Homework & Assignments

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

[ Phys 303 on Moodle]


  • 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.
  • Staple your pages together.
  • Circle or otherwise highlight answers and important intermediate results.
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

Notes / Answers

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

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.

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 $$\myv B = -\mu_0Ja\uv y \text{ for } z\gt +a;$$ $$\myv B = +\mu_0Ja\uv y \text{ for } z\lt -a;$$

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
: Magnetic field doing work?


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:

    { 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)



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 $\myv r.$

By superposition, the total dipole moment $\myv p$ is just the sum of the individual dipole moments, and the total average electric field $\myv E_\text{ave}$ must be just the sum of the individual-charge averages, so apparently (interpreting these now as total quantities, rather than individual point charge quantities): $

$$\myv E_\text{ave} = \frac{1}{4\pi \epsilon_0 R^3} \myv p.$$

part d

Run the same argument as in part a, but with the point charge outside the sphere:

What is the average value of the electric field (averaged over a sphere of radius $R$) due to a single point charge $q$ at a position outside the sphere? We can write this as a function of the position $\myv r)$ of the point charge, where this is now a vector outside the sphere, though we still want the average over the sphere of radius $R$:

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

Now, as a separate problem, we'd like to calculate the electric field at a position $\myv r)$ due to a uniform charge density $\rho= -q/{\cal V}$ in a sphere of radius $R$. This time, $\myv r)$ is a position outside the sphere with the constant charge density. Let's call it (once again) $\myv E_s$.

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

Once again, we can show that...

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

Once again we can use Gauss' law to evaluate $\myv E_s(\myv r).$

[You can do this!]

This time instead of the field being related to the dipole moment, we find that the average over a small sphere due to a single charge outside the sphere is the same as the field of that single charge at the center of the sphere.

So, we can use superposition again to say that the average of the electric field of many charges outside a sphere is just the same as the electric field at the center of the sphere summed up for all the external charges.


Problem 16 a: Use the superposition principle. You can create a cavity in your material with a uniform polarization $\myv P$ by placing a sphere with a uniform polarization $-\myv P$ at the desired position.

In Example 4.2 they do the calculation of the electric field at the center of a sphere with a uniform polarization.

In part b, figure out the charge due to polarization, $\myv P\cdot d\myv a$ at the end of the long-thin cavity, and figure out the electric field of that charge (or why it's not important).

In a big hunk of dielectric material, we have $\myv D_0 = \epsilon_0 \myv E_0 + \myv P$. Now we start hollowing out cavities. What if we hollow out a spherical cavity? Now, where we had material with polarization $\myv P$ before, now we have no polarization. Find the field $\myv E$ and the displacement $\myv D$ at the center of the cavity.

Taking the hint about adding a material of opposite polarization.... We can model the void as the sum of the original dielectric material which had polarization $\myv P$ plus a sphere with uniform polarization $-\myv P$. By superposition, the new electric field should be the original electric field $\myv E_0$ plus the field at the center of a sphere with uniform polarization $-\myv P.$

The field inside and outside of an isolated dielectric sphere with uniform polarization is the topic of Example 4.2. If the polarization is uniform, then there is no bound volume charge, but there is a bound surface charge proportional to $\myv P \cdot d \myv a \propto P \cos \theta$. According to example 4.2, the electric field inside a sphere with a uniform polarization $\myv P$ is uniform: $$\myv E = -\frac{1}{3 \epsilon_0} \myv P.$$

So, in our case, we'll take the original field $\myv E_0$ and add to that the electric field due to a sphere with polarization $-\myv P$, that is:

$$\myv E = \myv E_0 + \frac{1}{3 \epsilon_0} \myv P,$$

in terms of $\myv E_0$ and $\myv P.$

What about the displacement? $\myv D = \epsilon_0 \myv E + \myv P = \epsilon_0 \myv E$, since $\myv P=0$ in the cavity. Using our expression above for the electric field, this is: $$\begineq\myv D &=& \epsilon_0 (\myv E_0+\frac{1}{3 \epsilon_0} \myv P)=\epsilon_0 \myv E_0 +\frac{1}{3}\myv P\\ &=& \left[\epsilon_0 \myv E_0 + \myv P\right] -\frac{2}{3} \myv P.\endeq$$

The quantity in $[..]=\myv D_0$, so

$$\myv D = \myv D_0 -\frac{2}{3} \myv P.$$

What if we hollow out a long, needle-shaped cavity instead?

