# Homework & Assignments

Homework set #1 | #2 | #3 | #4 | #5 | #6 | #7 | #8 | #9 | #10

[ Phys 303 on Moodle]

### Guidelines

• You will always write up the homework problems yourself. But please do work together with others in the class on the assignments.
• Problems are from Griffiths' textbook, unless otherwise noted.

#### Written HW guidelines

• You'll be producing hand-written homework papers and Mathematica files.
• For written work, pencil is fine. Mechanical pencil is even better.
• You can show me your Mathematica work either by including a printout, or depositing a file (with a name that says what problem it contains) in our M:\classes folder, under your name.
• The answer alone is not enough. Show enough that I can see how you did the problem.
• Don't include crossed out "blind alleys" in your write ups. You might work out a problem on a scrap sheet of paper and then write up the problem once you've figured out how it works. The alternative is to use a pencil and a ¡polymer! eraser.
• Keep all the parts of each problem together: Don't do part a on page 2, then some other problems, then part b on page 7.
• Circle or otherwise highlight answers and important intermediate results.
2010-04-12
Chapter 7- 1,5,6,10,22
for Friday. 2010-03-31
Chapter 5- 18,26,33,34,39
Chapter 6- 3a,7,9,13ab 2010-03-19
Chapter 5- 13, 14, 15 2010-03-12
Chapter 12- 3,4,6,7,13 2010-03-03
Chapter 3- 33, 41 (after 41, read 4.2.3...)
Chapter 4- 4, 16ab, 17, 28, 32
Chapter 5- 1, 2ab 2010-02-17
Chapter 3- 2,30,31 2010-02-03 For Monday:
Ch 2- 22 (use $s=a$ as your reference point), 26, 30(b), 34, 36, 37, 45, 49(a-c), 50 2010-01-27
Ch 2- 9, 13, 17, 18 2010-01-20
Ch 1- 13, 29, 53, 62
Ch 2- 1-4

### Assignment #10 - due Friday, March 30

Chapter 6- 13a

Chapter 7- 1, 5 , 6

### Assignment #9 - due Monday, March 26

Chapter 5- 26,33,34,39
Chapter 6- 3,7,9,10

### Assignment #8 - due Friday, March 16

Chapter 12- 6, 10, 13

Chapter 5- 13, 14, 18

5.13
In part b) write your results in terms of the total current $I=\frac{2}{3}\pi k a^3$ for comparison with part a): For $r\gt a$ you get $B_\phi = \frac{\mu_0 I}{2\pi r}$ in both cases.

5.14
By the right-hand rule, the $\myv B$ field is pointing in the direction of $-\uv y$ for $z\gt 0$ and towards $+\uv y$ for $z\lt 0$. (And $B$=0 at $z=0$). So, applying Ampere's law to the loop shown, the only segment on which $\myv B\cdot d\myv l$ is non-zero is the top segment: $$\oint \myv B\cdot d\myv l = Bl=\mu_0 I_\text{enc}=\mu_0lzJ$$ So, from this, we find in the region $-a \lt z \lt +a$: $$\myv B = -\mu_0Jz\uv y$$ Outside of this region, $I_\text{enc}=\mu_0 laJ$, so $$\myv B = -\mu_0Ja\uv y \text{ for } z\gt +a;$$ $$\myv B = +\mu_0Ja\uv y \text{ for } z\lt -a;$$

5.18
Which surface you pick *does not matter*. Consider the volume which is bounded by two different surfaces which share the same boundary. As long as $\myv \grad\cdot \myv J=0$ then just as much charge flows into the volume through one surface as flows out through the other. That is, the flow of charge through both surfaces is the same. (See Theorem 2, Sect. 1.6.2).

The continuity equation says $\myv\grad\cdot\myv J=\frac{d\rho}{dt}$. Magnetostatics is defined by assuming $\frac{d\rho}{dt}=0$. So for magnetostatics the divergence of $\myv J$ is guaranteed to vanish everywhere.

