184_projects:f18_project_11_sol

The EM-Boar Tigers have been fought back for now but your team needs to get a message out that you all are under attack. Cellular and radio signals are being somehow blocked from an external source but you know that the annex building that is within sight of your compound has an emergency telegraph machine for just such attacks. You need to get a message to the telegraph operator. You notice that there is an old-fashioned camera that produces flashes via a capacitor. You decide that you might be able to communicate with the annex via Morse code and go to the roof to try and send your message.

You measure the capacitor and find that the capacitor plates are 2 cm by 3 cm and separated by a distance of 40 $\mu m$. When hooked up to a battery, the plates become fully charged with $Q = 1.5*10^{-7}C$. The team leader however is concerned that the electric field inside the capacitor is going to be too large and cause sparking between the plates and destroy your only chance at getting out a message.

With your team, build a model for the capacitor that allows you to calculate the electric field inside and outside the parallel plates. Make sure you have a diagram of the electric field, a graph of the electric field, and a graph of the electric potential to convince him that the electric field is either safe or unsafe so that you can get out the rescue message.

Learning Goals

  • Use Gauss's Law to calculate the electric field in between two charged parallel plates.
  • Explain why you picked your Gaussian surface and how it helped you simplify your calculations.
  • Describe how the magnitude and direction of the electric field changes both inside and outside the capacitor.
  • Explain the general steps that you take when using Gauss's Law.

Learning Issues

  • Watch for students plugging in numbers too early - especially here it is much better to keep the solution in terms of variables until the end. This helps the students to: check to see if their units are the end by examining the units of their variables, get some intuition about the general form of the solution, and then get some kind of numerical answer for their electric field.
  • Returning to the electric field - this is going to be kind of a harsh course correction from all of the magnetic field stuff the students have been working on fore the last while in class. We are coming back to electric fields to talk about a couple of mathematical tricks for working with electric fields and how they relate to magnetic fields. This is a good time to do it because now we have a little more intuition for what fields look like, which is a prerequisite for working with Gauss's Law.
  • Gaussian surfaces are imaginary, not real, purely mathematical constructs. They do NOT actually exist. Students often get confused thinking about imaginary surfaces that they build to do math with. We could pick any surface we want, but we look for two specific things to simplify the math:
    • We want a surface where $\vec{E}$ is parallel to $d\vec{A}$ to simplify the dot product.
    • We want a surface where the $\vec{E}$ is constant (magnitude and direction) everywhere on the surface.
    • Common choices are “pillboxes”, spheres, and cylinders because of symmetry.
  • $d\vec{A}$ is a unit vector (length 1) that points directly out from the surface. It points in a direction perpendicular to whatever Gaussian surface we draw.
  • The last page of notes for this section has an outline of the steps to follow if students are struggling to come up with a plan.

Timing Issues

There aren't a lot of extension questions for this problem. So, if your students finish early, make sure to really hammer the concepts and spend a lot of time going over the steps involved with Gauss's Law, creating Gaussian surfaces, etc.

  • Question: What do field lines look like from each plate? How do you know they point in that direction?
  • Expected Answer: See diagrams below. Should point from positive charge to negative charge.

Solution: One plate at a time

Start by looking only at a single plate of the capacitor:

1platefield.jpg

Reminder: $ \vec{E}$ points away from the positive charge - so points straight up/down from the plate as shown.

Assumptions with model
  • Plates are very large (compared to the separation of the plates).
  • Don't care about the field on edges of plate where they don't point straight up or straight down.
  • Uniform charge distribution: the same amount of charge is all over the plates.
  • Some assumption about V starting point. This will become important later when they start calculating the potential inside and outside of the plates.

Now we need to choose a Gaussian surface. Students can pick any prism they want as long as they understand how to do the calculation, however we will be using a cylinder because it will make the calculation easier. Our prism will only include the one plate and we will make the cylinder symmetric along plate (equal amount of cylinder above and below the plate).

1plategaussian.jpg

Here we have flux through both the top of the cylinder AND through the bottom of the cylinder. There is no flux through the side of the prism because $\vec{dA}$ is perpendicular to the $\vec{E}$, so no field enters or exits this side of the surface because $\vec{E} \cdot d\vec{A} = 0$ for any two perpendicular vectors.