Once again, we add a chunk of dielectric with a uniform polarization $-\myv P)$ to mimic the effect of a void.

So, where's the bound charge for a long, needle-like volume of dielectric?


Answer to electric field part is $\myv E = \myv E_0.$

Assignment #6

Chapter 2- 49, 56
Chapter 3- 2, 24 (just get the $\Phi(\phi)$ and $S(s)$ equations), 26

PMR- 1-d spreadsheet solution of Laplace's equation. Send a file rather than printing out.

Notes / Answers

2.56 - You may use the results of problem 2.32, that the electrostatic energy of a solid sphere of charge $q$ spread uniformly in a sphere of radius $R$ is $$W=\frac{1}{4\pi\epsilon_0}\left(\frac{3q^2}{5R}\right) $$

Don't miss the part of the problem that asks you to make a comparison with the actual age of the sun. Look that up! ...Or do you know roughly how old Earth is? That would let you come up with a lower-bound for the age of the sun.

Most of you got to the result that if the sun were only radiating the gravitational energy of the sun, that it should be exhausted in about 19 million years. The point was to make a comparison with the actual age of the sun.

I don't know how old the sun is, but Earth is approximately 4.6 billion years, and this is already greater than 19 million years by almost two orders of magnitude. $\Rightarrow$ the sun cannot be shining merely by harnessing its gravitational energy. (We know that there is lots of nuclear fusion going on.)

Of course the biggest source of energy for the sun is nuclear fusion, not gravitational energy.


Next year: Have students do a plot of V along some of these lines

A particle can only be in equilibrium where $V$ has a minimum. But this is only possible if $\del^2V/\del(x,y,z)^2 \gt 0$.

Where will the particle "leak" out of the box? It is possible to calculate the potential along various lines through the center of the box. (See this notebook). But even without doing the calculation, you can perhaps guess that because the centers of the faces are the places on the surface that are the farthest from any of the point charges, that this will be the "lowest energy" place on the surface for a positive charge, and the most likely location where charge could leak out.

See this notebook for some graphs of $V$ in different directions. It turns out that the potential is dropping along a line from the center to a face of the cube...

3.24 - Just take this problem to the point of finding the equations that $S(s)$ and $\Phi(\phi)$ must satisfy.

The final form of the potential (which you'll need for 3.26) is: $$\begineq V(s,\phi)&=&a_0+&b_0\ln s \\ & &+\sum_{k=1}^\infty & \left[ s^k(a_k\cos k\phi + b_k\sin k\phi)\right.\\ & & &\left.+s^{-k}(c_k\cos k\phi + d_k\sin k\phi) \right]\endeq$$

3.26 - For this problem, you'll use the general solution from problem 24, and for boundary conditions:

  • Equation 2.34: The potential is continuous across the surface at $s=R$: $$V_\text{out}=\left. V_\text{in}\right|^{s=R}.$$
  • Equation 2.36: There's a discontinuity in the electric field across the surface charge at $s=R$: $$-\epsilon_0\left(\frac{\del V_\text{out}}{\del s} - \frac{\del V_\text{in}}{\del s}\right)^{s=R}=\sigma.$$

Assignment #5

Chapter 2- 20, 22, 26, 30(b), 34, 36, 50
Moodle - Energy density question

Notes / Answers

Energy density [Moodle question] - the integral expression $W=\frac{\epsilon_0}2\int E^2\,d\tau$ is always positive and would account for the energy a.) to assemble point charges out of diffuse charge density, as well as b.) bring point charges in from infinity into proximity to other charges. The summation $frac 12\sum_i q_iV(\myv r_i)$ only takes into account the energy in b.)

2.22 - 5 pts extra credit if you first derived $E_s=\lambda/(2\pi\epsilon_0 s)$. [Problem 2.13]

Otherwise, you may have looked up the field $\myv E=\frac{1}{4 \pi \epsilon_0}\frac{2\lambda}{s}\uv s$. Instead of setting $V(s=\infty)=0$, this time set our zero point to $V(s=a)=0$. Once you solve this problem see what the consequences would have been of setting $V(s=\infty)=0$.

