Tues. Oct. 6, 2009Physics 208 Lecture 101 Last time… Fields, forces, work, and potential Electric forces and work + + Potential energy stored in electric.

Tues. Oct. 6, 2009 Physics 208 Lecture 10 1

Last time… Fields, forces, work, and potential

Electric forces and work+ +

Potential energy stored in electric field

Tues. Oct. 6, 2009 2

Work, KE, and potential energy

When 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

Tues. Oct. 6, 2009 3

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

Tues. Oct. 6, 2009 4

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

Tues. Oct. 6, 2009 5

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


Potential from electric field

dV largest in direction of E-field.

dV smallest (zero) perpendicular to E-field

€

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

Tues. Oct. 6, 2009 7

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 ∫

Tues. Oct. 6, 2009 8

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


Equipotential lines

Lines of constant potential In 3D, surfaces of constant potential


Topographic map

Each lines is constant elevation

Same as constant gravitational potentialgh (energy = mgh)

Height interval between lines constant


Electric field from potential

Spell out the vectors:

€

dW =r F ext • d

r s = −

r F Coulomb • d

r s ⇒

dV = −r E • d

r s

€

dV = − Exdx + Eydy + E zdz( )

€

Ex = −dV

dx, Ey = −

dV

dy, E z = −

dV

dz

Usually written

€

r E = −

r ∇V = −

dV

dx,dV

dy,dV

dz

⎛

⎝ ⎜

⎞

⎠ ⎟

€

r E = −

r ∇V = −

dV

dx,dV

dy,dV

dz

⎛

⎝ ⎜

⎞

⎠ ⎟

Said before that

This works for


Quick Quiz

Suppose the electric potential is constant everywhere. What is the electric field?

A) Positive

B) Negative

C) Increasing

D) Decreasing

E) Zero


Electric Potential - Uniform Field

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

€

dV = −r E • d

r s

€

=− EdxA

B

∫ = −E dxA

B

∫ = −E xB − xA( )

A

B

x

Here V depends only on x, not on y

€

⇒ VB −VA = −r E

A

B

∫ • dr s = −Eˆ x

A

B

∫ • dr s


Check of basic cases

Previous quick quiz: uniform potential corresponds to zero electric field

Linear potential corresponds to constant electric field

€

E = −∇V = −∇ constant( ) = 0

€

E = −∇V = −∇ −Ex( ) =∂

∂xEx,

∂

∂yEx,

∂

∂zEx

⎛

⎝ ⎜

⎞

⎠ ⎟= Eˆ x


Potential and charge

Have shown that a conductor has an electric potential, and that potential depends on its charge

For a charged conducting sphere:

+ +++

+++

++

++

€

V R( ) −V ∞( ) = kQ

R=

k

RQ

Electric potential proportional to total chargeElectric potential proportional to total charge


Quick Quiz

Consider 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

Electric potential of any conducting object proportional to its total charge.

€

V =1

CQ

C = capacitance Large capacitance: need lots of charge to change

potential Small capacitance: small charge can change

potential.


Capacitors

Where did the charge come from? Usually transferred from another conducting

object, leaving opposite charge behind

A capacitor consists of two conductors Conductors generically called ‘plates’ Charge transferred between plates

Plates carry equal and opposite charges Potential difference between plates

proportional to charge transferred Q


Definition of Capacitance Same as for single conductor

but ΔV = potential difference between plates Q = charge transferred between plates

SI unit of capacitance is farad (F) = 1 Coulomb / Volt This is a very large unit: typically use F = 10-6 F, nF = 10-9 F, pF = 10-12 F

€

ΔV =1

CQ


How was charge transferred? Battery has fixed electric potential difference

across its terminals Conducting plates connected to battery

terminals by conducting wires.

ΔVplates = ΔVbattery across plates

Electrons move from negative battery terminal to -Q plate from +Q plate to positive battery terminal

This charge motion requires work The battery supplies the work

ΔV

€

Q = CΔV


Requires work to transfer charge dq from one plate:

Work done to charge a capacitor

€

dW = ΔVdq =q

Cdq

€

dW = ΔVdq =q

Cdq

€

W =q

C0

Q

∫ dq =Q2

2C

€

W =q

C0

Q

∫ dq =Q2

2C

Work done stored as potential energy in capacitor

Total work = sum of incremental work

€

U =Q2

2C=

1

2QΔV =

1

2C ΔV( )

2

€

U =Q2

2C=

1

2QΔV =

1

2C ΔV( )

2


Example: Parallel plate capacitor

Charge Q moved from right conductor to left conductor

Charge only on inner surfaces Plate surfaces are charge

sheets, each producing E-field

+Q -Q

d

innerouter

€

E left + E right = η /2εo + η /2εo = η /εo

Uniform field between plates


Quick Quiz

Electric field between plates of infinite parallel-plate capacitor has a constant value /o. What is the field outside of the plates?

