Chapter 26A Chapter 26A -- CapacitanceCapacitanceA PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

Objectives: Objectives: After completing this After completing this module, you should be able to:module, you should be able to:

•• Define Define capacitancecapacitance in terms of charge and in terms of charge and voltage, and calculate the capacitance for a voltage, and calculate the capacitance for a parallel plate capacitorparallel plate capacitor given separation and given separation and area of the plates.area of the plates.

•• Define Define dielectric constantdielectric constant and apply to and apply to calculations of voltage, electric field calculations of voltage, electric field intensity, and capacitance.intensity, and capacitance.

•• Find the Find the potential energypotential energy stored in capacitors.stored in capacitors.

Maximum Charge on a ConductorMaximum Charge on a Conductor

EarthBattery Conductor

- - - - ---- -- - - - -e-e-

A A batterybattery establishes a difference of potential that can establishes a difference of potential that can pump electrons pump electrons ee-- from a from a groundground (earth) to a conductor(earth) to a conductor

There is a limit to the amount of charge that a conductor can hold without leaking to the air. There is a certain capacity for holding charge.

There is a limit to the amount of charge that a conductor can hold without leaking to the air. There is a certain capacity for holding charge.

CapacitanceCapacitance

The capacitance C of a conductor is defined as the ratio of the charge Q on the conductor to the potential V produced.

The capacitance C of a conductor is defined as the ratio of the charge Q on the conductor to the potential V produced.

EarthBattery Conductor

- - - - ---- -- - - - -e-e-

Capacitance: ; : QC Units Coulombs per voltV

Q, V

Capacitance in FaradsCapacitance in FaradsOne One farad (F)farad (F) is the capacitance is the capacitance CC of a conductor that of a conductor that holds one coulomb of charge for each volt of potential.holds one coulomb of charge for each volt of potential.

(C); (F)(V)

Q coulombC faradV volt

Example:Example: When 40 When 40 CC of charge are placed on a conof charge are placed on a con-- ductorductor, the potential is 8 V. What is the capacitance?, the potential is 8 V. What is the capacitance?

40 C8 V

QCV

C = 5 FC = 5 F

Capacitance of Spherical ConductorCapacitance of Spherical Conductor

+Qr

E and V at surface.

At surface of sphere:At surface of sphere:

2 ; kQ kQE Vr r

0

14

k

Recall:Recall:

04kQ QVr r

And:And: Capacitance:Capacitance: QCV

04Q QCV Q r

04C r

Capacitance, C

Example 1:Example 1: What is the capacitance of What is the capacitance of a metal sphere of radius 8 cm?a metal sphere of radius 8 cm?

r = 0.08 m

Capacitance, C

+Qr

Capacitance: C = 4Capacitance: C = 4

rr

2-12 C

N m4 (8.85 x 10 )(0.08 m)C

C = 8.90 x 10-12 FC = 8.90 x 10-12 F

Note: The capacitance depends only on physical para- meters (the radius r) and is not determined by either charge or potential. This is true for all capacitors.

Note: The capacitance depends only on physical para- meters (the radius r) and is not determined by either charge or potential. This is true for all capacitors.

(8.90 pF)(400 V)Q

Q = 3.56 nCQ = 3.56 nCTotal Charge on Conductor:Total Charge on Conductor:

Example 1 (Cont.):Example 1 (Cont.): What charge Q is What charge Q is needed to give a potential of 400 V?needed to give a potential of 400 V?

r = 0.08 m

Capacitance, C

+Qr

C = 8.90 x 10-12 FC = 8.90 x 10-12 F

; QC Q CVV

Note: The farad (F) and the coulomb (C) are extremely large units for static electricity. The SI prefixes micro , nano n, and pico p are often used.

Note: The farad (F) and the coulomb (C) are extremely large units for static electricity. The SI prefixes micro , nano n, and pico p are often used.

Dielectric StrengthDielectric StrengthThe The dielectric strengthdielectric strength of a material is that of a material is that electric intensity electric intensity EEmm for which the material for which the material becomes a conductor. (Charge leakage.)becomes a conductor. (Charge leakage.)

rQ

Dielectric

EEmm varies considerably with varies considerably with physical and environmental physical and environmental conditions such as pressure, conditions such as pressure, humidity, and surfaces. humidity, and surfaces.

