sep. 27, 2007physics 208 lecture 81 from last time… this friday’s honors lecture: biological...

Sep. 27, 2007 Physics 208 Lecture 8 1

From last time…

This Friday’s honors lecture:Biological Electric FieldsDr. J. Meisel, Dept. of Zoology, UW

Continuous charge distributions

y

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

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dr E

Motion of charged particles


Exam 1 Wed. Oct. 3, 5:30-7 pm, 2103 Ch (here)

Covers Chap. 21.6, 22-27+ lecture, lab, discussion, HW

Sample exams on website.Solutions on website.HW4 (assigned today) covers exam material.


Electric Flux Electric flux E through a surface:

(component of E-field surface) X (surface area) Proportional to

# E- field lines penetrating surface


Why perpendicular component? Suppose surface make angle surface normal

E = EA cos E =0 if E parallel A E = EA (max) if E A

Flux SI units are N·m2/C

€

r E = E||ˆ s + E⊥ˆ n

€

ˆ n

€

ˆ s

€

r A = A ˆ n

Component || surfaceComponent surface

Only component‘goes through’ surface

€

E =r E •

r A


Total flux E not constant

add up small areas where it is constant

Surface not flat add up small areas

where it is ~ flat

€

δEi = E iδA icosθ =

r E i • δ

r A i

Add them all up:

€

E =r E • d

r A

surface∫


q

Quick quizTwo spheres of radius R and 2R are centered on

the positive charges of the same value q. The flux through these spheres compare as

A) Flux(R)=Flux(2R)

B) Flux(R)=2Flux(2R)

C) Flux(R)=(1/2)Flux(2R)

D) Flux(R)=4Flux(2R)

E) Flux(R)=(1/4)Flux(2R)

q

R

2R


Flux through spherical surface Closed spherical surface,

positive point charge at center E-field surface,

directed outward E-field magnitude on sphere:

Net flux through surface: E constant, everywhere surface

€

E = 14πεo

qR2

R

€

E =r E • d

r A ∫ = EdA∫ = E dA∫ = EA = 1

4πεo

qR2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 4πR2( ) = q /εo


q

Quick quizTwo spheres of the same size enclose different

positive charges, q and 2q. The flux through these spheres compare as

A) Flux(A)=Flux(B)

B) Flux(A)=2Flux(B)

C) Flux(A)=(1/2)Flux(B)

D) Flux(A)=4Flux(B)

E) Flux(A)=(1/4)Flux(B)

2q

A B


Gauss’ law

net electric flux through closed surface = charge enclosed /

€

E =r E • d

r A ∫ = Qenclosed

εo


Using Gauss’ law Use to find E-field from known charge distribution.

Step 1: Determine direction of E-field

Step 2: Make up an imaginary closed ‘Gaussian’ surface Make sure E-field at all points on surface either

perpendicular to surface or parallel to surface

Make sure E-field has same value whenever perpendicular to surface

Step 3: find E-field on Gaussian surface from charge enclosed. Use Gauss’ law:

electric flux thru surface = charge enclosed / o

€

=r E • d

r A ∫ = Qin /εo


Field outside uniformly-charged sphere

Field direction: radially out from charge Gaussian surface:

Sphere of radius r Surface area where

Value of on this area:

Flux thru Gaussian surface:

Charge enclosed:

€

r E • d

r A ≠ 0 :

€

4π r2

€

r E • d

r A

€

E

€

Q

€

E 4π r2

Gauss’ law:

€

E 4π r2 = Q /εo ⇒ E = 14πεo

Qr2


Quick QuizWhich Gaussian surface could be used to calculate

E-field from infinitely long line of charge?

None ofthese

A. B. C. D.


E-field from line of charge Field direction: radially out from line charge Gaussian surface:

Cylinder of radius r Area where



Charge enclosed:

€

r E • d

r A ≠ 0 :

€

2πrL

€

r E • d

r A

€

E

€

λL

€

E2πrL

Gauss’ law:

€

E2π rL = λL /εo ⇒ E = 12πεo

λr

Linear charge density λ C/m


Quick QuizWhich Gaussian surface could be used to calculate

E-field from infinite sheet of charge?

Any ofthese

A. B. C. D.


Field from infinite plane of charge Field direction: perpendicular to plane Gaussian surface:

Cylinder of radius r Area where



Charge enclosed:

€

r E • d

r A ≠ 0 :

€

2π r2

€

r E • d

r A

€

E

€

ηπ r2

€

E2πr2

Gauss’ law:

€

E2π r2 = ηπ r2 /εo ⇒ E = η2εo

Surface charge η C/m2


Properties of Conductor in Electrostatic Equilibrium

In a conductor in electrostatic equilibrium there is no net motion of charge Property 1: E=0 everywhere inside the conductor

Conductor slab in an external field E: if E 0 free electrons would be accelerated

These electrons would not be in equilibrium

When the external field is applied, the electrons redistribute until they generate a field in the conductor that exactly cancels the applied field.Etot = E+Ein= 0

Ein

Etot =0


Induced dipole What charge is

induced on sphere to make zero electric field?

+


Conductors: charge on surface only Choose a gaussian surface inside (as

close to the surface as desired)

There is no net flux through the gaussian surface (since E=0)

Any net charge must reside on the surface (cannot be inside!)

E=0


Field’s Magnitude and Direction

E-field always surface: Parallel component of E would put

force on charges Charges would accelerate This is not equilibrium

Apply Gauss’s lawField lines

sep. 27, 2007physics 208 lecture 81 from last time… this friday’s honors lecture: biological...

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