Only flux is through top & bottom surfaces where $ \vec{dA}$ and $ \vec{E}$ are parallel or pointing anti-parallel to one another.

Because $\vec{E}$ and $d\vec{A}$ are parallel/anti-parallel($\theta = 0, 180$). $$\int \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}$$ $$\int \vec{E} \cdot \vec{dA} = \int |\vec{E}||\vec{dA}|cos(0)=\int E*dA$$ Because $E$ is constant (constant magnitude and direction) along the top/bottom faces of the Gaussian surface: $$\int E*dA = E*\int dA = E_{top}*A_{top}+E_{bot}*A_{bot}$$

Since the cylinder is symmetric above and below the plate, $E_{top}$ = $E_{bot}$, so this simplifies to $E(A_{top}+A_{bot})$. Then because the area of the plates are the same, we get: $E(2A)$.

$$E(2A) = \frac{Q_{enclosed}}{\epsilon_0}$$

In order to calculate the enclosed charge, we need to consider the charge distribution. Since we assumed that the amount of charge was constant across the entire plate, then we can define $\sigma$, the charge density, to be the total amount of charge on the entire plate divided by the area of the plate

$$\sigma = \frac{Q_{tot}}{A_{tot}}$$

The enclosed charge is thus the charge density times the amount of area of the plate that is enclosed by our Gaussian surface. Think about this like a cookie cutter. We draw our cylinder such that the two halves (top and bottom) meet together in the middle at the plate. The amount of area “enclosed” is the little circle of the plate that gets “cut out” by the two halves of the clinder coming together.

$$Q_{enc}=\sigma A_{enc}$$

$$E(2A) = \frac{\sigma A_{enc}}{\epsilon_0}$$

Since the area of the circle that we enclose on the plate is the same as the area of the top and bottom circles on the cylinder, our Gaussian surface, we know that

$$A = A_{enc}$$

This simplifies our expression for the electric field to be

$$E = \frac{\sigma}{2\epsilon_0}$$

This is the electric field above and below just this one plate that we are working with. Notice that the field is constant and does not depend on distance from the plate that we are. Or, another way of saying that is it doesn't matter how large we draw our Gaussian surface, the electric field will be the same. For the negative plate, we would expect the same answer only with a negative charge density:

$$E = \frac{-\sigma}{2\epsilon_0}$$

We can then use superposition to get the total electric field inside the plates. Here it helps to draw the picture: 5b_diagram_solution.jpg

From the diagram, the electric field inside is twice the magnitude of a single plate (the fields add together), whereas outside the electric field cancels. $$E_{outside}=0$$ $$E_{inside}=\frac{\sigma}{\epsilon_0}$$

The electric fields cancel outside of the plates because the expression that we got for the electric field of a single plate does not depend on distance from the charges. Thus, when we look at the region outside of the positive plate (top), there is a contribution from the positive plate, and even though we could be much further away from the negative plate comparatively, we get the exact same contribution to the electric field but in the opposite direction. Superposition tells us that these electric fields have to cancel outside of the plates, but is constant inside of the plates.

Solution: Two plates together

This solution hinges on students knowing already that the electric field is constant from a large parallel plate. The first step in solving any Gauss's law problem is an intuition about what the field looks like so that one can draw a proper Gaussian surface. Gauss's law then takes our intuition for the direction of the field and lets us solve for the magnitude. They may remember this from the capacitor notes or from the coding problem early in the semester. If so, they can follow this solution path - otherwise it is better to have them work with one plate first then use superposition to add the second plate.

The first step is to draw out the electric field from both plates - here we've drawn it as negative charge on the bottom and positive at the top.

Electron cloud around positive nucleus

Reminder: $ \vec{E}$ points away from the positive charge and towards the negative charge.

Above and below the plates, the electric field of the positive charge cancels out the field of the negative charge (arrows of equal magnitude point in opposite directions. So, $ \vec{E_{above}} = \vec{E_{below}} = 0$. Additionally, the $ \vec{E_{between}}$ is strong.

Now we need to choose a Gaussian surface. Students can pick any prism they want as long as they understand how to do the calculation, however we will be using a cylinder because it will make the calculation easier. Our prism will only include the one plate because it we include neither or both plates then the total charge contained will be 0.

