Gauss's Law 02

OPENING QUESTION: Write down the various permutations of Gauss's Law on one sheet of paper on a single sheet of paper (we'll use this to write, rewrite and otherwise edit our own personal working model for all things GAUSS)

On another blank sheet a paper write down a list of questions that come to mind about when and how to apply Gauss's Law (write them in full sentences please, not in shorthand)

As we go through various permutations of Gaussian Surfaces and such, you'll need to edit/rewrite your model (page 1) answer questions you've posed (page 2) and add new questions that arise (page 2 again)

Remember, this is brand new stuff... being baffled is ok, we'll make good progress today (promise!)

OBJECTIVE:  I will continue working on formulating a model for using Gauss's Law during today's class.

WORDS/FORMULAE FOR TODAY

Terms:

  • Gauss's Law: Electric flux through an enclosed area is equal to the total charge enclosed divided by the permeability of free space:

ΦE = q/εo

  • Electric Flux (efield magnitude) x (area) but there's a lot more to it than that!
  • Test Charge: A mathematical construct-- a charge that does not exert any influence on surrounding particles but IS influenced by other electric fields.
  • Conductors - materials where electrons can roam
  • Insulators - materials that keep their electrons close to home
  • coulomb - a unit of electrical charge (see below)

Constants:

  • e-mass = 9.1 x 10-31 kg
  • charge of an electron = 1.60 x 10-19 coulombs (C)
  • ke = 1/4πεo = <huh?> = 8.987 x 109 Nm2/C2
  • εo = perceptivity of free space = 8.854 x 10-12 C2/Nm2

Formulae:

      • Electrical Force: Fe = (keq1q2)/r2
      • Electrical Field: E = (keq)/r2 = Fe / qo
      • Electric Flux: ΦE = EA = EAcos(θ) =Ecos(θ)dA =E ∙ dA = q/εo

WORK O' THE DAY

Now let's add yet ONE MORE item to your overflowing plate of Gauss!

There is an important geometric difference between these two images... can you find it?

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Notice that this image has Efield lines passing through it...the charge itself is OUTSIDE the Gaussian Surface. That means the electric flux (ΦE) through the surface is zero (we'll talk about that in more detail very shortly)!

This image shows a Gaussian Surface completely ENCLOSING the existing charge.

Such an ENCLOSED charge means we have to modify our integral notation to show we are integrating over a surface that itself is ENCLOSING the charge. That means we replace our comfy ol' integral signwith a sign that shows we are integrating over a closed surface:.

The GOOD news is that we do the integration exactly as before.

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So... let's recap our permutations of Gauss's Law using our more accurate integral notation:

ALWAYS start with a basic recollection of Gauss's Law:

The electric flux through an enclosed space is equal to the sum of the enclosed charge divided by permativity of free space

Which we express mathematically as:

ΦE =q/εo

By itself, that is pretty boring stuff. However, recollect that we have another more power equation:

ΦE =EcosθdA

Setting those two formulae equal gives us:

q/εo =EcosθdA

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Let' start with that equation as our *Standard Model*.

You'll need to and apt/develop your own model further from now on.

For now, let's take a fairly easy example and and apply our model to that situation.

Consider a cube with electric field lines moving parallel to the x-axis in the positive x direction as shown below:

How do we calculate the electric flux through that cube?

1) Please start by adding rules to your page 1 that you think you should apply. Any questions that arise should go on page 2.

2) Without consulting your group (yet), suggest an answer to the problem and provide backup to your response.

3) Compare your work with your group and make additions/deletions to your model. Cross out (but do not erase) questions that you answered and add new questions that arise.

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We can answer this qualitatively as follows:

Gauss's Law applies to charges enclosed by a Gaussian Surface. The cube does not enclose any charges therefore it has zero electric flux.

We can answer the question with a bit of math as follows:

The efield lines enter the Gaussian Surface (the cube) and an angle of 180o to the area vector of the left side of the cube and exit 0o to the area vector on the right side of the cube. cos0o and cos180o will cancel giving us zero electric flux.

There are no electric field lines moving through the top/bottom (yz plane) or the other two sides (xz plane) so there is zero electric flux there.

Therefore there is zero electric flux.

Please adjust your model and questions once again. Add new questions, cross out old ones and make changes to your model as appropriate.

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Take a look at the following images and note any specific observations that you think will be helpful in evaluating the electric FIELD on the surfaces S1, S2 and S3 of image #1, and the surface of image #2

Image #1
Image #2

Gaussian Surface S1 is VASTLY preferable because the area vector AND the efield vectors are constant making use of our basic formula tres easius:

q/εo =∮EcosθdA

E is constant so:

q/εo =E∮cosθdA

Because the Gaussian Surface is a sphere, all Efield lines will make a nice, pleasing 0 degree exit through the sphere so:

q/εo =E∮dA

So now let's integrate dA (which is kinda nice too!)

q/εo =EA

We also realize that the surface area of a sphere is pretty basic:

A = 4πr2

Which gives us:

q/εo =E4πr2

Isolating for E:

E = q/(4πεor2)

Recollecting that ks = 1/4πεo:

E = ksq/(r2)

Which looks MIGHTY familiar, at least it should....

 

Notice that this surface does NOT enclose any charges! Therefore it violates the basic rule of Gauss's Law that charges MUST be enclosed. So the surface shown in image #2 has ZERO electric flux.

Mathematically, the angle that each efield line makes coming into the surface is the exact opposite of the angle coming out of the surface. That means we have a cos180 going in and cos0 sorta thing (or similar) coming out so once again, ZERO electric flux.

Ok, so I know what yer thinkin':

"We just derived the equation for finding an E field... how is that helpful?"

Indeed...!

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Let's revisit our favorite "Long Thin Wire" scenario. Let's say we have such a beastie with uniform charge density *yay*.

  • What is so important about it being a "Thin" wire?
  • What is so important about it being a "Long" wire (this is particularly important in EM scenarios)?
  • Why is "uniform density" so important?
  • What sort of Gaussian Surface might we choose to analyze such a beastie?

Something like this, mayhap?

  • Why would this scenario FAIL if the Gaussian Surface was conducting?
  • Why would a Gaussian Surface FAIL if this were a thin wire with uniform charge density that was NOT "long"?

Please make any appropriate adjustments to your page 1 and page 2 as appropriate.

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Now let's try some reverse logic (ie, eg, for exampe, whatever) a scenario that looks promising but turns out to be inaccurate and misleading... work with your group to see if you can determine the bogartry of the situation:

Let's say we have a +5.0 μC charge and a -13.25 μC charge and a +7.2 μC charge in close proximity and we want to find the value of the Efield at 1.0 m distant.

Sketch the charges thusly:

Now choose a Gaussian Surface surrounding those charges.

Now we add up all the charges and solve the problem yay? Nay!

Hint: Use our animation tool to show the efield lines for such a situation.

Does that help?

Let's try some fill-in-the-blank:

We can use Gauss's Law to find the ________________ in the above situation but we CANNOT use it to find the ______________. That's because Gauss's Law allows us to sum up the charges when we find the _____________ but it also requires a high degree of ________________ when we use that law to find _____________; and in this case that high degree of symetry is clearly lacking which means the Efield is NOT _______________ which means we've got a *nasty* integral to deal with and we have to punt!

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Please work on the following problems with your crew:

Objective problems 8, 9 & 11. Conceptual problem # 7. Problems 7 & 8