Physics 2102 Physics 2102 Lecture 3Lecture 3

Gauss’ Law IGauss’ Law IMichael Faraday

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

WeakE-Field

Number of E-Lines Through Differential

Area “dA” is a Measure of Strength

dA

Angle Matters 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:

= E•A = 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 pointing OUTWARDS

– 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 NORMAL to both bottom and top cap

• E is PARALLEL to curved surface everywhere

• So: = + + R2E + 0 - R2E = 0!

• Physical interpretation: total “inflow” = total “outflow”!

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/C everywhere 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 centered on the charge?

q

r

Electric Flux: ExampleElectric Flux: Example

€

dr A = + dA( )ˆ r

€

rE • d

r A = EdAcos(180°) = −EdA

€

rE = −

q

r2ˆ r

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

€

=r

E ⋅dr A ∫ = −E dA = −

kq

r2

⎛

⎝ ⎜

⎞

⎠ ⎟∫ 4πr2( )

= −q

4πε0

4π( ) = −q /ε0 Gauss’ Law:Special Case!

(Outward!)

Surface Area Sphere

(Inward!)

Gauss’ Law: General CaseGauss’ Law: General Case

• Consider any ARBITRARY CLOSED surface S -- NOTE: this does NOT have to be a “real” physical object!

• The TOTAL ELECTRIC FLUX through S is proportional to the TOTAL CHARGE ENCLOSED!

• The results of a complicated integral is a very simple formula: it avoids long calculations!

€

≡r

E ⋅dr A =

±q

ε0Surface

∫

S

(One of Maxwell’s 4 equations!)

ExamplesExamples

=⋅≡Surface 0ε

qAdErr

Gauss’ Law: Example Gauss’ Law: Example Spherical symmetrySpherical symmetry

• Consider a POINT charge q & pretend that you don’t know Coulomb’s Law

• Use Gauss’ Law to compute the electric field at a distance 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

0εq=

r

q E

24|||| rEAE ==

2204

||r

kq

r

qE ==

ε

Gauss’ Law: Example Gauss’ Law: Example Cylindrical symmetryCylindrical symmetry

• Charge of 10 C is uniformly spread over a line of length L = 1 m.

• Use Gauss’ Law to compute magnitude of E at a perpendicular distance of 1 mm from the center of the line.

• Approximate as infinitely long line -- E radiates outwards.

• Choose cylindrical surface of radius R, length L co-axial with line of charge.

R = 1 mmE = ?

1 m

Gauss’ Law: cylindrical Gauss’ Law: cylindrical symmetry (cont)symmetry (cont)

RLEAE |||| ==

00 ελ

εLq

==

Rk

RRL

LE

λελ

ελ

||00

===

• Approximate as infinitely long line -- E radiates outwards.

• Choose cylindrical surface of radius R, length L co-axial with line of charge.

R = 1 mmE = ?

1 m

Compare with Example!Compare with Example!

− +

=2/

2/2/322 )(

L

Ly

xa

dxakE λ

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

224

2

Laa

Lk

+= λ

2/

2/222

L

Laxa

xak

−⎥⎦

⎤⎢⎣

⎡

+= λ

a

k

La

LkEy

λλ

==

SummarySummary

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

• Gauss’ law provides a very direct way to compute the electric flux.