From last time

More on electric potential and connection to E-fieldHow to calculate E-field from VCapacitors and Capacitance

Today:

Gauss’ lawWork, potential energy Electric potentialconnection between E-field and potential

This week Honor

• Prof. S. Westheroff on THE HIGHEST ENERGY COSMIC RAYS with the Pierre Auger observatory

Electric Potential

!

"U /q #V = Electric potential

Work done by E-field produced by source charge Q on a test charge q to move it along a path:

!

Wel = "#U =r F Coulomb $ d

r s = q

r E $%% d

r s = q

r E $ d

r s %

Potential energy of 2 charges:

Electric potential of a point charge Q:

!

UQq r( ) = keQq

r

!

Vq r( ) =UQq r( )q

= keQ

r

Superposition: electric potential of dipole

+Q -Q

x=+ax=-a

Superposition of

• potential from +Q

• potential from -Q

+ =

y

Eg: V(y=0) = kQ/a-kQ/a=0In general: for a group of point charges

!

V = keqi

rii

"

!

V =kQ

r

Electric Potential of a continuous charge distribution

Consider a small charge element dq. The potential at some point due to this charge element is

Total potential:

This value of V used the reference that V = 0 at infinite distance

Quick QuizThe total work required for YOU to assemble the set of

charges as shown below is:

+Q

+Q

! Q5 m

5 m5 m

1. positive2. zero3. negative

Wext = !U = Ufinal - Uinitial <0 =>

U decreases.

1 2

3

!

W1 = 0

W2 = kQ("Q)

d

W3 = kQQ

d+ k

Q("Q)

d

Total work = "kQ

2

d

!

=QV12

= 0

Work to put Q at pos 1

Work to put -Q at pos 2where V1 = kQ/r

!

= "QV1

Work to put Q at pos 3where V12 = 0

Electric field from Potential

Differential form:

Spell out the vectors:

Each component:

One only symbol:

Work done = qoV

B-q

oV

A =

!

"r F

Coulomb( ) • dr l

A

B

# = -Wel

!

dWel =r F Coulomb • d

r s = "dU = "qdV #

r F Coulomb $ d

r s = q

r E • d

r s = "qdV

!

dV = "r E • d

r s

!

dV = " Exdx + Eydy + Ezdz( )

!

Ex = "dV

dx, Ey = "

dV

dy, Ez = "

dV

dz

!

r E = "

r # V = "

dV

dx,dV

dy,dV

dz

$

% &

'

( )

Work to move a charge q0 from A to B and overcome the Coulomb repulsive force dure to Q:

!V=VB-VA

potential difference

Quick QuizSuppose the electric potential is constant everywhere. What is the electric field?

A) Positive

B) Negative

C) Increasing

D) Decreasing

E) Zero

!

dV = "r E • d

r s

Explain

!

r E

V=Vo

!

d

r l

!

V = Vo"

r E d

r l

!

V = Vo

+r E d

r l

!

d

r l

!

dV = "r E • d

r l

E-field can be used to find changes in V

!

d

r l

!

V =Vo

Potential changes largest in direction of E-fieldZero changes in direction perpendicular to E-field

Equipotential lines

Lines of constant potential are perpendicular at any point to E-field

Quick Quiz

How does the electric potential outside a uniforminfinite sheet of positive charge vary with distancefrom the sheet?

A. Is constant

B. Increasing as (distance)1

C. Decreasing as (distance)1

D. Increasing as (distance)2

E. Decreasing as (distance)2

Notice the E field is uniform!

!

dV = "r E • d

r l

A B

xd x

E+++++++

ExplainOnly one dimension => easy problem! Remember:

!

r E || d

r s E cnst

!

Ex

= E = "dV

dx#

$V = VB"V

A= "

r E • d

r x

A

B

% = " Edx

A

B

% = "E(xB" x

A) = "Ed

A B

xdx

E

Constant E-field corresponds to linearlydecreasing (in direction of E) potential

Particle gains kinetic energy equal to thepotential energy lost

!

dV = "r E • d

r s

+++++++

+

!

Wel = "#U = "q#V = #K

Potential of spherical conductor

• Another 1D problem

• charge is on surface

• E-field is in radial direction difficultpath

easypath

Easy because is samedirection as E,

!

E• ds = E dr = Edr

At any distance r =>

Inside the sphere E = 0 so all points have the same potential of the surfaceV(R) = kQ/R

+Q

!

V (r) = kQdr

r2

= kQ

rr

"

#

R

!

V (") #V (r) = #r E • d

r s = #

r

"

$ Edr

r

"

$

Quick QuizTwo conducting spheres of diff radii connected by long conducting wire.

What is approximately true of Q1, Q2?

A) Q2>Q1

B) Q2<Q1

C) Q2=Q1

R1R2

Q1 Q2

Explain! Since both must be at the same potential,

!

kQ1

R1

=kQ

2

R2

"Q1

Q2

=R1

R2

Surface charge densities?

!

# =Q

4$R2"

#1

#2

=R2

R1

Charge proportionalto radius

Surface charge densityproportional to 1/R

Electric field?

Since ,

!

E =#

2%o

Local E-field proportional to 1/R(1/radius of curvature)

The surface chargedensity and the E-field are larger at

sharp points!

Varying E-fields outside a conductor

Expect larger electric fields near the small end. Electric field approximately proportional to 1/(local radius of curvature).

Large electric fields at sharp points because E-field proportional to charge density and charge density is larger for smaller radius of curvature

Fields can be so strong that air ionized and ions accelerated.

Any conductor quizConsider this conducting object. When it has total

charge Qo, its electric potential is Vo. When it has

charge 2Qo, its electric potential

A. is Vo

B. is 2Vo

C. is 4Vo

D. depends on shape

Capacitance and capacitors! Electric potential of any conducting object

proportional to its total charge.

!

V =1

CQ

A capacitor consists of two conductors:

•Conductors generically called ‘plates’

•Charge transferred between plates

•Plates carry equal and opposite charges

•Potential difference between plates !Vproportional to charge transferred Q

!

"V =1

CQ

! The SI unit of capacitance is the farad (F)

! 1 Farad = 1 Coulomb / Volt

! This is a very large unit: typically use

µF = 10-6 F, nF = 10-9 F, pF = 10-12 F

Parallel plate capacitor

"=Q/A surface charge density

A=surface area!

Eleft + Eright =" /2#o +" /2#o =" /#o

-Q+Q

d

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

Eleft

Eright

Uniform E-fieldNotice outside the plates E = 0

Work done = qoV

B-q

oV

A =

!

"r F

Coulomb( ) • dr l

A

B

#

Work to move q from positive to negative plate =

!

C ="oA

dThis is a geometrical factor

+ -

E+

E-

E-

E+

E-

E+

!

q(V" "V+) = "qr E • d

r s

0

d

# = "qEd $%V = Ed =&

'0

d = Qd

'0A

(

) *

+

, - =

Q

C

!

"V =V+ #V#