1

From Last Time:   Electric field of a single charge   Electric field from multiple charges

  Superposition   Ex: calculate electric field of dipole

Oct. 1, 2009 Physics 208 Lecture 9 1

Today:   Continuous charge distributions

  Electric field of line, ring, plane   Conductors : electrostatic equilibrium   Electrostatic potential

Exam 1 results

  Average = 66%   Curve

  Ave at B/BC boundary

  Contributes to course grade as curved score.

  Exams returned in Monday discussion

Oct. 1, 2009 Physics 208 Lecture 9 2

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5

10

15

20

25

20 30 40 50 60 70 80 90 100

Physics 208 Exam 1

Cou

nt

Exam % Score

Mean = 66%

A

AB

B

BC

C

D

F

From Last Time:   Electric field of a single charge   Electric field from multiple charges

  Superposition   Ex: calculate electric field of dipole

Oct. 1, 2009 Physics 208 Lecture 9 3

Today:   Continuous charge distributions

  Electric field of line, ring, plane   Conductors : electrostatic equilibrium   Electrostatic potential

Oct. 1, 2009 Physics 208 Lecture 9 4

Charge Densities

  Volume charge density: when a charge is distributed evenly throughout a volume   ρ = Q / V dq = ρ dV

  Surface charge density: when a charge is distributed evenly over a surface area   η = Q / A dq = η dA

  Linear charge density: when a charge is distributed along a line   λ = Q / dq = λ d

Oct. 1, 2009 Physics 208 Lecture 9 5

Infinite line of charge

An infinite line of charge has a uniform charge density λ Coulombs / meter. What direction is the electric field at point x?

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

x A.

B. C.

D.

E.

Oct. 1, 2009 Physics 208 Lecture 9 6

E-field: infinite line of charge

  Charge density λ Coulombs/meter

y

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

€

dQ = λdx

€

d E

r

x €

r2 + x 2€

θ

€

dEy = cosθdE

=14πεo

dQ

r2 + x 2( )2 cosθ

=14πεo

dQr2 + x 2

rr2 + x 2

=14πεo

rλdxr2 + x 2( )

3 / 2

2

Oct. 1, 2009 Physics 208 Lecture 9 7

Add all these up

  Units? €

Ey = 2 14πεo

rλdxr2 + x 2( )

3 / 20

∞

∫

=λ2πεo

xdxr r2 + x 2( )1/ 2

0

∞

=

€

λ2πεor

€

= 2k λr

€

[k] =

€

[λ] =

€

N ⋅m2 /C2

€

C /m

€

[r] = m

Oct. 1, 2009 Physics 208 Lecture 9 8

Non-infinite (finite) line of charge

What direction is the E-field above the end of the line charge?

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

A

B

C D

E

Oct. 1, 2009 Physics 208 Lecture 9 9

Ring of uniform positive charge

y

x

z

€

Ez = k zQz2 + R2( )3 / 2

€

Ez

Which is the graph of on the z-axis?

A)

B)

C)

D)

E)

z

Oct. 1, 2009 Physics 208 Lecture 9 10

Quick quiz

We have an infinite sheet of charge of uniform charge density η. The electric field

A.  increases with distance from plane

B.  decreases with distance from plane

C.  is independent of distance from plane

D. changes direction with distance from plane

Oct. 1, 2009 Physics 208 Lecture 9 11

E-field of plane of charge

  Text does this as adding up rings of charge.

€

Eplane =η2εo

€

η = Surface charge density

Oct. 1, 2009 Physics 208 Lecture 9 12

Electric fields and forces

  Original definite of E-field was (Coulomb Force) / charge

  E-field produced force on charged particle.

€

F = q

E

3

Oct. 1, 2009 Physics 208 Lecture 9 13

Force on charged particle

  Electric field produces force qE on charged particle

  Force produces an acceleration a = FE / m

  Uniform E-field (direction & magnitude) produces constant acceleration

  Positive charge   accelerates in direction of the field

  Negative charge   accelerates in direction opposite the electric field

Oct. 1, 2009 Physics 208 Lecture 9 14

Motion of charged particle

  If no other forces, positive charge accelerates in direction of E-field.

  But many systems have drag forces (e.g. molecules in a liquid, etc)

  Drag force is complex, but usually depends on velocity.

  Particle reaches terminal velocity, determined by force balance

Oct. 1, 2009 Physics 208 Lecture 9 15

Application: Gel Electrophoresis   Charged macromolecules in ‘gel’ with applied E-field   Electric force: FE = qE

Drag force: FD ~ -cv   v = qE / c   Speed depends on �charge and drag (molecule size)

  Sometimes too complex to interpret   Protein electrophoresis:�

soak in detergent to give proteins�all the same charge density.

