From last time(s)…

Oct. 2, 2008 1

From last time(s)…

Finish conductors in electrostatic equilibrium Work, energy, and (electric) potential Electric potential and charge Electric potential and electric field.

Gauss’ law Conductors in electrostatic equilibrium

Today…

Oct. 2, 2008 2

Exam 1 Scores

Class average = 76%

(This is 84/110)

Curve: B / BC boundary is 76%

Your score postedat learn@uw

Oct. 2, 2008 3

Conductor in Electrostatic Equilibrium

In a conductor in electrostatic equilibrium there is no net motion

of charge 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

Oct. 2, 2008 4

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

Oct. 2, 2008 5

E-Field 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 law at surface

€

ΦETot = EA

€

Qencl = ηA

€

⇒ E = η /εo

€

rE || this surface

€

⇒ ΦE = 0

€

rE ⊥this surface

€

⇒ ΦE = EA

€

rE = 0

this surface

Oct. 2, 2008 6

Summary of conductors everywhere inside a conductor

Charge in conductor is only on the surface

surface of conductor

€

rE = 0

€

rE ⊥

----

-

-

+ +++++

Oct. 2, 2008 7

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 . =1

2mv 2 = W = qEd

Oct. 2, 2008 8

Work and kinetic energy Work-energy theorem:

Change in kinetic energy of isolated particle = work done

€

dW =r F ⋅d

r s = Fdscosθ

Total work

€

ΔK = dWstart

end

∫ =r F • d

r s

start

end

∫

In our case,

€

rF = q

r E

Oct. 2, 2008 9

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. 2, 2008 10

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 = −qr E ⋅Δ

r r

€

Ur r ( ) = 0 at

r r = (0,0,0)

€

Ur r ( ) = −q

r E ⋅

r r

Oct. 2, 2008 11

Electric potential V

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

For uniform electric field,

€

ΔVr r ( ) =

ΔUr r ( )

q=

−qr E ⋅Δ

r r

q= −

r E ⋅Δ

r r

This is only valid for a uniform electric field

Oct. 2, 2008 12

Quick QuizTwo 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. 2, 2008 13

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. 2, 2008 14

Potential from electric field

Potential changes largest in direction of E-field.

Smallest (zero) perpendicular to E-field

€

dV = −r E • d

r l

€

dV = −r E • d

r l

€

dr l

€

rE

V=Vo

€

V = Vo −r E d

r l

€

V = Vo +r E d

r l

€

dr l

€

dr l

€

V = Vo

Oct. 2, 2008 15

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 =r F Coulomb • d

r s ∫ = q

r E • d

r s ∫ = q

r E • d

r 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

€

= r

E • dr s ∫

Oct. 2, 2008 16

Electric potential of point charge

Electric field from point charge Q is

What is the electric potential difference?

€

rE =

kQ

r2ˆ r

€

ΔV =r E • d

r s

start

end

∫ = kQ

r2dx

rinitial

rfinal

∫

= −kQ

r rinitial

rfinal

= kQ

rinital

− kQ

rfinal

Define

€

V r( ) = kQ

rfor point charge

€

V r = ∞( ) = 0 Then

Oct. 2, 2008 17

Electric Potential of point charge

Potential from a point charge Every point in space has a

numerical value for the electric potential

y

x

€

V =kQ

r

Distance from ‘source’ charge +Q

+Q

Oct. 2, 2008 18

Potential energy, forces, work

U=qoV Point B has greater potential

energy than point A Means that work must be done

to move the test charge qo

from A to B. This is exactly the work to

overcome the Coulomb repulsive force.

Ele

ctric

pot

entia

l ene

rgy=

qoV

A

B

qo > 0Work done = qoVB-qoVA =

€

− r

F Coulomb( ) • dr l

A

B

∫

Differential form:

€

qodV = −r F Coulomb • d

r l

Oct. 2, 2008 19

V(r) from multiple charges

Work done to move single charge near charge distribution. Other charges provide the force, q is charge of interest.

q

q1

q2

q3

€

U =r F • d

r s ∫ =

r F q1

+r F q 2 +

r F q3( ) • d

r s ∫

= Uq1r1( ) + Uq 2 r2( ) + Uq 3 r3( )

= kq1q

r+ k

q2q

r+ k

q3q

r

= q kq1

r+ k

q2

r+ k

q3

r

⎛

⎝ ⎜

⎞

⎠ ⎟

= q Vq1 r( ) + Vq2 r( ) + Vq 3 r( )( )Superposition of individual electric potentials

€

V r( ) = Vq1 r( ) + Vq 2 r( ) + Vq 3 r( )

Oct. 2, 2008 20

Quick Quiz 1 At what point is the electric potential zero for this

electric dipole?

+Q -Q

x=+ax=-aA

B

A. A

B. B

C. Both A and B

D. Neither of them

Oct. 2, 2008 21

Superposition: the dipole electric potential

+Q -Q

x=+ax=-aSuperposition of• potential from +Q• potential from -Q

Superposition of• potential from +Q• potential from -Q

+ =

V in plane