5a_sol_4.jpg

No flux through top half of the prism because $ \vec{E_{above}} = 0$. No flux through the side of the prism because $\vec{dA}$ is perpendicular to the $\vec{E}$, so no field enters or exits the surface.

Only flux is through bottom surface where $ \vec{dA}$ and $ \vec{E}$ are parallel.

Because $\vec{E}$ and $d\vec{A}$ are parallel($\theta = 0$). $$\int \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}$$ $$\int \vec{E} \cdot \vec{dA} = \int |\vec{E}||\vec{dA}|cos(0)=\int E*dA$$ Because $\vec{E}$ is constant everywhere on our surface, we can pull it out of the integral $$\int E dA = E \int dA = E*A$$

Now, plugging this back into Gauss's Law

$${E}{A} = \frac{Q_{enclosed}}{\epsilon_0}$$

Using the same process outlined for just the single plate, we can represent the charge per unit area, $\sigma$, as the following

$$\sigma = \frac{Q_{tot}}{A_{tot}}$$

And use that to calculate the charge enclosed by the Gaussian surface

$$Q_{enc}=\sigma A_{enc}$$

Substituting back in to Gauss's Law

$$EA = \frac{\sigma A_{enc} }{\epsilon_0}$$

And using the argument from the other solution about the area enclosed as it relates to the face of the cylinder which is our Gaussian surface

$$A = A_{enc}$$

Solving for the electric field, we get the same result as our other solution

$$E = \frac{\sigma}{\epsilon_0}$$

E is constant between the plates, which matches the other solution, but required us to use more intuition to picture the electric field of the two plates first.

For both solutions:

  • Question: Draw graphs for what the magnitude of the electric field and potential look like in regions outside of the two plates, between the two plates, and then outside of the two plates again.
  • Expected Answer: See below.
  • Question: Why positive slope on V graph?
  • Expected Answer: Because potential is the NEGATIVE derivative of the field. We have to take the negative sign into account when we look at the area under the electric field graph.

Graphs: (note, these graphs are from original context where we were talking about clouds as the model instead of capacitor plates, but the graphs look exactly the same, just with different axes labels and annotations)

graphs for E and V

Derivation for graphs if required:

To find potential, students should remember that: $$\Delta{V} = -\int \vec{E} \cdot d\vec{h}$$

5a_sol_3.jpg

(You integrate over the distance between the two plates because that's where the E-field is between)

$$\Delta{V} = -\int_{h_{top}}^{h_{bottom}} -\frac{\sigma}{\epsilon_0} \hat{y} \cdot d\vec{h} \hat{y}$$

$\hat{y}$ would be positive if integrating from bottom to top.

$$\Delta{V} = \int_{h_{top}}^{h_{bottom}} \frac{\sigma}{\epsilon_0} d{h} $$

$\frac{\sigma}{\epsilon_0}$ is a constant, so we can pull it out of the integral

$$\Delta{V} = \frac{\sigma}{\epsilon_0} \int_{h_{top}}^{h_{bottom}} dh $$

Integrating over $dh$ we get

$$\Delta V = \frac{\sigma}{\epsilon_0} h$$ between the two plates. Potential has to be a continuous function (not important), so while we know what the potential between the plates is, we don't know what the potential outside of the plates is since this same method will return $\Delta V = 0$ because our electric field is zero outside of the plates. Thus, we say that the bottom plate is at some potential $V_{bottom}$, which we don't know, and the top plate is at some potential $V_{top}$, which we also don't know. The potential then increases linearly as we move from the bottom plate to the top plate.

Discussion Prompts

  • Question: How did you choose your Gaussian surface? (Why put it on 1 plate?)
  • Expected Answer: Want a surface with flat areas so that the area vector is parallel to the electric field.
  • Question: What would happen if you enclosed both plates in your Gaussian surface?
  • Answer: If you choose to include both plates then the total charge enclosed is zero and it won't help you find the field between the plates.
  • Question: Why is Q enclosed on 1 plate? What is the area of Qenc?
  • Expected Answer: Because of how the Gaussian surface is chosen. The area of Qenc should be the same as the shape of the prism.
  • Question: What direction is $\vec{E}$? How can you tell from your equation?
  • Expected Answer: Gauss's Law doesn't give the direction (it disappears in the dot product) - you have to add the direction (by adding a unit vector) back to the field to get the electric field vector.