  Result, small proteins move faster

Steady state: FE+FD=0

Oct. 1, 2009 16

Conductors: electrostatic equilibrium

Because charges are mobile in conductor :   E-field inside conductor = 0

  Non-zero E-field causes charges to move   Stop at surface, generate canceling E-field

  Local net charge resides at surface   Repulsive force between charges pushes them away from

each other – stop at surface   E-field at locally perpendicular to surface

  A parallel component would cause charge motion   Has magnitude

€

E⊥surf =η /εo

Surface charge density

Oct. 1, 2009 Physics 208 Lecture 9 17

Charge distribution on conductor   What charge is

induced on sphere to make zero electric field?

+

18

Electrogenic fish

  Dipole + nearby conducting object

Some fish generate charge separation - electric dipole.

Dipole is induced in nearby (conducting) fish

Small changes detected by fish.

4

Oct. 1, 2009 19

Summary of conductors   everywhere inside a conductor

  Charge in conductor is only on the surface

  surface of conductor

€

E = 0

€

E ⊥

- - - -

-

-

+ + + + + +

Oct. 1, 2009 20

Electric forces, work, and energy   Consider positive particle charge q, mass m at rest

in uniform electric field E   Force on particle from field   Opposite force on particle from hand

  Let particle go - it moves a distance d   How much work was done on particle?   How fast is particle moving?

+ + v=0 v>0 €

W = Fd = qEd

€

ΔK .E .= 12mv 2 =W = qEd

Oct. 1, 2009 21

Work and kinetic energy   Work-energy theorem:

  Change in kinetic energy of isolated particle = work done

 

€

dW = F ⋅ d s = Fdscosθ

  Total work  

€

ΔK = dWstart

end

∫ = F • d s

start

end

∫

In our case,

€

F = q

E

Oct. 1, 2009 22

Electric forces, work, and energy   Same particle, but don’t let go

  How much force does hand apply?

  Move particle distance d, keep speed ~0   How much work is done by hand on particle?   What is change in K.E. of particle?

+ +

€

F = qE

€

W = Fd = qEd

€

ΔK .E .= 0

Conservation of energy? W stored in field as potential energy

Oct. 1, 2009 23

Work, KE, and potential energy   If particle is not isolated,

€

Wexternal = ΔK + ΔUWork done on system Change in

kinetic energy Change in electric potential energy

Works for constant electric field if

  Only electric potential energy difference   Sometimes a reference point is chosen

  E.g.   Then for uniform electric field

€

ΔU = −q E ⋅ Δ r

€

U r ( ) = 0 at r = (0,0,0)

€

U r ( ) = −q

E ⋅ r

Oct. 1, 2009 24

Electric potential V

  Electric potential difference ΔV is the electric potential energy / unit charge = ΔU/q

  For uniform electric field,

€

ΔV r ( ) =

ΔU r ( )

q=−q E ⋅ Δ r

q= − E ⋅ Δ r

This is only valid for a uniform electric field

5

Oct. 1, 2009 25

Quick Quiz Two points in space A and B have electric potential

VA=20 volts and VB=100 volts. How much work does it take to move a +100µC charge from A to B?

A.  +2 mJ

B.  -20 mJ

C. +8 mJ

D. +100 mJ

E.  -100 mJ

Oct. 1, 2009 26

Check for uniform E-field

+ +

Push particle against E-field, or across E-field

Which requires work? Constant electric potential in this direction

Decreasing electric potential in this direction

Increasing electric potential in this direction

Oct. 1, 2009 27

Potential from electric field

  Potential changes largest in direction of E-field.

  Smallest (zero) perpendicular to E-field

€

dV = − E • d

€

d

€

E

V=Vo

€

V = Vo − E d

€

V = Vo + E d

€

d

€

d

€

V =Vo

Oct. 1, 2009 28

Electric potential: general

  Electric field usually created by some charge distribution.   V(r) is electric potential of that charge distribution

  V has units of Joules / Coulomb = Volts

€

ΔU = F Coulomb • d

s ∫ = q

E • d s ∫ = q

E • d s ∫

Electric potential energy difference ΔU proportional to charge q that work is done on

€

ΔU /q ≡ ΔV = Electric potential difference

Depends only on charges that create E-fields

€

= E • d s ∫

Oct. 1, 2009 29

Electric potential of point charge

  Electric field from point charge Q is

  What is the electric potential difference?

€

E = kQ

r2ˆ r

€

ΔV = E • d s

start

end

∫ = k Qr2

dxrinitial

rfinal

∫

= −k Qr rinitial

rfinal

= k Qrinital

− k Qrfinal

Define

€

V r( ) = k Qr

for point charge

€

V r =∞( ) = 0 Then

Oct. 1, 2009 30

Electric Field and equipotential lines for + and - point charges

  The E lines are directed away from the source charge

  A positive test charge would be repelled away from the positive source charge

The E lines are directed toward the source charge

A positive test charge would be attracted toward the negative source charge

Blue dashed lines are equipotential