Gauss's Law 01

OPENING QUESTION: Draw a sketch to show why the electric field inside a conductor = 0?

OBJECTIVE:  I will begin working on formulating a model to apply Gauss's Law during today's class.



  • 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)


  • 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 = permitivity of free space = 8.854 x 10-12 C2/Nm2


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


Recall that:

Electric fields lines are a visual representation of the strength of the electrical force a positively charged *TEST PARTICLE* experiences at some distance away from that particle.

Today we'll learn about electric FLUX.

Electric flux is a measure of the total number of electric field lines present in the TOTAL area surrounding a charge or charges (*whew*). YAY! That means more Greek Letters!!!


Yup.... Φ is phi (upper case) not phi as in phi (φ) lower case

In words: electric flux is equal to the electrical field strength times the area in which that field is passing.

But wait, there's more!

We need to consider the ANGLE at which the efield lines pass through some defined area. To do that, we include an AREA VECTOR. The area vector is a tad tricky since it is perpendicular to the bit of area under discussion:

Soooooooooo... when calculating electric flux, we must include not just the efield lines, but also consider the angle those lines make to the area vector as shown below:


The units for electrical flux are, not surprisingly:


So... I know what yer thinking: "Hey Mr W, that smells an awful lot like a dot product".....

and yes, you'd be exactly right so let's include THAT:

ΦE = EAcos(θ)

But now (no doubt!) you're thinking.... "wait a sec, what if there be many elines that we'd need to evaluate all at once per some unit area, wouldn't that be a sum-of-all-the-efields kinda thing... and ain't that gonna lead to... YES.... integration?"

But of course...

ΦE =Ecos(θ) dA

Simplist case where elines are in the same direction as the area vector so θ = 0 and:

ΦE =E ∙ dA

But wait.... what if the area is part of a structure that actually encloses some volume of space, can it still work????

Indeed! But we'll talk about the nifty surface integral (ΦE =E dA) tomorrow...


Let your brain get kinda queasy at this point, recall our various definitions of work:

Work = fd = fdcos(θ) =fcos(θ)dr

etc ad nauseum.... so we DO have precedence!


Gauss's Law can get confusing in a very big fat hurry... so, we'll apply our favorite engineering concept of K.I.S.S. (to Keep It Simple, Stupid!)

Let's consider a charge, sitting around, minding it's own business with it's electric field lines nice and symetric as shown below:

Now lelt's say we want to investigate the electric field at some distance away from the charge. We KNOW that symetry makes life easier (we're not quite sure why... yet) so we choose a sphere for our Gaussian Surface.

WHOA... I know what yer thinkin'...

"Hey Mr W, a sphere ain't no surface!"

Quite so! However a sphere does have a surface, so BOO back atcha.

Anywho....Take a gander at the image below. In terms of efields and especially electric flux. What about this spherical shape comes in particularly handy? (please discuss)

ANSWER: If we choose a sphere as our Gaussian Surface, we can be assured that the angle that each electric field line makes with the surface of our Gaussian Surface is 180o, which makes our math easier but getting rid of that pesky cosine:

ΦE =ECosθ ∙ dA

ΦE =E ∙ dA

Although I'm generally not a huge fan of derivations, this one is pretty dark cool so let's take a gander:

Initial Equation:

ΦE =E ∙ dA

We KNOW that the electric field is constant so let's back that out of our equation thusly:

ΦE = E∫dA

The integral lof dA is pretty much ducksoup, so let's take that integral now:


We know the basic equation for an E field so let's substitute for that:

ΦE =[(keq)/r2]A

The area of a sphere is easy enough (4πr2) so let's substitute that in next:

ΦE =[(keq)/r2][4πr2]

Any thoughts on how to proceed? Talk with your group please!



Yes indeed! We know that:

ke = 1/4πεo

So let's substitute that puppy in next:

ΦE =[(1/4πεo)q)/r2][4πr2]

MUCH carnage ensues:

ΦE =[(1/εo)q)/r2][r2]

Giving us this MOST excellent result (Gauss's Law!):

ΦE =q/εo

The total electric flux out of ANY closed surface is equal to the TOTAL CHARGE enclosed divided by the permittivity of free space.)

Take a look at THIS graphic:

What is that image trying to teach us?

Why would we NOT choose the irregular surfaces for our calculations?


That might not seem so incredibly helpful, however if we realize:

E ∙ dA= q/εo

We have a powerful tool for evaluating Efields... provided that we do what??? (please discuss!)




YES... symetry is massively important here.

Let's try some basic problems. Use Gauss's Law to find the electric field at some position 'r' away from a 1.0 C charge.


Now let's say we have a +1.0 C charge and a -1.0 C charge seperated by a distance of 5.0 x 10-6 meters. Use Gauss's Law to show that the E field 1.0 meter away is zero.


Work through objective problems 8 and 10 with your group.... (now if you would please)

Closing Question: Why would the following shape be a POOR choice for a Gaussian Surface?

Why would an object with a conducting surface be a poor choice for a Gaussian Surface?

Please work on the following problems with your crew:

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