PHYSICS 1B – Fall 2009

Electricity&

Magnetism

Professor Brian Keating

SERF Building. Room 333

Capacitor combinations

Capacitors connected in seriesand parallel

Electrical circuit elements

Circuit diagramVoltage source resistor capacitor

Conductor

VR

C

+-

one capacitor

+q

-q+

-

CΔV Vc=ΔV

qCV

=!

Two Capacitors in Parallel

equivalentcapacitance

=!eqqCV

C1 C2ΔV

q

q1q2 Ceq

q

qΔV

What single capacitor has the same properties as two capacitors in parallel?

=

PHYSICS 1B – Fall 2009

Electricity&

Magnetism

Professor Brian Keating

SERF Building. Room 333

Capacitors

Two Parallel Capacitors

C1 C2ΔV

1 2V V V! = ! = !

1 2eq

C V C VqCV V

! + != =! !

equivalentcapacitance

=!eqqCV

= + = ! + !1 2 1 2q q q C V C V

q

q1 q2

1 2eqC C C= +

Parallel Capacitors

= + +1 2 ........eq NC C C C

For N capacitors in parallel

Ceq is the sum of capacitances

Like a larger capacitor, larger area

5µF 3µFΔV

10µF

Find the equivalent capacitance

µ

= + +

= + + =1 2 3

5 3 10 18eq

eq

C C C C

C F

A. 15 uF

B. 17 uF

C.18 uF

D.20 uF

5µF 3µFΔV

10µF

Find the equivalent capacitance

µ

= + +

= + + =1 2 3

5 3 10 18eq

eq

C C C C

C F

Two Capacitors in Series

C1

C2

ΔV

q +q-q

+q-q

q

q=!eqqCV

What is the equivalentcapacitance?

ΔV1

ΔV2

the charge on bothcapacitors in seriesis q

Two Capacitors in Series

C1

C2

ΔV

q +q-q

+q-q

q

q

=!eqqCV

q=q1=q2

1 2

1 1 ( )eq

V q qC q q C C

!= = +

! = ! + ! = +1 21 2

q qV V VC C

1 2

1 1 1

eqC C C= +

For N capacitors in series

= + +1 2

1 1 1 1.........eq NC C C C

Capacitors in series

Ceq is smaller than the smallest capacitance.You store less charge on series capacitorsthan you would on either one of them alonewith the same voltage!

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

- - - - - - - - - - -

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

- - - - - - - - - - -

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

C

ΔV ΔV/2

ΔV/2

C1=C2=Ceq =

The equivalent capacitance is less than thecomponent capacitances

C1

C2

2CC

Ceq <C1 or C2

Physical ArgumentTake a parallel plate capacitor and place a thin metal plate with the same area in the middle of the gap.

the component capacitancesare larger than the total

34. Find the equivalent capacitance.

X

X

Ceq= 4.00+2.00+6.00=12.00 µF

= + = =1 1 1 4 1

24 8 24 6seriesC

Cseries=6µF

34. Find the charge on each capacitor.

X

XCseries =6µF

= !q C V

C1

C4

C3

C2

q1=C1ΔV=4x10-6(36)=1.44x10-4 C

q2 =C1ΔV=2x10-6(36)=0.72x10-4 C

q3=q4=CseriesΔV=6x10-6(36)=2.16x10-4C

34.Find the voltage drop across eachcapacitor.

ΔV1=ΔV2=36V

ΔV1 ΔV2

ΔV3

ΔV4

! =qVC !

!" = = =

4

3 63

2.16 10 9.024 10

q xV V

C x!

!" = = =

4

4 64

2.16 10 278 10

q xV V

C x

series capacitorsThe larger C has the smaller voltage drop

16.9 Dielectrics, Energy

Dielectric constant-effect on capacitanceEnergy stored in a capacitorEnergy density (depends on E2)Biological Membranes

Dielectric material – insulators such as paper, glassplastic, ceramic. Used in the gap in capacitors.

“Dielectric Strength” - is the electric field at which conduction occurs through the material

dielectric material

-q

+q

Electric Fields in Dielectric FilledCapacitors

Vacuum

dielectric material

ΔVo

oVV!"

" =

Effects of a dielectric material inserted into a capacitor,with charge q

-q

-q

+q

+q

Potential due tocharge qdecreases by κ

ΔV

κ= dielectric constant (dimensionless)

Dielectric Properties of Selected Materials

Dielectric Strength Material dielectric constant, κ (Volt/m)

Vacuum 1.000000 -----Air 1.00059 2x106

Polystyrene 2.3 24x106

Paper 3.4 16x106

Pyrex 5.6 14x106

Water 80 -------

ΔVo

-q

-q

+q

+q

ΔV

How does the capacitance change?

Co

C=o

q qV V

!=

" "

oC C!=Capacitance increases

Originally

Add dielectric

ΔVo

-q

-q

+q

+q

ΔV

How does the E field change?

E= oVVd d!

""=

oEE!

