vec{E}$ is zero (and stable) when ${\vec{p}}_n\parallel \vec{E}$. So the needle lines up with $\vec{E}$.

Far field of $\vec{p}=2Qd\hat{x}$ (Griffiths 3.4.4, coordinate-free form):
$\vec{E}=\frac{p}{4\pi {\epsilon }_0{r}^3}\left[3(\hat{x}\cdot \hat{r})\hat{r}-\hat{x}\right]$
Let $c=\hat{x}\cdot \hat{r}=x/r$. Then $\vec{E}\cdot \hat{x}\propto 3c^2-1$ and $|\vec{E}|\propto \sqrt{9c^2-6c^2+1}=\sqrt{1+3c^2}$:
$\cos C=\frac{3c^2-1}{\sqrt{1+3c^2}}=\frac{3x^2-r^2}{r\sqrt{r^2+3x^2}}$
At $(3,0,4)$ with $r=5$: $\cos C=2/(5\sqrt{52})=0.0555$, so $C\approx 86.8^{\circ }$.

For the 2017 version (charges on the $z$-axis), swap $x\to z$.

X5 · Dielectric boundary

Beginner

Vacuum for $z>0$, dielectric with ${\epsilon }_r=3$ for $z<0$, no free surface charge. Just above the interface, ${\vec{E}}_1$ points into the dielectric at 30° from the normal. Find the angle and magnitude of ${\vec{E}}_2$, and the bound surface charge density.

Solution

Boundary conditions: $E_{1\parallel }=E_{2\parallel }$ and $D_{1\perp }=D_{2\perp }$ (no free charge). So
$\tan {\theta }_2={\epsilon }_r\tan {\theta }_1=3\tan 30^{\circ }=\sqrt{3}\Rightarrow {\theta }_2=60^{\circ }$
$E_{2\parallel }=E_1/2$ and $E_{2\perp }=E_1\cos 30^{\circ }/3=0.289E_1$, giving $|{\vec{E}}_2|=0.577E_1$.

$\vec{P}={\epsilon }_0({\epsilon }_r-1){\vec{E}}_2$. The dielectric’s outward normal is $+\hat{z}$ and $E_{2z}=-0.289E_1$:
${\rho }_{sb}=\vec{P}\cdot \hat{z}=-\frac{\sqrt{3}}{3}{\epsilon }_0{E}_1\approx -0.577{\epsilon }_0{E}_1$
Griffiths 4.3.3, 4.4.1

X6 · Conservative or not

Beginner

$\vec{F}=(2xy+z)\hat{x}+x^2\hat{y}+x\hat{z}$. Is it conservative? If it is, find the work from $(0,0,0)$ to $(1,2,3)$ without doing a line integral.

Solution

$\nabla \times \vec{F}=(0-0)\hat{x}+(1-1)\hat{y}+(2x-2x)\hat{z}=0$
Curl-free everywhere, so it’s conservative. Find $f$ with $\nabla f=\vec{F}$: $f=x^{2}y+xz$.
Work $=f(1,2,3)-f(0,0,0)=2+3=5$. Circulation around any closed loop is 0 (Stokes).

X7 · Coordinate curves and surfaces

Beginner

a) Describe the spherical surface $\theta =\pi /4$, the cylindrical surface $\phi =\pi /3$, and the curve $\gamma =2,z=1$.
b) Write the outward $d\vec{a}$ on a sphere $r=A$, a cylinder $\gamma =a$, and the cone $\theta ={\theta }_0$.

Solution

(a) $\theta =\pi /4$ is a cone around the $+z$-axis with a 45° half-angle. $\phi =\pi /3$ is a half-plane hinged on the $z$-axis. $\gamma =2,z=1$ is a circle of radius 2 in the plane $z=1$.

(b)

• Sphere: $A^2\sin \theta d\theta d\phi \hat{r}$
• Cylinder: $ad\phi dz\hat{\mathbf{\gamma }}$
• Cone: $r\sin {\theta }_{0}drd\phi \hat{\mathbf{\theta }}$ (same geometry as the 2018 Midterm 2 capacitor)

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Past-exam answer checks

Do these papers without notes first, then compare. I worked these out myself. They aren’t an official key.

Paper	Question	Final answer
2017 & 2018 #1	$V=x^{3}y-y{z}^2+z^5$ at $(3,-1,2)$	$\vec{E}=27\hat{x}-23\hat{y}-84\hat{z}$ V/m
2017 #2	Shielding, $R=20$ cm	+1 nC on the outer surface, $V\approx 44.9$ V
2018 #2	Shielding, $R=50$ cm	${\rho }_s=1\text{nC}/(4\pi \cdot 0.25)\approx 0.318$ nC/m². Cavity shape doesn’t matter.
2017 #3	Sheets ±5 nC/m²	$\vec{E}=0$ for $z>20$ cm and $z<3$ cm; $+564.7\hat{z}$ V/m between
2017 #4	${\epsilon }_r=B(A/r{)}^2$ sphere	$\vec{D}=\frac{Q}{4\pi r^2}\hat{r}$ everywhere. Inside: $\vec{E}=\frac{Q}{4\pi {\epsilon }_{0}B{A}^2}\hat{r}$ (constant magnitude), $V=\frac{Q}{4\pi {\epsilon }_{0}A}+\frac{Q(A-r)}{4\pi {\epsilon }_{0}B{A}^2}$. ${\rho }_{sb}(A)=\frac{Q}{4\pi A^2}(1-\frac{1}{B})$
2017 #5	Needle, dipole on $z$	$\cos C=\frac{3z^2-r^2}{r\sqrt{r^2+3z^2}}$, or $\tan C=\frac{3z\sqrt{x^2+y^2}}{2z^2-x^2-y^2}$
2017 #6	Torus, $\cos 3\beta$ density	$Q=0$
2018 #3	$3\gamma \hat{\mathbf{\gamma }}+6\phi \gamma \hat{\mathbf{\phi }}+5z\hat{z}$	$\nabla \cdot \vec{F}=6+6+5=17$, flux out of unit cube $=17$
2018 #4	Uniform $F_0\hat{x}$ on semicircle	$x=A\cos u,y=z=\frac{A}{\sqrt{2}}\sin u$. Integral $=-2A{F}_0$. Conservative (curl 0).
2018 #5	${\rho }_0\frac{A-\gamma }{A}$ cylinder	Inside $E_{\gamma }=\frac{{\rho }_0}{{\epsilon }_0}\left(\frac{\gamma }{2}-\frac{{\gamma }^2}{3A}\right)$; outside $E_{\gamma }=\frac{{\rho }_0{A}^2}{6{\epsilon }_0\gamma }$
2018 #6	Shell potential on $z$-axis, outside	$V(z)=\frac{{\rho }_{s0}{A}^2}{{\epsilon }_{0}z}$

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Sources: ENSC 316-1267 ME1 study guide and the 2017–2018 sample exams. Official reading is Notaros Ch. 1.1–1.19 and 2.1–2.4. Section numbers here are Griffiths 4th edition.