# Flux Capacitor (Schematic) Physics 2102 Lecture 3phys.lsu.edu/~jdowling/PHYS21024SP07/lectures/  · PDF fileFlux Capacitor (Schematic) ... 4πR2E= −40π Nm2/C ... (180¡)=!EdA r

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• Physics 2102Physics 2102Lecture 3Lecture 3

1791-1867

Version: 1/22/07

Flux Capacitor (Schematic)

Physics 2102

Jonathan Dowling

• What are we going to learn?What are we going to learn?A road mapA road map

Electric charge Electric force on other electric charges Electric field, and electric potential

Moving electric charges : current Electronic circuit components: batteries, resistors, capacitors Electric currents Magnetic field

Magnetic force on moving charges Time-varying magnetic field Electric Field More circuit components: inductors. Electromagnetic waves light waves Geometrical Optics (light rays). Physical optics (light waves)

• What? What? The Flux! The Flux!STRONGE-Field

WeakE-Field

Number of E-LinesThrough Differential

Area dA is aMeasure of Strength

dA

AngleMatters Too

• Electric Flux: Planar SurfaceElectric Flux: Planar Surface

Given: planar surface, area A uniform field E E makes angle with NORMAL to

plane

Electric Flux: = EA = E A cos

Units: Nm2/C Visualize: Flow of Wind

Through Window

E

AREA = A=An

normal

• Electric Flux: General SurfaceElectric Flux: General Surface For any general surface: break up into

infinitesimal planar patches

Electric Flux = EdA Surface integral dA is a vector normal to each patch and

has a magnitude = |dA|=dA CLOSED surfaces:

define the vector dA as pointingOUTWARDS

Inward E gives negative flux Outward E gives positive flux

E

dA

dA

EArea = dA

• Electric Flux: ExampleElectric Flux: Example

Closed cylinder of length L, radius R Uniform E parallel to cylinder axis What is the total electric flux through

surface of cylinder?(a) (2RL)E(b) 2(R2)E(c) Zero

Hint!Surface area of sides of cylinder: 2RLSurface area of top and bottom caps (each): R2

L

R

E

(R2)E(R2)E=0What goes in MUST come out!

dA

dA

• Electric Flux: ExampleElectric Flux: Example

Note that E is NORMALto both bottom and top cap

E is PARALLEL tocurved surface everywhere

So: = 1+ 2 + 3= R2E + 0 - R2E= 0!

Physical interpretation:total inflow = totaloutflow! 3

dA

1dA

2 dA

• Electric Flux: ExampleElectric Flux: Example Spherical surface of radius R=1m; E is RADIALLY

INWARDS and has EQUAL magnitude of 10 N/Ceverywhere on surface

What is the flux through the spherical surface?

(a) (4/3)R2 E = 13.33 Nm2/C

(b) 2R2 E = 20 Nm2/C

(c) 4R2 E= 40 Nm2/C

What could produce such a field?

What is the flux if the sphere is not centeredon the charge?

• q

r

Electric Flux: ExampleElectric Flux: Example

dr A = + dA( ) r

r E d

r A = EdAcos(180) = !EdA

r E = !

q

r2

r

Since r is Constant on the Sphere RemoveE Outside the Integral!

! =r E "d

r A # = \$E dA = \$ kq

r2

% & '

( ) * # 4+r2( )

= \$q

4+,0

4+( ) = \$q /,0 Gauss Law:Special Case!

(Outward!)

Surface Area Sphere

(Inward!)

• GaussGauss Law: General Case Law: General Case

Consider any ARBITRARYCLOSED surface S -- NOTE:this does NOT have to be areal physical object!

The TOTAL ELECTRIC FLUXthrough S is proportional to theTOTAL CHARGEENCLOSED!

The results of a complicatedintegral is a very simpleformula: it avoids longcalculations!

!"r E # d

r A =

q

\$0Surface

%

S

(One of Maxwells 4 equations!)

• ExamplesExamples

! ="#\$Surface 0

%

• GaussGauss Law: Example Law: ExampleSpherical symmetrySpherical symmetry

Consider a POINT charge q & pretend that youdont know Coulombs Law

Use Gauss Law to compute the electric field at adistance r from the charge

Use symmetry: draw a spherical surface of radius R centered

around the charge q E has same magnitude anywhere on surface E normal to surface

0!

q="

rq E

24|||| rEAE !=="

2204

||r

kq

r

qE ==

!"

• GaussGauss Law: Example Law: ExampleCylindrical symmetryCylindrical symmetry

Charge of 10 C is uniformly spreadover a line of length L = 1 m.

Use Gauss Law to computemagnitude of E at a perpendiculardistance of 1 mm from the center ofthe line.

Approximate as infinitely longline -- E radiates outwards.

Choose cylindrical surface ofradius R, length L co-axial withline of charge.

R = 1 mmE = ?

1 m

• GaussGauss Law: cylindrical Law: cylindricalsymmetry (cont)symmetry (cont)

RLEAE !2|||| =="

00!

"

!

Lq==#

Rk

RRL

LE

!

"#

!

"#

!2

22||

00

===

Approximate as infinitely longline -- E radiates outwards.

Choose cylindrical surface ofradius R, length L co-axial withline of charge.

R = 1 mmE = ?

1 m

• Compare with Example!Compare with Example!

!" +

=

2/

2/

2/322 )(

L

L

yxa

dxakE #

if the line is infinitely long (L >> a)

224

2

Laa

Lk

+

=!

2/

2/

222

L

Laxa

xak

!"#

\$%&

'

+

= (

a

k

La

LkEy

!! 22

2==

• SummarySummary

Electric flux: a surface integral (vector calculus!);useful visualization: electric flux lines caught by thenet on the surface.

Gauss law provides a very direct way to computethe electric flux.

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