A. /o

B. /2o

C. - /2o

Δ. /4o

E. 0


What is potential difference?

-Q+Q

d

+++

+

+

+++

+

++

+++

+

---

-

-

---

-

--

---

-

Potential difference = V+-V-

= - (work to move charge q from + plate to - plate) / q

€

=−−qEd( ) /q

ΔV = Ed = ηd /εo = Qd

εoA

⎛

⎝ ⎜

⎞

⎠ ⎟

€

ΔV = V+ −V− = Qd

εoA

⎛

⎝ ⎜

⎞

⎠ ⎟

€

rE


What is the capacitance?

+Q

-Q

d

€

ΔV = V+ −V− = Qd

εoA

⎛

⎝ ⎜

⎞

⎠ ⎟

€

ΔV = Q /C

€

C =εoA

dThis is a geometrical factor

€

U =1

2C ΔV( )

2=

1

2

εoA

dEd( )

2=

1

2Ad( )εoE 2

Energy stored in parallel-plate capacitor

€

U / Ad( ) =1

2εoE 2

Energy density


Human capacitors Cell membrane:

‘Empty space’ separating charged fluids (conductors)

~ 7 - 8 nm thick In combination w/fluids, acts as

parallel-plate capacitor

Cytoplasm

Extracellular fluid

Plasma membrane

100 µm


Modeling a cell membrane

Charges are +/- ions instead of electrons

Charge motion is through cell membrane (ion channels) rather than through wire

Otherwise, acts as a capacitor ~0.1 V ‘resting’ potential

Cytoplasm

Extracellular fluid

Plasma membrane

Na+ Cl-

K+ A-

- - - - - -

+ + + + + +

7-8 nm

ΔV~0.1 V

Ionic charge at surfaces of conducting fluids

Capacitance:

€

oA

d=

8.85 ×10−12 F /m( )4π 50 ×10−6 m( )2

8 ×10−9 m= 3.5 ×10−11F = 35 pF

100 µm sphere surface area

~ 3x10-4 cm2

~0.1µF/cm2


Cell membrane depolarization

Cell membrane can reverse potential by opening ion channels.

Potential change ~ 0.12 V

Ions flow through ion channels Channel spacing ~ 10xmembrane thickness (~ 100 channels / µm2 )

How many ions flow through each channel?

Cytoplasm

Extracellular fluid

Plasma membrane

Na+ Cl-

K+ A-

- - - - - -+ + + + + +

7-8 nm

ΔV~0.1 V

+ + + + + +- - - - - -

ΔV~-0.02 V

(100 channels/µm2)x4π(50 µm)2=3.14x106 ion channels

Charge xfer required ΔQ=CΔV=(35 pF)(0.12V) =(35x10-12 C/V)(0.12V)

= 4.2x10-12 Coulombs1.6x10-19 C/ion -> 2.6x107 ions flow

Ion flow / channel =(2.6x107 ions) / 3.14x106 channels ~ 7 ions/channel


Cell membrane as dielectric

Membrane is not really empty

It has molecules inside that respond to electric field.

The molecules in the membrane can be polarized

Cytoplasm

Extracellular fluid

Plasma membrane

Na+ Cl-

K+ A-

- - - - - -

+ + + + + +

7-8 nm

Dielectric: insulating materials can respond to an electric field by generating an opposing field.


Effect of E-field on insulators If the molecules of the dielectric are non-polar

molecules, the electric field produces some charge separation

This produces an induced dipole moment

+

-

+

-

E=0E


Dielectrics in a capacitor An external field can polarize

the dielectric

The induced electric field is opposite to the original field

The total field and the potential are lower than w/o dielectric E = E0/ and V = V0/

The capacitance increases C = C0

E0

Eind


Cell membrane as dielectric

Without dielectric, we found 7 ions/channel were needed to depolarize the membrane. Suppose lipid bilayer has dielectric constant of 10. How may ions / channel needed?

Cytoplasm

Extracellular fluid

Plasma membrane

Na+ Cl-

K+ A-

- - - - - -

+ + + + + +

7-8 nm

C increases by factor of 1010 times as much charged needed to reach potential

A. 70

B. 7

C. 0.7

Tues. Oct. 6, 2009Physics 208 Lecture 101 Last time… Fields, forces, work, and potential Electric forces and work + + Potential energy stored in electric.

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