Possible problems that I'll assign or use from Chapter 5 (avoid these when choosing problems to write up)- 14, 16, 17, 26,34,37,39

## Assignment #10

Chapter 6- 3, 9 Chapter 7- 1, 5, 17

Problem 7.1 - Conductivity = $\sigma$ in a region between two concentric metal spheres, where $$\myv J = \sigma \myv E$$

Let there be a charge $Q$ on the inner shell of radius $a$. Then the electric field outside the shell (but $r \lt b$) is $$\myv E=\frac{Q}{4\pi\epsilon_0r^2}\uv r$$ The voltage difference is $$\begineq V_a-V_b &=& -\int_a^b \myv E\cdot d\myv r=-\frac{Q}{4\pi\epsilon_0}\int_a^b \frac{1}{r^2}dr\\ &=&\frac{Q}{4\pi\epsilon_0}\left(\frac 1a-\frac 1b\right)\\ \endeq$$ $$\begineq I&=&\oint_{\cal S}\myv J\cdot d\myv a\\ &=&\sigma\oint_{\cal S}\myv E\cdot d\myv a=\sigma \frac{Q_\text{enc}}{\epsilon_0}=\sigma \frac{Q}{\epsilon_0} \endeq$$

To find the resistance, $R=\Delta V/I$,...
We solve the expression for $\Delta V$ for $Q$... $$Q = \frac{(V_a-V_b)4\pi\epsilon_0}{\left(\frac 1a-\frac 1b\right)}$$ Putting this into our expression for current: $$I=\frac{4\pi\sigma}{ \left(\frac 1a-\frac 1b\right)}.$$ Now, we can take the ratio of voltage to current... $$R=\Delta V/I=\frac{1}{4\pi\sigma} \left(\frac 1a-\frac 1b\right).$$ and in the limit $b \gg a$ this approaches... $$R\to \frac{1}{4\pi\sigma a}.$$

For two submerged balls ... the effective resistance is twice this. Once as the current gets "far away" from the first ball, and once as the current from "far away" ends up on the second ball. So the net current will be $$I=V/2R=V2\pi\sigma a.$$

## Assignment #9

Chapter 5- 14, 16, 23, 34

## Assignment #8

Chapter 4: 23
PMR Lucite problem (see below)
Chapter 5- 1, 2ab, 8
Chapter 12- 3, 4, 7
Moodle
: Magnetic field doing work?

#### Notes

Lucite problem: When a beam of high energy electrons bombards an insulating material such as lucite, the electrons penetrate into the material and remain trapped inside. Consider a 0.1 microampere beam that bombarded 25 cm${}^2$ of lucite (dielectric constant $K=3.2$) for 1 second. Essentially all the electrons were trapped about 6 mm below the surface in a region about 2 mm thick. The block was 12 mm thick.

Assume a uniform density for the trapped electrons in the 2 mm thick layer, and neglect edge effects (that is, all quantities vary only in one direction). Now, assume also that both faces are in contact with grounded ($V=0$) conducting plates. Calculate the following:

1. What is the bound charge density in the charged region?
2. What is the bound charge density at the surface of the lucite?
3. Sketch graphs of $D$, $E$, $V$ as functions of position inside the dielectric. Show that the potential at the center of the sheet of charge is about 4 kilovolts.
4. What is the electric field intensity in the charge-free region?
5. What is the energy stored in the block? What is the danger that the block will explode?
6. How would the curve of $V$ be affected if the sheet of electrons were closer to one face of the block than to the other?
1. A sketch, $x$ and $y$ not to the same scale as $z$ is below. Let's say that the 2 mm layer with electrons is centered at the origin, at $z=0$, extending in the $x-$ and $y-$directions. So the grounded plates are at $z=\pm 6$ mm.

The electrons implanted in the 2mm-thick region represent free charge. Since 1 Amp = 1 Coulomb / second, the total free charge is $$Q_f=0.1\mu A*1s=1\times 10^{-7} \text{ C}.$$ The volume of the central region is
$\tau=$2 mm * 25 cm^2=$\tau = 2\times 10^{-3} 25\times(10^{-2})^2=$$5.0\times 10^{-6}\text{ m^3}. The free charge density in the central region is \rho_f=\frac{Q_f}{\tau} = -\frac{1\times 10^{-7}}{ 5.0\times 10^{-6}}$$= -0.02 \text{ C/m^3}.$
This is negative because we put electrons in there.

