1P05 -

Class 05: Outline

Hour 1:Gauss’ Law

Hour 2:Gauss’ Law

2P05 -

Six PRS Questions On Pace and Preparation

3P05 -

Last Time:Potential and E Field

4P05 -

E Field and Potential: Creating

A point charge q creates a field and potential around it:

2ˆ;e e

q qk V kr r

= =E r Use superposition for systems of charges

;B

B A AV V V V d= −∇ ∆ ≡ − = − ⋅∫E E s

They are related:

5P05 -

E Field and Potential: Effects

q=F EIf you put a charged particle, q, in a field:

To move a charged particle, q, in a field:

W U q V=∆ = ∆

6P05 -

Two PRS Questions:Potential & E Field

7P05 -

Gauss’s Law

The first Maxwell Equation A very useful computational technique

This is important!

8P05 -

Gauss’s Law – The Idea

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

9P05 -

Gauss’s Law – The Equation

0Ssurface

closed εin

Eqd =⋅=Φ ∫∫ AE

Electric flux ΦE (the surface integral of E over closed surface S) is proportional to charge inside the volume enclosed by S

10P05 -

Now the Details

11P05 -

Electric Flux ΦE

Case I: E is constant vector field perpendicular to planar surface S of area A

∫∫ ⋅=Φ AE dE

E EAΦ = +

Our Goal: Always reduce problem to this

12P05 -

Electric Flux ΦE

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

cosE EA θΦ =

∫∫ ⋅=Φ AE dE

13P05 -

PRS Question:Flux Thru Sheet

14P05 -

Gauss’s Law

0Ssurface

closed εin

Eqd =⋅=Φ ∫∫ AE

Note: Integral must be over closed surface

15P05 -

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

16P05 -

Area Element dA: Closed Surface

For closed surface, dA is normal to surface and points outward

( from inside to outside)

ΦE > 0 if E points out

ΦE < 0 if E points in

17P05 -

Electric Flux ΦE

Case III: E not constant, surface curved

Ed dΦ = ⋅E A

S

Ad E

∫∫ Φ=Φ EE d

18P05 -

Example: Point ChargeOpen Surface

19P05 -

Example: Point ChargeClosed Surface

20P05 -

PRS Question:Flux Thru Sphere

21P05 -

Electric Flux: Sphere Point charge Q at center of sphere, radius r

E field at surface:

20

ˆ4

Qrπε

=E r

Electric flux through sphere:

20S

ˆ ˆ4

Q dArπε

= ⋅∫∫ r rS

E dΦ = ⋅∫∫ E A

rA ˆdAd =0

Qε

=22

0

44

Q rr

ππε

=20 S4

Q dArπε

= ∫∫

22P05 -

Arbitrary Gaussian Surfaces

0Ssurface

closed εQdE =⋅=Φ ∫∫ AE

For all surfaces such as S1, S2 or S3

23P05 -

Applying Gauss’s Law1. Identify regions in which to calculate E field.2. Choose Gaussian surfaces S: Symmetry3. Calculate

4. Calculate qin, charge enclosed by surface S5. Apply Gauss’s Law to calculate E:

0Ssurface

closed εin

Eqd =⋅=Φ ∫∫ AE

∫∫ ⋅=ΦS

AE dE

24P05 -

Choosing Gaussian SurfaceChoose surfaces where E is perpendicular & constant.

Then flux is EA or -EA.

Choose surfaces where E is parallel.Then flux is zero

OR

EA

EA−

EExample: Uniform Field

Flux is EA on topFlux is –EA on bottomFlux is zero on sides

25P05 -

Symmetry & Gaussian Surfaces

Symmetry Gaussian Surface

Spherical Concentric Sphere

Cylindrical Coaxial Cylinder

Planar Gaussian “Pillbox”

Use Gauss’s Law to calculate E field from highly symmetric sources

26P05 -

PRS Question:Should we use Gauss’ Law?

27P05 -

Gauss: Spherical Symmetry+Q uniformly distributed throughout non-conducting solid sphere of radius a. Find E everywhere

28P05 -

Gauss: Spherical Symmetry

Symmetry is Spherical

Use Gaussian Spheres

rE ˆE=

29P05 -

Gauss: Spherical SymmetryRegion 1: r > aDraw Gaussian Sphere in Region 1 (r > a)

Note: r is arbitrary but is the radius for which you will calculate the E field!

