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W02D2Gauss’s Law

Announcements

Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus

PS 2 due Week Three Tuesday Tues at 9 pm in boxes outside 32-082 or 26-152

W02D3 Reading Assignment Course Notes: Chapter Course Notes: Sections 3.6, 3.7, 3.10

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Outline

Electric Flux

Gauss’s Law

Calculating Electric Fields using Gauss’s Law

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Gauss’s Law

The first Maxwell Equation!

A very useful computational technique to find the electric field when the source has ‘enough symmetry’.

E

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Gauss’s Law – The Idea

The total “flux” of field lines penetrating any of these closed surfaces is the same and depends only on the amount of charge inside

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Gauss’s Law – The Equation

Electric flux (the surface integral of E over

closed surface S) is proportional to charge

enclosed by the volume enclosed by S

ΦE =rE⋅d

rA

closedsurface S

“∫∫ =qenclosed

ε0

ΦE

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Electric Flux Case I: E is a uniform vector field perpendicular to planar surface S of area A

ΦE =+EA

∫∫Φ AE

dE

Our Goal: Always reduce problem to finding a surface where we can take E out of integral and get simply E*Area

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Case II: E is uniform vector field directed at angle to planar surface S of area A

ΦE =EAcosθ

Electric Flux

ΦE =

rE ⋅d

rA∫∫

drA =dAn̂

n̂

θ

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Concept Question: FluxThe electric flux through the planar surface below (positive unit normal to left) is:

+q -qn̂

1. positive.

2. negative.

3. zero.

4. Not well defined.

Concept Question Answer: Flux

The field lines go from left to right, opposite the assigned normal direction. Hence the flux is negative.

Answer: 2. The flux is negative.

+q -qn̂

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Open and Closed Surfaces

A rectangle is an open surface — it does NOT contain a volume

A sphere is a closed surface — it DOES contain a volume

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Area Element: Closed Surface

Case III: not uniform, surface curved

For closed surface, is normal to surface and points outward ( from inside to outside)

if points out

if points in

dΦE > 0

dΦE < 0

drA

dΦE ≡rE⋅d

rA

ΦE = dΦE“∫∫ =

rE⋅d

rA“∫∫

rE

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Group Problem Electric Flux: Sphere

Consider a point-like charged object with charge Q located at the origin. What is the electric flux on a spherical surface (Gaussian surface) of radius r ?

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Arbitrary Gaussian Surfaces

True for all surfaces such as S1, S2 or S3

Why? As area gets bigger E gets smaller

ΦE =rE⋅d

rA

closedsurfaceS

“∫∫ =Qε0

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Gauss’s Law

Note: Integral must be over closed surface

ΦE =rE⋅d

rA

closedsurface S

“∫∫ =qenclosed

ε0

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Concept Question: Flux thru Sphere

The total flux through the below spherical surface is

+q+q

1. positive (net outward flux).

2. negative (net inward flux).

3. zero.

4. Not well defined.

Concept Question Answer: Flux thru Sphere

We know this from Gauss’s Law:

No enclosed charge no net flux. Flux in on left cancelled by flux out on right

Answer: 3. The total flux is zero

+q+q

ΦE =rE⋅d

rA

closedsurface S

“∫∫ =qenclosed

ε0

Concept Question: Gauss’s Law

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The grass seeds figure shows the electric field of three charges with charges +1, +1, and -1, The Gaussian surface in the figure is a sphere containing two of the charges. The electric flux through the spherical Gaussian surface is

1. Positive

2. Negative

3. Zero

4. Impossible to determine without more information.

Concept Question Answer: Gauss’s Law

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Answer 3: Zero. The field lines around the two charged objects inside the Gaussian surface are the field lines associated with a dipole, so the charge enclosed in the Gaussian surface is zero. Therefore the electric flux on the surface is zero.

Note that the electric field E is clearly NOT zero on the surface of the sphere. It is only the INTEGRAL over the spherical surface of E dotted into dA that is zero.