2.26 - Integrate over all the charge directly, using as integration variable the distance $\rr$ along the surface of the cone.

Solution: All the charge on the circle shown is at the same distance $\rr$ from point a, and distance $\rr'$ from point b. The radius of the circle is $r(\rr)=\rr/\sqrt 2$. So, the total charge $dq$ on the circle is $$dq=\sigma 2\pi r\,d\rr=2\pi \rr\sigma\,d\rr/\sqrt 2.$$ The potential at point a is: $$\begineq V(a)&=&\frac{1}{4\pi \epsilon_0}\int \frac{dq}\rr =\frac{1}{4\pi \epsilon_0}\int_0^{\sqrt 2 h}\frac{2\pi\rr\sigma}{\sqrt 2 \rr }d\rr \\ &=&\frac{1}{4\pi \epsilon_0}\int_0^{\sqrt 2 h}\frac{2\pi\sigma}{\sqrt 2 }d\rr\\ &=&\frac{1}{4\pi \epsilon_0} \frac{2\pi\sigma}{\sqrt 2 }(\sqrt 2 h)=\frac{\sigma h}{2\epsilon_0} \endeq $$

To find the potential at point b: $$V(b)=\frac{1}{4\pi \epsilon_0}\int \frac{dq}{\rr'} =\frac{1}{4\pi \epsilon_0}\int_0^{\sqrt 2 h}\frac{2\pi\rr\sigma}{\sqrt 2 \rr' }d\rr, $$ we need to find $\rr'(\rr)$. Consider the colored right triangle. Pythagoras says: $$\rr'^2=(h-r)^2+r^2= \left(h-\frac{\rr}{\sqrt 2}\right)^2+\left(\frac{\rr}{\sqrt 2}\right)^2.$$ After expanding $$\rr'=\sqrt{h^2-\sqrt 2h\rr+\rr^2}.$$ You can substitute this into the expression for $V(b)$ and we now have a (complicated) integral over $\rr$ to solve. The result (Mathematica!) is $$V(b)=\frac{\sigma h}{2\epsilon_0}\ln(1+\sqrt 2).$$

Therefore the potential difference is $$V(a)-V(b)=\frac{\sigma h}{2\epsilon_0}\[1-\ln(1+\sqrt 2)\].$$

2.34 - This problem uses the results of problem 2.21 which we did in class.

2.36 - The only region where $\myv E_1\cdot\myv E_2 \neq 0$ is $r\gt b$.

2.50 - To get the electric field from the potential (which only depends on $r$) $$\myv E = -\myv\grad V=-A\uv r\frac{\del}{\del r}\left(\frac{e^{-\lambda r}}{r}\right)=Ae^{-\lambda r}(1+\lambda r)\frac{\uv r}{r^2}.$$

To find the charge density, we'll use $\myv \grad\cdot\myv E = \rho/\epsilon_0$ and also product rule #5 which tells us how to expand $\myv \grad\cdot (f\myv A)$: $$\rho = \epsilon_0\myv \grad \myv E = \epsilon_0 A\{ e^{-\lambda r}(1+\lambda r)\myv \grad \cdot \frac{\uv r}{r^2} +\frac{\uv r}{r^2} \cdot \myv \grad\left(e^{-\lambda r}(1+\lambda r) \right) \}.$$

Now, $\myv \grad\cdot(\uv r/r^2)=4\pi\delta^3(\myv r)$. Furthermore, according to 1.88, $f(x)\delta(x)=f(0)\delta(x)$. So, the first term in the $\{...\}$ above becomes $e^{- 0}(1+0)4\pi\delta^3(\myv r)=4\pi\delta^3(\myv r)$.

The second term in the $\{...\}$ is $(\uv r/r^2)\cdot \uv r (-\lambda^2re^{-\lambda r})=-\lambda^2e^{-\lambda r}/r$. So the charge density is $$\rho(\myv r)=\epsilon_0 A\{ 4\pi\delta^3(\myv r)-\frac{\lambda^2}{r}e^{-\lambda r} \}.$$

The total charge is $$Q=\int_{\text{all space}} \rho(\myv r)\, d\tau.$$ When you work out the integrals, the answer is 0.

Assignment #4

Chapter 1- 61(a), 62(a), 63(a)
Chapter 2- 1, 3, 12, 18

Notes / Answers

1.61(a) - It's really important on this one to keep track of which quantities are vectors and which are scalars.