=

Electric field decreases (when not connected to a battery)

Permittivity is increased

Compared to vacuum

o! "!=

o

q qEA A!" "

= =

ΔVo

-q

-q

+q

+q

ΔV

Example: A parallel plate capacitor consists of metal sheets(A= 1.0m2) separated by a Teflon sheet (κ=2.1) with a thicknessof 0.005 mm. (a) find the capacitance. (b) Find the maximumvoltage. The maximum electric field across Teflon is 60x106

V/m. – this is its dielectric strength.

12

3

6

2.1(8.8 10 )(1.0)0.005 10

3.7 10

oA xCd x

C x F

!" #

#

#

= =

=

(a)Κ=2.1

A=0.25m2

d=0.005m

A parallel plate capacitor consists of metal sheets(A= 0.25m2)separated by a Teflon sheet (κ=2.1) with a thickness of 0.005mm. (a) find the capacitance. (b) Find the maximum voltage.The maximum electric field across Teflon is 60x106

V/m.(dielectric strength)

6 3max 60 10 (0.005 10 ) 300dsV E d x x V!" = = =

(b)

Molecular basis for dielectric constant

Dipolar molecules.

e.g. water

+

-

Eare oriented inan E field

Oriented molecules decrease the net charge nearthe plates

The E field in the Capacitor is reduced

!= oEE

Polarization of Dielectric

Dielectric Screening

High dielectric constant of water allows ions to dissociate

Na+ Cl-

+

+-

+- +-+-+ -

+-

+-

+-

+-

+ --

!= =

80o oV V

V

Find the potential energy in electron volts for theinteraction of Na+ and Cl- separated by 0.5 nm in water.

19 2

12 9

2119

(1.6 10 )4 4 (80)(8.8 10 )(0.5 10 )

15.8 101.6 10

Na cl

o

q q xPE

r x x

eVPE x Jx

x J

!"# !

$

$ $

$$

$= =

% &= $ ' (

) *

PE= -0.036 electron Volts

comparable to thermal energies (kinetic energy ofthe ion) about 0.025 eV at room temperature

Energy stored in a capacitor.

q=0

q=0

Workdone tocharge acapacitor

ΔV=0

+q

-q

ΔVq

221 1

2 2= = !

qW C VC

So work depends on the square of q or ΔV

A parallel plate capacitor consists of metal sheets (A= 1m2)separated by a teflon sheet (κ=2.1) with a thickness of 0.005mm. Find the maximum energy that can be stored.

C

C=3.7 x10-6F ΔVmax=300V

2 6 2

1

1 1(3.7 10 )(300)2 21.7 10

Energy C V x

Energy x J

!

!

= " =

=

Insert the dielectric material with dielectric constant κ into the capacitor keeping the voltage source connected. Find C,q,E, PE

-qo

+qoΔVo

ΔVo

q

-q

Co

C=

E=

q=

oC!

o oCV C V q! != =

oo

VV Ed d

!!= =

PE= 2 21 12 2 o oC V C V PE! !" = " =

(quick quiz 16.6)

Energy Density in a Capacitor

Suppose you wanted to store a large amount of energyin a capacitor with a given volume of 1m3 using TeflonAs the dielectric (dielectric strength of 60x106 V/m).What is the maximum energy that could be stored?

212

Energy CV=

oAC

d!"

=

22 2

2

1 1 1 ( )2 2 2

o oo

A Ad VEnergy V E volume

d d!" !"

!"= = =

212 o

Energy Evolume

!"= The energy density dependsonly on the E field squared.

2

12 6 2

4

1 ( )21(2.1)(8.8 10 )(60 10 ) (1)23.4 10

oEnergy E volume

Energy x x

Energy x J

!"

#

=

=

=

For comparison the energy content of burning

1 gallon of gasoline is 1.3x108 J

Chemical energy has a higher energy density.

For a 1 m3 capacitor atthe maximum voltage.

For the maximum electric field = dielectric strength

Capacitance of Biological Membranes

Axon - Nerve cells

+

+

+ + ++ ++

-

- ----

Potential difference across the membrane

membrane

Nerve transmission – involves a discharge of membrane potential

The low dielectric portion of a biological membranehas a thickness of 2.0 nm. Assume that it has adielectric constant of 2.5 (silicone oil) find thecapacitance of 1m2 of membrane.

12

9

2

2.5(8.8 10 )(1)2.0 10

1.1 10

xC

xC x F

!

!

!

=

=

High Capacitance

Compare to 3.7x10-6 F for1m2 the Teflon capacitor.

Biological membranes –Capacitance

oAAC

d d!""

= =

lowdielectric

high dielectric

A nerve cell has a potential across it of 60 mV. Find the density of charges on the membrane that can give rise to this potential

ΔV=60 mV

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

37

960 10 3 10 /2 10

V xE x V m

d x

!

!

"= = =

This is close to the dielectric strength

12 7

4 2

2.5(8.8 10 )(3 10 )

6.6 10 /

o o

o

qE

A

E x x

x C m

!"# "#

! "#

!

$

$

= =

= =

=

This corresponds to an ion density of 4.1x10-3 ions /nm2

or a distance between ions of about 16 nm. A small numberof excess charges.