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Choosing Gaussian Surface

Desired E: Perpendicular to surface and uniform on surface. Flux is EA or -EA.

Other E: Parallel to surface. Flux is zero

ΦE =rE⋅d

rA

closedsurfaceS

“∫∫ =qenclosed

ε0

True for all closed surfaces

Useful (to calculate electric field ) for some closed surfaces for some problems with lots of symmetry.

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Symmetry & Gaussian Surfaces

Source Symmetry Gaussian Surface

Spherical Concentric Sphere

Cylindrical Coaxial Cylinder

Planar Gaussian “Pillbox”

Desired E: perpendicular to surface and constant on surface. So Gauss’s Law useful to calculate electric field from highly symmetric sources

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Virtual ExperimentGauss’s Law Applet

Bring up the Gauss’s Law Applet and answer the experiment survey questions

http://web.mit.edu/viz/EM/visualizations/electrostatics/flux/closedSurfaces/closed.htm

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Applying Gauss’s Law1. Based on the source, identify regions in which to calculate electric field.

2. Choose Gaussian surface S: Symmetry

3. Calculate

4. Calculate qenc, charge enclosed by surface S

5. Apply Gauss’s Law to calculate electric field:

rE⋅d

rA

closedsurface S

“∫∫ =qenclosed

ε0

ΦE =

rE ⋅d

rA

S“∫∫

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Examples:Spherical Symmetry

Cylindrical SymmetryPlanar Symmetry

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Group Problem Gauss: Spherical Symmetry

+Q uniformly distributed throughout non-conducting solid sphere of radius a. Find everywhere.

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Concept Question: Spherical Shell

We just saw that in a solid sphere of charge the electric field grows linearly with distance. Inside the charged spherical shell at right (r<a) what does the electric field do?

a

Q

1. Zero

2. Uniform but Non-Zero

3. Still grows linearly

4. Some other functional form (use Gauss’ Law)

5. Can’t determine with Gauss Law

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Concept Question Answer: Flux thru Sphere

Spherical symmetry Use Gauss’ Law with spherical surface.

Any surface inside shell contains no charge No flux

E = 0!

Answer: 1. Zero

a

Q

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Demonstration Field Inside Spherical Shell

(Grass Seeds):

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Worked Example: Planar Symmetry

Consider an infinite thin slab with uniform positive charge density . Find a vector expression for the direction and magnitude of the electric field outside the slab. Make sure you show your Gaussian closed surface.

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Gauss: Planar Symmetry

Symmetry is Planar

Use Gaussian Pillbox

rE =±Ex̂

x̂Note: A is arbitrary (its size and shape) and should divide out Gaussian

Pillbox

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Gauss: Planar Symmetry

NOTE: No flux through side of cylinder, only endcaps

Total charge enclosed: qenclosed= A

=E 2A( ) =

qenclosed

ε0

= Aε0

ΦE =

rE⋅d

rA

S“∫∫ =E dA

S“∫∫ =EAEndcaps

E =

2ε0

++++++++++++

E

E

x

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Concept Question: Superposition

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Three infinite sheets of charge are shown above. The sheet in the middle is negatively charged with charge per unit area , and the other two sheets are positively charged with charge per unit area . Which set of arrows (and zeros) best describes the electric field?

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Concept Question Answer: Superposition

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Answer 2 . The fields of each of the plates are shown in the different regions along with their sum.

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Group Problem: Cylindrical Symmetry

An infinitely long rod has a uniform positive linear charge density .Find the direction and magnitude of the electric field outside the rod. Clearly show your choice of Gaussian closed surface.

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Electric Field for Charged Infinite Plane

1) Dipole: E falls off like 1/r3

1) Spherical charge: E falls off like 1/r2

1) Line of charge: E falls off like 1/r

(infinite)

4) Plane of charge: E uniform on

(infinite) either side of plane