For air: Em = 3 x 106 N/C for spherical surfaces and as low as 0.8 x 106 N/C for sharp points. For air: Em = 3 x 106 N/C for spherical surfaces and as low as 0.8 x 106 N/C for sharp points.

Example 2:Example 2: What is the maximum charge What is the maximum charge that can be placed on a spherical surface that can be placed on a spherical surface one meter in diameter? (R = 0.50 m)one meter in diameter? (R = 0.50 m)

rQ

Em = 3 x 106 N/C

Maximum Q

Air

2

2 ; mm

E rkQE Qr k

22

6 2NC

9 NmC

(3 x 10 )(0.50 m)9 x 10

Q

Maximum charge in air:Maximum charge in air: Qm = 83.3 CQm = 83.3 C

This illustrates the large size of the coulomb as a unit of charge in electrostatic applications. This illustrates the large size of the coulomb as a unit of charge in electrostatic applications.

Capacitance and ShapesCapacitance and ShapesThe charge density on a surface is significantly The charge density on a surface is significantly affected by the affected by the curvaturecurvature. The density of charge . The density of charge is greatest where the curvature is greatest.is greatest where the curvature is greatest.

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

+++++++++

++

+++

+

Leakage (called corona discharge) often occurs at sharp points where curvature r is greatest. Leakage (called corona discharge) often occurs at sharp points where curvature r is greatest.

2m

mkQEr

Parallel Plate CapacitanceParallel Plate Capacitance

d

Area A+Q

-Q

You will recall from GaussYou will recall from Gauss’’ law that law that EE is also:is also:

and Q VC EV d

For these two parallel plates:

0 0

QEA

QQ is charge on either is charge on either plate. plate. AA is area of plate.is area of plate.

0

V QEd A

AndAnd 0Q ACV d

Example 3.Example 3. The plates of a parallel The plates of a parallel plate capacitor have an area of 0.4 mplate capacitor have an area of 0.4 m22

and are 3 mm apart in air. What is the and are 3 mm apart in air. What is the capacitance?capacitance?

3 mmd

A

0.4 m2

0Q ACV d

2

2-12 2C

Nm(8.85 x 10 )(0.4 m )

(0.003 m)C

C = 1.18 nFC = 1.18 nF

Applications of CapacitorsApplications of Capacitors

+++++++

-

--

---- A

Variable Capacitor

Changing Area0

ACd

d

Changing d

Microphone

QVC

A A microphonemicrophone converts sound waves into an converts sound waves into an electrical signal (varying voltage) by changing electrical signal (varying voltage) by changing dd..

TheThe tunertuner in a radio is a in a radio is a variable capacitorvariable capacitor. The changing . The changing area area A A alters capacitance until desired signal is obtained.alters capacitance until desired signal is obtained.

Dielectric MaterialsDielectric MaterialsMost capacitors have a Most capacitors have a dielectric materialdielectric material between between their plates to provide greater their plates to provide greater dielectric strengthdielectric strength and and less probability for electrical discharge.less probability for electrical discharge.

The separation of dielectric charge allows more charge The separation of dielectric charge allows more charge to be placed on the platesto be placed on the plates——greater capacitancegreater capacitance C > CC > Coo ..

++++++

------

AirAir

CCoo

EEoo++++++

------

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

C > CC > Coo

E < E < EEoo++++++

------

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

DielectricDielectric

reduced reduced EE

• Smaller plate separation without contact.

• Increases capacitance of a capacitor.

• Higher voltages can be used without breakdown.

• Often it allows for greater mechanical strength.

•• Smaller plate separation without contact.Smaller plate separation without contact.

•• Increases capacitance of a capacitor.Increases capacitance of a capacitor.

•• Higher voltages can be used without Higher voltages can be used without breakdown.breakdown.

•• Often it allows for greater mechanical strength.Often it allows for greater mechanical strength.