You can use Eq 4.39 (Griffiths) to get the bound charge: $$\rho_b=-\left(\frac{\chi_e}{1+\chi_e}\right)\rho_f = -\left(\frac{K-1}{K}\right)\rho_f$$ Since the dielectric constant of lucite is $3.2=K\equiv \epsilon_r=1+\chi_e$. Putting this value for $K$ into the formula above we get
$\rho_b=+(2.2/3.2)\rho_f = +0.0138\text{ C/m^3 }.$

2. Perhaps the easiest way to get the surface charge density is to recognize that, well, if the lucite started out neutral, then the total surface charge (split evenly between the top and bottom of the block of lucite) is just equal and opposite to the total bound charge.

There is no bound charge outside of the region of free charge (see equation 4.39 again). The bound charge in a box of volume $At$, where $t=0.002$m (the thickness of the charged region) is $Q_b=\rho_b At$. So, the charge density on one of the surfaces would be
$\sigma_b=-\frac{Q_b}{2A}=-\frac{\rho_bAt}{2A}=-\frac{\rho_bt}{2}=-1.38\times 10^{-5}\text{C/m^2}$.
Since the bound charge in the middle is positive, this surface charge density has to be negative.

[We should get the same answer if we figure the polarization $\myv P$, and then take $\sigma_b =\myv P\cdot \uv n$. We'll try that later, when we have $\myv P$ in hand.]

3. $\myv D$ depends only on $\rho_f$. Gauss's law for $\myv D$ is $$\oint \myv D\cdot d\myv a= Q_{f\text{ enc}}$$ The block is wider than it is thick (12 mm). "Ignoring edge effects" means we'll approximate $\myv D$ as not varying in the $x$- and $y$-directions, and only depends on $z$, and is directed in the $\uv z$ direction, such that $$\myv D=D(z)\uv z.$$ In the middle of my sandwich, I have just as much free charge above as below, so $D(z=0)=0$.

Now I make a little Gaussian box, with top and bottom area $A$. The bottom is located at the origin $z=0$, where $D(0)=0$, and the top extends to a height $z$, where I'd like to find $D(z)$. On the sides of the box there will be no perpendicular component of $\myv D$. The area with free charge lies between $z=\pm$ 1 mm. I'll call $h=1$ mm. Then Gausses law for the $\myv D$ states: $$\oint \myv D\cdot d\myv a=DA= Q_{f\text{ enc}}.$$

As long as $z\lt h$, $Q_{f\text{ enc}}=\rho_fAz$, so $D=\rho_f z=0.02z$ C/m^2.

When $z\gt h$, $Q_{f\text{ enc}}=\rho_fAh$ , so $D=\rho_f h=2\times 10^{-5}$C/m^2 (no $z$ dependence). So, here's the sketch:

So, $\myv D$ is pointing towards the free charge (electrons), this means $D(z)$ is negative for $z\gt 0$ and positive for $z \lt 0$.

Now, $\myv E =E(z)\uv z= \myv D/\epsilon=(D(z)/\epsilon)\uv z$. Since the dielectric constant of Lucite is 3.2, that means $\epsilon=3.2\epsilon_0$. The graph of $E$ looks qualitatively identical to the graph of $D$, but with the maximum value at $h\rho_f/(3.2*\epsilon_0)=7.06\times 10^5$ V/m.

Electric potential? We know that $V(z=-6mm)=0=V(z=+6m)$. To find $V(z)$ we can integrate the electric field:$$V(z)=\int_{-6mm}^z -E(z')dz'.$$ This must be dropping linearly in the charge free region $z\lt -h$, and then it's curved (like a parabola) in the charged region, and increases linearly in the charge-free region $z\gt h$. (Sketch below).

It's easy to integrate the voltage in the charge-free region $z\lt -h$ where $E$ is constant, up to $z=-h$: that's going to be $E*5$ mm = -3.5 kV. Sure, close enough to 4 kv (question d)

4. In the charge-free region the electric field is constant with value:
$E=7.06\times 10^5$ V/m.

Alternate way to get the bound charge density, surface charge density...To find the polarization, use the definition of the displacement field to get: $$\myv P = \myv D-\epsilon_0\myv E = (D(z)-\epsilon_0E(z))\uv z=P(z)\uv z$$

We figured out that $E(z)=D(z)/\epsilon$, so $P(z)=D(z)-\epsilon_0(D(z)/\epsilon)= D(z)\left(1-1/3.2\right)=0.688D(z)$. So, the graph of $P(z)$ has the same shape as $D(z)$.