Evaluation Questions

  • Question: What would be the difference if we want the electric field close to the plate versus in the middle of the two plates?
  • Expected Answer: The electric field between plates DOES NOT depend on distance, so they would be the same.

Extension Questions

  • Question: What would be the E-field for 1 plate? How do you know?
  • Expected Answer: $$E = \frac{\sigma}{2\epsilon_0}$$
  • Question: What would be the capacitance of this capacitor?
  • Expected Answer: $C = \frac{\epsilon_0 A}{h}$ This relates to the electric potential over the charge (which you can find from the electric field), or $Q = CV \rightarrow C = \frac{Q}{V} = \frac{\sigma A_{enc}}{\frac{\sigma h}{\epsilon_0}} = \frac{A \epsilon_0}{h}$ where $h$ is the separation between the plates, sometimes labeled in the notes as $d$.

Testing Observations:

  • Students struggled with figure out to model the cloud as two parallel plates. We added the line about area in the problem statement to hopefully help with this.

Your team has been hired by the Umbrella Corporation to evaluate the design of a “major experiment.” At present, it is planned that the experimental apparatus will be contained by a 6 meter wide cylindrical metal pipe that is 2 km long near Laredo, Texas. The company that has been chosen to manufacture the pipe is recommending a pipe with walls that are at least 25 cm thick to protect the apparatus inside from lightning strikes. Umbrella Corporation is concerned because the project is already over budget and a 10 cm thick pipe would cost 3 times less.

Your team has been hired to determine the electrical effects both inside and outside the pipe if it is struck by lightning and to make a recommendation to Umbrella executives. Your recommendation should include diagrams and graphs to communicate your point.

Learning Goals

  • Use Gauss's Law to calculate the electric field inside and outside a charged cylinder
  • Explain why you picked your Gaussian surface and how it helped you simplify your calculations.
  • Explain the general steps that you take when using Gauss's Law to solve a problem.
  • Explain what would change about your solution if the pipe were metal vs plastic.

Learning Issues

  • Watch for students plugging in numbers too early - especially here it is much better to keep the solution in terms of variables until the end. This helps the students to: check to see if their units are the end by examining the units of their variables, get some intuition about the general form of the solution, and then get some kind of numerical answer for their electric field.
  • Returning to the electric field - this is going to be kind of a harsh course correction from all of the magnetic field stuff the students have been working on fore the last while in class. We are coming back to electric fields to talk about a couple of mathematical tricks for working with electric fields and how they relate to magnetic fields. This is a good time to do it because now we have a little more intuition for what fields look like, which is a prerequisite for working with Gauss's Law.
  • Gaussian surfaces are imaginary, not real, purely mathematical constructs. They do NOT actually exist. Students often get confused thinking about imaginary surfaces that they build to do math with. We could pick any surface we want, but we look for two specific things to simplify the math:
    • We want a surface where $\vec{E}$ is parallel to $d\vec{A}$ to simplify the dot product.
    • We want a surface where the $\vec{E}$ is constant (magnitude and direction) everywhere on the surface.
    • Common choices are “pillboxes”, spheres, and cylinders because of symmetry.
  • $d\vec{A}$ is a unit vector (length 1) that points directly out from the surface. It points in a direction perpendicular to whatever Gaussian surface we draw.
  • The last page of notes for this section has an outline of the steps to follow if students are struggling to come up with a plan.
  • Focus on when it is good to use Gauss's Law and when it is not the best plan
    • It's good for symmetric situations where our charge is spread out evenly on lines, planes, cylinders, spheres, etc. The symmetric charge distribution with some nice geometry helps us to reason what the field looks like and also makes the math doable.
    • It is not good for fields that change magnitude or direction over one of our three standard surfaces (pillbox, sphere, and cylinder). This is basically almost all real-world problems, but we can make assumptions to build models where the math is doable.
    • Gauss's Law is ALWAYS true, but that doesn't mean we know how to do it or it's useful. But is ALWAYS holds.

Timing Issues

The lightning strike is a little confusing for the students. If they get hung up on this and 20-25 minutes has passed, push them to only think of the lightning strike as a source of charge. It's how the charge gets deposited onto our cylinder.