30P05 -

Gauss: Spherical SymmetryRegion 1: r > aTotal charge enclosed qin = +Q

S

E dA EA= =∫∫S

E dΦ = ⋅∫∫ E A

( )24E rπ=

2

0 0

4 inE

q Qr Eπε ε

Φ = = =

204

QErπε

= 20

ˆ4

Qrπε

⇒ =E r

31P05 -

Gauss: Spherical Symmetry

Gauss’s law:

QarQ

a

rqin ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

= 3

3

3

3

3434

π

π

Region 2: r < aTotal charge enclosed:

( )24E E rπΦ =

304

Q rEaπε

= 30

ˆ4Q r

aπε⇒ =E r

OR inq Vρ=

3

30 0

inq r Qaε ε

⎛ ⎞= = ⎜ ⎟

⎝ ⎠

32P05 -

PRS Question:Field Inside Spherical Shell

33P05 -

Gauss: Cylindrical SymmetryInfinitely long rod with uniform charge density λFind E outside the rod.

34P05 -

Gauss: Cylindrical SymmetrySymmetry is Cylindrical

Use Gaussian Cylinder

rE ˆE=

Note: r is arbitrary but is the radius for which you will calculate the E field!

is arbitrary and should divide out

35P05 -

Gauss: Cylindrical Symmetryλ=inqTotal charge enclosed:

( )0 0

2 inqE r λπε ε

= = =

S SE d E dA EAΦ = ⋅ = =∫∫ ∫∫E A

02E

rλπε

=0

ˆ2 rλπε

⇒ =E r

36P05 -

Gauss: Planar SymmetryInfinite slab with uniform charge density σFind E outside the plane

37P05 -

Gauss: Planar Symmetry

Symmetry is Planar

Use Gaussian Pillbox

xE ˆE±=x̂

GaussianPillbox

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

38P05 -

Gauss: Planar SymmetryAqin σ=Total charge enclosed:

NOTE: No flux through side of cylinder, only endcaps

( )0 0

2 inq AE A σε ε

= = =

S SE Endcapsd E dA EAΦ = ⋅ = =∫∫ ∫∫E A +

+++++++++++ σ

E E

x

A

{ }0

ˆ to rightˆ- to left2

σε

⇒ = xE x02

E σε

=

39P05 -

PRS Question:Slab of Charge

40P05 -

Group Problem: Charge SlabInfinite slab with uniform charge density ρThickness is 2d (from x=-d to x=d). Find E everywhere.

x̂

41P05 -

PRS Question:Slab of Charge

42P05 -

Potential from E

43P05 -

Potential for Uniformly Charged Non-Conducting Solid SphereFrom Gauss’s Law

20

30

ˆ,4

ˆ,4

Q r Rr

Qr r RR

πε

πε

⎧ >⎪⎪= ⎨⎪ <⎪⎩

rE

r

B

B AA

V V d− = − ⋅∫E sUse

( )0

BV V=

− ∞Region 1: r > a

204

r Q drrπε∞

= −∫0

14

Qrπε

=

Point Charge!

44P05 -

Potential for Uniformly Charged Non-Conducting Solid Sphere

( ) ( )R r

RdrE r R drE r R

∞= − > − <∫ ∫

Region 2: r < a( )

0

DV V=

− ∞

2

20

1 38

Q rR Rπε⎛ ⎞

= −⎜ ⎟⎝ ⎠

2 30 04 4

R r

R

Q Qrdr drr Rπε πε∞

= − −∫ ∫

( )2 23

0 0

1 1 14 4 2

Q Q r RR Rπε πε

= − −

45P05 -

Potential for Uniformly Charged Non-Conducting Solid Sphere

46P05 -

Group Problem: Charge SlabInfinite slab with uniform charge density ρThickness is 2d (from x=-d to x=d). If V=0 at x=0 (definition) then what is V(x) for x>0?

x̂

47P05 -

Group Problem: Spherical ShellsThese two spherical shells have equal but opposite charge.

Find E everywhere

Find V everywhere (assume V(∞) = 0)