Using the hint, we look at a vector $\myv v=\myv c T=c_xT(x,y,z)\uv x+c_yT(x,y,z)\uv y +c_zT(x,y,z)\uv z$, which is the product of a constant vector $\myv c$ and a scalar field $T(x,y,z)$. The fundamental theorem for divergences is our starting point: $$\begineq \int_{\cal V}\myv\grad\cdot \myv v\,d\tau &=& \oint_{\cal S}\myv v \cdot d\myv a\\ \endeq $$ .
$$ \int_{\cal V}\myv \grad T \,d\tau = \oint_{\cal S} T(x,y,z) d\myv a$$ which is the result we are to show.

1.62(a) - Find the vector area $\myv a = \int_{\cal S}d\myv a$ where the surface ${\cal S}$ to integrate over is half of a sphere of radius $R$.

Here, $d\myv a$ is a vector perpendicular to the surface of our half sphere (centered on the origin), that is in the radial direction $\uv r$: $$d\myv a=dl_\theta\,dl_\phi\,\uv r=R^2\sin\theta\,d\theta\,d\phi\,\uv r.$$

You should start by making an explicit argument that $a_x = \int da_x = 0$ and $a_y=\int da_y=0$. Then, you *only* have to worry about how to compute the $z$-component: $\left(\int d\myv a\right)_z=\int d\myv a \cdot \uv z.$

Putting this together: $$d\myv a = R^2\sin\theta\,d\theta\,d\phi\,\uv r$$

Now think about the Cartesian components of $d\myv a$ with the center of the bowl, as shown, on the $z-$axis. Draw $d\myv a$ at some position $r, \theta, \phi$ on the surface of the bowl. Note that at the point at $\phi + 180^o$ on the other side of the bowl, the $x$- and $y$-components of $d\myv a$ will be exactly opposite. So, that when we integrate over the whole surface, these components will sum to zero. Ah-ha--we only need to worry about the $z$-component. That $z$-component is $$\begineq (d\myv a)_z&=&R^2\sin\theta\,d\theta\,d\phi\,\uv r\cdot \uv z\\ &=&R^2\sin\theta\,d\theta\,d\phi\,\cos\theta \endeq$$

Now we're ready to integrate that $z$-component over the surface of the bowl $$\begineq \myv a =(\myv a)_z \uv z&=& \uv z \int_0^{\pi/2} d\theta\int_0^{2\pi} d\phi R^2\sin\theta\cos\theta \\ &=& \uv z R^2 2\pi \int_0^{\pi/2} d\theta \,\sin\theta\cos\theta\\ &=& \uv z R^2 2\pi \left[ \frac{1}{2}\sin^2\theta\right]_0^{\pi/2} \\ &=& \uv z R^2 2\pi \frac{1}{2}=\pi R^2 \uv z. \endeq$$

1.63a - With $\myv v=\frac{1}{r}\uv r$, you will find that $\myv \grad \cdot \myv v=\frac{1}{r^2}$.

Next, the idea is to calculate separately the two integrals of the divergence theorem for a spherical volume $\cal V$ of radius $R$ bounded by a surface $\cal S$: $$\oint_{\cal S} \myv v\cdot d\myv a = \int_{\cal V} \myv \grad \cdot \myv v\,d\tau.$$ For the function $\myv v'=1/r^2$ these two were *not* equal, leading us to believe that there is a dirac delta function (at the origin). But for this $\myv v$, you will find that both integrals are equal (value $4\pi R$) and so there is no dirac delta function.

2.1 - Hint: Can you think of the arrangement with one charge missing as the sum of a symmetrical arrangement of charges plus an additional charge (of a different sign)?
c) Symmetry sez net force = 0.

d) Using the principle of superposition, think of this as [symmetrical arrangement] + [charge of opposite sign at position of 'missing' charge]. So the net force is that due to one opposite charge at the position of the hole.


Assignment #3

Chapter 1- 26c, 28, 30, 32, 34, 38, 39

PMR- spot assignment on Mathematica and the gaussian approx to $\delta(x)$.

