184_projects:e_field_in_cap_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.
  • 184_projects/e_field_in_cap_sol.txt
  • Last modified: 2021/08/19 19:21
  • by dmcpadden