Alternate Solution

Students can solve this problem assuming that the cylinder housing the experiment is either a conductor or an insulator.

When describing the charge distribution, students can used $Q_{enc} = \lambda dl$ or $Q_{enc} = \sigma dA$. Either will work depending on if they are treating the cylinder as a “line” of charge or if they are working with 2D geometry. Both will result in the same answer.

To start, we need to figure out where the charge is on the pipe and what the electric field looks like from that charge. We know that the lightning is going to strike the pipe, which is going to make the pipe charged.

We will start by making the following assumptions:

  • The pipe is made of a metal/conductor
  • The lightning deposits a charge Q on the conductor, which doesn't discharge
  • We will ignore the lightning otherwise
  • The pipe is infinitely (or really really) long compared to its diameter, so we can ignore the ends and focus on the electric field somewhere in the middle of the pipe

Students should look up or assume a charge from the lightning bolt - somewhere in the 15-300 C range.

Learning issues
  • Students get hung up on the lightning, do not let them spend more than 20 minutes discussing the lightning as is it not super important. It's just how we deposit charge onto the cylinder

Because we say the pipe is metal, the excess charge will spread to the surface of the pipe - so there would be electric field pointing radially towards (if negative charge assumed) outside of the pipe. Students may remember that this means the electric field inside the pipe must be zero (because it is a conductor). If not, they can show this with Gauss's Law (below). Either way, they should use Gauss's Law to prove the electric field inside of the pipe is zero and show what the electric field outside of the pipe is.

Because the pipe is metal, we also assume that

  • The charge is uniformly distributed along the surface.

The electric field is constant/uniform direction near the center of the pipe, so we will pick our Gaussian surface to be near the middle of the pipe so we can ignore the ends/edge effects.

Electric field outside the pipe

We'll start by finding the electric field outside the pipe.

Step 1: Draw the electric field lines and determine a good Gaussian surface The electric field lines here point toward the cylinder radially at the center of the pipe, so we will a Gaussian surface that is a cylinder. This means that the electric field lines are parallel to the dA vectors along the curved part of the Gaussian surface and perpendicular to the dA's for the end caps of the cylinder.

Step 2: Find the electric flux through the Gaussian surface The flux through the Gaussian cylinder here is given by: $$\Phi_{tot}=\int \vec{E}_{top} \cdot \vec{dA}_{top}+ \int \vec{E}_{bottom} \cdot \vec{dA}_{bottom}+\int \vec{E}_{side} \cdot \vec{dA}_{side}$$

Because E and dA are perpendicular for both the top and bottom surfaces, the dot product for those terms will be zero, leaving: $$\Phi_{tot}=\int \vec{E}_{side} \cdot \vec{dA}_{side}$$

Since $E_{side}$ is parallel to $dA_{side}$, the dot product becomes multiplication. $$\Phi_{tot}=\int {E}_{side} {dA}_{side}$$

Because E is constant at every point on the side of the cylinder, the E can come out of the integral, which gives: $$\Phi_{tot}={E}_{side} \int {dA}_{side} = E_{side}*A_{side}=E*2\pi*s*l$$

where A is the area of the cylinder side $2\pi*s*l$.

Step 3: Find $Q_{enc}$ Here we can use a proportionality (since the charge is uniformly distributed) to solve for $Q_{enc}$: $$\frac{Q_{enc}}{L_{enc}}=\frac{Q_{tot}}{L_{tot}}$$ Since we called the height of the Gaussian cylinder l then $L_{enc}=l$, $Q_{tot}=Q$, and $L_{tot}=L$ (where L is the total length of the pipe).

So $Q_{enc}$ is then: $$Q_{enc}=\frac{Q}{L}l$$

Step 4: Piece together and solve for electric field Now we can plug these pieces into Gauss's Law: $$\Phi_{tot}= \frac{Q_{enclosed}}{\epsilon_0}$$ $$E 2\pi s l=\frac{Q*l}{L\epsilon_0}$$ $$E = \frac{Q}{2 \pi s L \epsilon_0}$$

So the electric field dies off as $1/s$, or like 1/distance we are from the outside of the cylinder.