In the charged region, $P(z)=0.688\rho_f z$.

Outside of the charged region, $P(z)$ is either $+0.688\rho_f h$ or $-0.688\rho_fh$.

The bound charge is $\rho_b=\myv\grad\cdot \myv P$. Since $\myv P=P(z)\uv z$, the divergence is $\frac{d}{dz}P(z)$: In the charge-free region, $\rho_b=0$, and in the charged region $\rho_b=0.688\rho_f=+0.014$C/m^3. It's positive to counteract the negative free charge.

The surface charge is $\sigma_b =$$\myv P(z=\pm 6\text{ mm})\cdot \uv n=-0.688\rho_fh=$$-1.4\times 10^{-5}$C/m^2.

5. The energy density in a linear dielectric is $u=\frac{1}{2}\myv E\cdot\myv D =\frac{1}{2\epsilon}\myv D\cdot\myv D$. The E-M field energy in the block is twice the energy in the volume from $z=0$ to $z=6$mm. The volume element $d\tau=z*25$cm^2. $$\begineq U&=&25\text{cm^2}*2*\frac{1}{2\epsilon}\int_0^{6\text{mm}} D(z)^2\,dz\\ &=&\frac{0.0025}{3.2*\epsilon_0}\left(\int_0^{1\text{mm}}(\rho_fz)^2\,dz+\int_{1\text{mm}}^{5\text{mm}}(\rho_f h)^2\,dz\right)\\ &=&\frac{0.0025\rho_f^2}{3.2*\epsilon_0}\left(\left.\frac{z^3}{3}\right|_0^{0.001}+\left.h^2z\right|_{0.001}^{0.006}\right)\\ &=&\frac{0.0025\rho_f^2}{3.2*\epsilon_0}\left( (0.001m)^3/3 + (0.001m)^2(0.005m)\right)\\ &=&1.88\times 10^{-4} J \endeq$$ The energy density works out to 0.063 J/L. Compare to energy densities of fuels and batteries most of which are in excess of MJ/L. So we conclude there is little risk of explosion.

Problem 5.2

You can use the general solution (equation 5.6 in Griffiths) and fit the constants to the boundary conditions (that is, the initial velocity)

The solution in part a is no net acceleration, so particle continues in a straight line along $y$ axis.

We need to find a solution in part b with the particle starting at the origin, with $\myv v(0)=(E/2B)\uv y$.The two equations of 5.6 are... $$y(t)=C_1\cos\omega t+C_2\sin\omega t+(E/B)t+C_3.$$ $$z(t)=C_2\cos\omega t-C_1 \sin\omega t + C_4.$$ The condition that $y(0)=0$ and $z(0)=0$ implies that... $$C_1=-C_3;\ \ \ C_2= -C_4.$$ The condition that $\dot z(0)=0$ implies that $C_1=0=C_3$.

The condition that $\dot y(0)=(E/2B)$ implies that... $$(E/2B)=C_2\omega+E/B$$ That is, $C_2=-(E/2B\omega)=-C_4$.

Putting these constants into the general solution, we get $$y(t)=\frac{E}{2 B\omega}$2\omega t-\sin\omega t$$$ $$z(t)=\frac{E}{2 B\omega}$1-\cos\omega t$$$

You can do a parametric plot in Mathematica to see the trajectory. Let the $y$-axis represent $z(t)$ and the $x$-axis represent $y(t)$. If the "units" of each axis are $E/(2\omega B)$, then this Mathematica command will plot the trajectory:

ParametricPlot[
{ 2*t-Sin[t],   1-Cos[t] },
{t, 0, 8*Pi}
]

Problem 12.3

Part a:

According to Galileo, speeds add like: $$v_G=v_1+v_2.$$ According to Einstein, speeds add like: $$v_E=\frac{v_1+v_2}{1+v_1v_2/c^2}=v_G\frac{1}{1+v_1v_2/c^2}$$ Since $(v_1v_2/c^2)\ll 1$, we can approximate the fraction using the binomial expansion and write $v_E$ as $$v_E\approx v_G(1-v_1v_2/c^2)$$ So, the fractional difference between Galileo's and Einstein's expressions is $$\frac{v_G-v_E}{v_G}\approx \frac{v_G-v_G(1-v_1v_2/c^2)}{v_G}= v_1v_2/c^2$$ With $v_1$=60 mph, $v_2$=5 mph, $c=3\times 10^8$m/sec*(3600 sec/hr)*(1 mile/1600 m)=$6.75\times 10^8$mi/hr, the percentage difference is: $$100 * (60)(5)/(6.67\times 10^8)^2=6.6*10^{-14}\%$$

## Assignment #7

Chapter 3 32, 36, 47
Chapter 4- 1, 7, *16a (only a)

3.47

#### part a

This problem requires that you think rather carefully about how you label positions. Often I use $\myv \rr=\myv r-\myv r'$ to indicate the separation vector between say, sources of fields, and the location where you wish to evaluate the field. In this problem you need to think carefully how you use $\myv r$ and $\myv r'$. I think it's more helpful to write things in terms of $\myv r$ and $\myv r'$ directly, and not use $\myv \rr$.

To find the average electric field inside a sphere due to a single point charge $q$ at position $\myv r$, we are going to use $\myv r'$ to locate the position where we evaluate the field $\myv E(\myv r')$ due to $q$, and then integrate over all positions $\myv r'$. The field at $\myv r'$ is proportional to $q/\rr^2$: $$\myv E(\myv r')=\frac{1}{4\pi\epsilon_0}\frac{q*(\myv r' -\myv r)}{|\myv r'-\myv r|^3}.$$ So, the average of the electric field over the volume ${\cal V}$ is $$\myv E_\text{ave}(\myv r) = 1/ {\cal V} \int_{\cal V} \frac{1}{4\pi \epsilon_0} \frac{q*(\myv r' - \myv r)}{|\myv r'-\myv r|^3} d \tau'.$$

If this sphere has a radius $R$, then ${\cal V} = \frac{4}{3}\pi R^3$.

To continue with this first part, you don't actually have to solve the integral. You just have to show that when you write down how to calculate the other situation, that you get the same answer.

Another problem: Imagine we have a sphere with a constant charge density $\rho$, and we'd like to know the field inside this sphere at a position $\myv r$. To get that, we sum up the field due to each little chunk of charge $dq = \rho d \tau' at \myv r'$ like this (the 's' subscript indicates the constant charge density sphere):

$$\myv E_s(\myv r)=\int_{\cal V} \frac{1}{4\pi \epsilon_0} \frac{\rho (\myv r-\myv r')}{|\myv r-\myv r'|^3} d \tau'.$$

If it happens that $\rho = -\frac{q}{\cal V}$:

$$\myv E_s= -\frac{1}{\cal V}\int_{\cal V} \frac{1}{4\pi \epsilon_0} \frac{q (\myv r-\myv r')}{|\myv r-\myv r'|^3} d \tau'.$$

Now, $|\myv r-\myv r'| = |\myv r'-\myv r|$, and $\myv r-\myv r' = - (\myv r'-\myv r)$, so we see that:

$$\myv E_\text{ave} = \myv E_s.$$

#### part b

You worked the "Electric field due to a constant charge density inside a sphere" problem already. It was problem 2.12 (Assignment #4). Using Gauss' law you should have shown that inside the sphere, the field is...

$$\myv E_s(\myv r) = \frac{\rho}{3 \epsilon_0} \myv r.$$

Putting in our constant charge density...

$$\myv E_s(\myv r) = \frac{1}{3{\cal V} \epsilon_0} q \myv r= \frac{1}{4\pi \epsilon_0 R^3} \myv p.$$

Where $\myv p \equiv q \myv r.$

#### part c

Now, $\myv E_s(\myv r)$ is the same as the average field $\myv E_\text{ave}(\myv r)$ due to a single point charge at $\myv r.$