OBJECTIVE:  I will be able calculate a NET flux through a surface after today's classs

WORDS/FORMULAE FOR TODAY

Terms:

• 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)
• k_e = 8.987 x 10⁹ Nm²/C²

Formulae:

• Electrical Force: F_e = (k_eq₁q₂)/r²
• Electrical Field: E = F_e / q_o where q_o  = a positive test charge
• Electrical Flux (generally) : Φ = ∮E ∙ dA or ∮Ecosθ dA where θ is the angle between the electrical field line vector and the area vector
• Gauss's Law (for an object enclosing 1 or more charged particles)

  Φ = q/ε_o

WORK O' THE DAY: 

Keep in mind that the Flux is a measure of the number of field lines present passsing through the surface of some sort of material.

Flux is NOT the number of field lines passing PER unit area (that's why we integrate over the WHOLE surface of the object in question)

HERE is a very reasonable explanation of flux (courtesy of the University of Tennessee at Knoxville)

Here's a bit more courtesy of lecture notes from the University of Boulder (http://www.colorado.edu/physics/phys1120/phys1120_fa09/LectureNotes/Gauss.doc)

The flux F has a the following geometrical interpretation:
flux  µ  the number of electric field lines crossing the surface. 

Think of the E-field lines as rain flowing threw an open window of area A.  The flux is a measure of the amount of rain flowing through the window.  To get a big flux, you need a large E, a large A, and you need the area perpendicular to the E-field vector, which means the area vector A is parallel to E.  (In the rain analogy, you need the window to be facing the rain direction.)

 

 

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GAUSS'S LAW has two basic componants:

1) The total electrical field flux through a (non charged) enclosed surface is always zero (what goes in, MUST come out)

2) The total electrical flux trhough a closed surface containing a charged particle (and therefore a charged surface) is q/ε_o

Gauss's Law works best with objects of high symetry (cubes, spheres, rods etc...), we won't be dealing with objects who's symetry is such that Gauss's Law does not apply, however.

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I RARELY have you derive one formula from another, but in this case I'm going to make an exception:

Starting with the formula for determining electric flux and from (ahem) what you know about the geometry of a sphere.... derive a formula for the electric flux of a uniform electric field passing through a spherical region surrounding a charged particle.

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Answer:

Φ = ∮E ∙A dA

HOWEVER, a sphere is a highly semetric so A is constant.

We were told the electrical field is uniform, so E is constant:

Because the field lines entering or leaving a charged surface will be perpindicular to the surface of the sphere our 'dot' product goes away leaving us with:

Φ = EA

E = k_eq/r²

A for the surface of a sphere = 4πr²

sooooo running that together we get:

Φ = q_in/ε_o

Which is Gauss's Law

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From what you know about Gauss's Law, discuss the following with your groupies and be preparedto DEFEND your answer to the class:

Consider the case of a point charge q surrounding by a spherical gaussian surface:

Describe what happens to the TOTAL FLUX through that surface if:

A) The charge is trippled

B) The radius of the sphere is doubled

C) The surface is changed from a sphere to a cube

D) The charge is moved to another location inside the sphere

answers are also on page 730

Takeaways for today:

1. Net flux through a Gaussian surface is zero when that surface DOES NOT enclose any charges or the algebraic sum of those enclosed charges is zero.
2. The net flux through a Gaussian surface enclosing charges who's algabraic sum is NOT zero is => Φ = q_in/ε_o
3. Keep in mind WHEN to use the various flavors of Gauss's Law

HOMEWORK: 24.7, 8, 9 on page 740