Electric field inside pipe

Students may remember that this means the electric field inside the pipe must be zero (because it is a conductor). If not, they can show this with Gauss's Law. If all the charge is on the surface of the pipe, then when we draw the Gaussian surface on the inside of the pipe, there would be no $Q_{enc}$. Thus, the electric field would also be zero.

In any case, the thickness of the pipe doesn't matter and the company should go with the cheaper one if all they are worried about is shielding the field.

Discussion Prompts

  • Question: What steps did you take to solve this problem?
  • Expected Answer: Find the charge on the cylinder, draw a gaussian surface, find the flux through the surface, find the charge enclosed by the surface, solve for electric field.
  • Question: Where did you use symmetry in this problem?
  • Expected Answer: To simplify the flux (only need the flux through the side) and in the integral (E-field is constant along the side so it can come out of the integral).
  • Question: What is different about the field inside the pipe vs outside the pipe? What implications does this have for your pipe if lightning hits?
  • Expected Answer: Inside the field is zero, where as outside the field is not zero and stronger the closer you are to the pipe. This means no matter how thick your pipe is the inside won't have an electric field.
  • Question: What is similar/different about the problem today compared to the plates that you did on Tuesday?
  • Expected Answer: E-field is constant between the plates, the field here dies off like 1/r. The process is the same but the electric field looks different.

Evaluation Questions

  • Question: How did you simplify the problem? What assumptions did you make? How did those help you?
  • Expected Answer:
  • Question: When is Gauss's Law appropriate to use? When is better to not do Gauss's Law?
  • Expected Answer: Gauss's Law is only good for very symmetric cases (long lines of charge, long cylinders of charge, large planes, spheres). If you have literally anything else, it's better to do an integral or use a computer to compute the integral

Extension Questions

  • Question: What happens if the pipe develops a crack in it? Where does the charge go and is that a problem?
  • Expected Answer: The charge would still spread out along the surface because it is a conductor; however, with a crack the symmetry is ruined. There are no longer all of the charges needed to perfectly cancel the electric field in the middle of the pipe, so your equipment may now be in danger.
  • Question: What would the graphs of the electric potential and electric field look like around the pipe?
  • Expected Answer: E-field should go like 1/r outside of the pipe, then drop to zero inside the pipe. Potential should go like $-1/r^2$ outside the pipe and BE CONSTANT inside the pipe. This can be interesting talking point - just because the field is zero does NOT mean that the potential is zero.]
  • Question: What would change about your solution if the pipe were an insulator instead of a conductor (or vice-versa if students solved the insulator first)?

Electric Field for an Insulator

The steps and assumptions for this solution are nearly identical to the solution for a conductor, but the charge distribution is a little bit different. We will still assume that the pipe is really long compared to its diameter, so we will ignore the edge effects and look at a region towards the middle of the pipe. We also assume that the lightning strike deposits some amount of negative charge onto the pipe.

The ambiguity here is that depending on the kind of insulator and how good of an insulator the material is, the charge distribution could look very different. If the material were highly insulated, then the charge would be confined (more or less) to the area where it was deposited because it would not be able to dissipate through the material well. For the sake of being able to solve this problem, however, let's assume that the charge gets distributed through the material evenly. Unlike the conductor, however, the charge does not only live on the surface of the material, but it gets spread out throughout the volume of the material. So, the excess charge deposited by the lightning bolt is spread evenly throughout the thickness of the pipe walls. We will follow the same steps to solve for the electric field as we did for the conductor.

Step 1: Draw the electric field lines and determine a good Gaussian surface The electric field lines here point toward the cylinder radially at the center of the pipe, so we will a Gaussian surface that is a cylinder. This means that the electric field lines are parallel to the dA vectors along the curved part of the Gaussian surface and perpendicular to the dA's for the end caps of the cylinder. We wanted to choose a Gaussian surface where there are apparent symmetries and $\vec{E}$ is either parallel or perpendicular to $d\vec{A}$ and $\vec{E}$ has a constant magnitude on the entire Gaussian surface. We will define our Gaussian cylinder to be slightly smaller than the pipe so that some of the thickness of the pipe in enclosed. That way, we can get a picture for what the electric field looks like inside of the walls of the pipe. If our Gaussian cylinder has a radius larger than the radius of the pipe, than the amount of charge that we enclose is exactly the same as it was in the conductor scenario and the result is the same. The more interesting case is what happens in the thickness.