Notes & Answers

1.26c - Consider the scalar function $$T=e^{-5x}\sin 4y\cos3z.$$ The Laplacian of this function is: $$\grad^2 T = \frac{\partial^2 T}{\partial x^2} +\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}.$$ Calculating each second derivative separately: $$\frac{\partial^2 T}{\partial x^2} = (-5)(-5)e^{-5x}\sin 4y\cos3z=25 T.$$ $$\frac{\partial^2 T}{\partial y^2} = (4)(-4)e^{-5x}\sin 4y\cos3z=-16 T.$$ $$\frac{\partial^2 T}{\partial x^2} = (-3)(3)e^{-5x}\sin 4y\cos3z=-9 T.$$ Adding them up: $$\grad^2 T = (25 -16 -9) T = 0$$

1.28 - The proof involves writing out the components of the cross product of the gradient of a function in Cartesian coordinates, and then using the fact that, for any function of several variables, the order of (partial) differentiation does not matter.

1.30 - In the first part "upward" is the positive direction, so $$d\myv a = dx\,dy\,\uv z$$ You have to solve the double integral... $$\begineq \int_0^2\int_0^2 \myv v(x,y,z=0) \cdot dx\,dy\,\uv z &=&\int_0^2\int_0^2 v_z(x,y,z=0) \, dx\,dy\\ &=&\int_0^2\int_0^2 -3y dx\,dy\ \\ &=&\int_0^2 -3y\,dy \left(\int_0^2 dx \right)\\ &=&\int_0^2 -6y\,dy=-12. \endeq$$ The four edges around the "bottom of the box" in Fig. 1.23 form the boundary line of both:

  • the surface made up of the five squares of example 1.7, and
  • the surface consisting of the single, bottom square of problem 30.

The flux "up" through the top five squares (Example 1.7) was 20. You just figured that the flux "up" through the bottom was -12. Not the same, so apparently the surface integrals of this function do not depend just on the boundary line.

[Really, it's only guaranteed to be the same if the vector field $\myv v$ is the curl of some other vector function, that is, if $\myv v= \myv \grad \times \myv A$. So, we could conclude that there's *no way* that we could write $\myv v = 2xz \uv x+(x+2)\uv y +y(z^2-3)\uv z$ as the curl of some other vector function $\myv A$.]

As for the total flux out of the box, we need to add the flux out through the surfaces of Ex 1.7., (20) to the flux outward through the bottom surface of the box.

To find the outward flux, we do the same integral as in the first part, but this time the surface normal is $d \myv a=-dx\,dy\,\uv z$, so it's just (-1) times the previous answer.

The flux going "down" through that face is +12. The total flux out of the box is then $$\Phi_\text{net}=20+12=32.$$

1.32c - Use $x$ as your integration variable. You have expressions for $y(x)$ and $z(x)$. Go ahead and express all the components of $d\myv l$ in terms of $dx$.

1.34 - For the surface integral, you'll have $d\myv a=dy\,dz\,\uv x$.

1.38 - Work these out by trigonometry starting with Figure 1.36. Make a couple sketches to illustrate your derivations of the unit vectors.

1.39 - Work these out using the spherical-polar forms for the divergence, $\myv \grad \cdot$, and the volume differential, $d\tau$. Oh! We already did part b) in class, but write it down again as review.

Assignment #2

Chapter 1- 12, 13, 15c, 16, 18c


12 - You were asked about the peak position relative to South Hadley. The answer, peak at (-2,3), means 2 miles west and 3 miles north of South Hadley.

13 - Since $x$ and $x'$ are different, independent variables, $\frac{\del x'}{\del x}=0$.
9/10 for running through all the gradients correctly. But 10/10 if you recognized that you could write the last result in terms of $\rr$.

Assignment #1

Chapter 1- 2, 3, 5, 10(abc)

PMR- "What is direction?" [type in your answer on Moodle.]


10(ab) - Say that the vector $\myv V$ has components $\{l,m,n\}$ in the original coordinate system. Give its components $\{l',m',n'\}$ in the new coordinate system in terms of $l$, $m$, and $n$.

10(c) You might want to do this one with a particular set of vectors $\myv A \times \myv B = \myv V=\{l,m,n\}$. In the inverted system, find the components of $\myv A'$ and $\myv B'$, and then find their cross product, $\myv V'$, using equation 1.14. Now see how the components of $\myv V'$ relate to $\myv V$.