Step 2: Find the electric flux through the Gaussian surface The flux through the Gaussian cylinder here is given by: $$\Phi_{tot}=\int \vec{E}_{top} \cdot \vec{dA}_{top}+ \int \vec{E}_{bottom} \cdot \vec{dA}_{bottom}+\int \vec{E}_{side} \cdot \vec{dA}_{side}$$

Because E and dA are perpendicular for both the top and bottom surfaces, the dot product for those terms will be zero, leaving: $$\Phi_{tot}=\int \vec{E}_{side} \cdot \vec{dA}_{side}$$

Since $E_{side}$ is parallel to $dA_{side}$, the dot product becomes multiplication. $$\Phi_{tot}=\int {E}_{side} {dA}_{side}$$

Because E is constant at every point on the side of the cylinder, the E can come out of the integral, which gives: $$\Phi_{tot}={E}_{side} \int {dA}_{side} = E_{side}*A_{side}=E*2\pi*s*l$$

where A is the area of the cylinder side $2\pi*s*l$.

Step 3: Find $Q_{enc}$ Here we can use a proportionality (since the charge is uniformly distributed) to solve for $Q_{enc}$. Now, unlike the conductor, we can't just consider the charge distribution on the surface of the pipe. Here, since the charge is distributed throughout the bulk of the material (which is an assumption that we made about the kind of insulator we are working with), we have to consider how much of the volume of the pipe we are enclosing. So, we look at volume ratios: $$\frac{Q_{enc}}{V_{enc}}=\frac{Q_{tot}}{V_{tot}}$$

Let's say that the outer wall of the pipe has a radius $b$ and an inner radius $a$. The thickness of the pipe is given as $b-a$. If we draw our Gaussian surface so that it encloses some amount of the bulk of the pipe, then the fraction of the total volume enclosed is

$$\frac{Q_{enc}}{\pi s^2 l - \pi a^2 l}=\frac{Q_{tot}}{\pi b^2 L - \pi a^2 L}$$

$\pi b^2 L - \pi a^2 L$ represents the total volume of the thickness of the pipe (difference between the volume of a cylinder with a radius equal to that of the outer wall minus the volume from a cylinder with a radius equal to that of the inner wall). $\pi s^2 l - \pi a^2 l$ is the amount of the thickness of the pipe our Gaussian cylinder encloses and can be derived using the same concept. Here, $L$ is the height of the pipe itself (2 km) and $l$ is the height of our Gaussian cylinder.

$$\frac{Q_{enc}}{\pi l (s^2 - a^2)}=\frac{Q_{tot}}{\pi L (b^2 - a^2)}$$

Now, we solve for our enclosed charge

$$Q_{enc} = Q_{tot} \frac{s^2 - a^2}{b^2 - a^2} \frac{l}{L}$$

Step 4: Piece together and solve for electric field Now we can plug these pieces into Gauss's Law: $$\Phi_{tot}= \frac{Q_{enclosed}}{\epsilon_0}$$ $$E 2\pi s l= Q_{tot} \frac{s^2 - a^2}{b^2 - a^2} \frac{l}{L} \frac{1}{\epsilon_0}$$ Simplifying this, we get $$E = Q_{tot} \frac{s^2 - a^2}{b^2 - a^2} \frac{1}{2 \pi s L \epsilon_0}$$

Inside of the bulk of the material, the electric field depends on how much of the thickness of the pipe we enclose. As $s \rightarrow b$, the expression for the electric field outside of the pipe becomes the same as it was for the conductor. If we are still concerned about the contents of the pipe, then we could draw the Gaussian cylinder to have a radius smaller than the inside wall, $r < a$, but here we see we still enclose no charge. So, the electric field inside of the inner wall of the pipe is still zero and thickness does not matter.

Testing Observations:

  • Students struggled with the lightning piece and how the lightning charged the pipe. Don't let them focus on this for too long.
  • We can ask students to graph the potential and electric fields as functions of distance. Ask about the implications inside and outside due to a lightning strike. We can ask what happens if the pipe develops a crack in it (where does the charge go and is that a problem?)
  • 184_projects/f18_project_11_sol.txt
  • Last modified: 2018/11/15 17:56
  • by dmcpadden