Advantages of DielectricsAdvantages of Dielectrics

Insertion of DielectricInsertion of Dielectric

++++++

Co Vo Eo

+Q

-Q

++

+Q

-Q

DielectricAir

Permittivity increases.> o

Capacitance increases.C > Co

Voltage decreases.V < Vo

Field decreases.E < Eo

Insertion of a dielectric

Same QQ = Qo

C V E

Dielectric Constant, KDielectric Constant, KThe The dielectric constant Kdielectric constant K for a material is the for a material is the ratio of the capacitance ratio of the capacitance CC with this material as with this material as compared with the capacitance compared with the capacitance CCoo in a vacuum.in a vacuum.

Dielectric constant: K = 1 for Air

Dielectric constant: K = 1 for Air0

CKC

K can also be given in terms of voltage K can also be given in terms of voltage VV, , electric field intensity electric field intensity EE, or permittivity , or permittivity ::

0 0

0

V EKV E

The Permittivity of a MediumThe Permittivity of a MediumThe capacitance of a parallel plate capacitor The capacitance of a parallel plate capacitor with a dielectric can be found from:with a dielectric can be found from:

0 0 or or A AC KC C K Cd d

The constant The constant

is the is the permittivitypermittivity of the medium of the medium which relates to the density of field lines.which relates to the density of field lines.

2

2-12 C

0 0 Nm; 8.85 x 10K

Example 4: Example 4: Find the capacitance Find the capacitance CC and the and the charge charge QQ if connected to if connected to 200200--VV battery. battery. Assume the dielectric constant is Assume the dielectric constant is K = 5.0K = 5.0. .

2 mmd

A

0.5 m2

5(8.85 x 10-12C/Nm2)

44.25 x 1044.25 x 10--1212 C/NmC/Nm22

2

2-12 2C

Nm(44.25 x 10 )(0.5 m )

0.002 mACd

C = 11.1 nFC = 11.1 nF

Q if connected to V = 200 V?Q if connected to V = 200 V?

Q = CV = (11.1 nF)(200 V)Q = CV = (11.1 nF)(200 V) Q = 2.22 CQ = 2.22 C

Example 4 (Cont.): Example 4 (Cont.): Find the field Find the field E E between between the plates. Recall the plates. Recall Q = 2.22 Q = 2.22 CC; ; VV = 200 V= 200 V. .

44.25 x 1044.25 x 10--1212 C/NmC/Nm22

' : QGauss law EA

2

2

-6

-12 2

2.22 x 10 C(44.25 x 10 )(0.5 m )C

Nm

E

E = 100 N/CE = 100 N/C

Since Since V = 200 VV = 200 V, the same result is found , the same result is found if if E = V/dE = V/d is used to find the field.is used to find the field.

2 mmd

A

0.5 m2

200 V

Example 5:Example 5: A capacitor has a capacitance of A capacitor has a capacitance of 66FF with air as the dielectric. A battery charges the with air as the dielectric. A battery charges the capacitor to capacitor to 400 V400 V and is then disconnected. What and is then disconnected. What is the new voltage if a sheet of mica (is the new voltage if a sheet of mica (K = 5K = 5) is ) is inserted? What is new capacitance inserted? What is new capacitance C C ??

0 0

0

; V VCK VC V K

400 V ;5

V V = 80.0 VV = 80.0 V

C = C = KcKcoo = = 5(6 5(6 F)F)

C = 30 FC = 30 F

VVoo = 400 V= 400 V

Mica, K = 5

Air dielectricAir dielectric

Mica dielectricMica dielectric

Example 5 (Cont.):Example 5 (Cont.): If the If the 400400--VV battery is battery is reconnected after insertion of the mica, what reconnected after insertion of the mica, what additionaladditional charge will be added to the plates charge will be added to the plates due to the increased due to the increased CC??

QQ00 = C= C00 VV00 = = (6 (6 F)(400 V)F)(400 V)

Q = 9.60 mCQ = 9.60 mC

VVoo = 400 V= 400 V

Mica, K = 5

Air CAir Coo = 6 = 6 F

Mica C = 30 Mica C = 30 F

QQ00 = 2400 = 2400 CC

Q = CV = Q = CV = (30 (30 F)(400 V)F)(400 V)

Q = 12,000 Q = 12,000 CC

Q = 12,000 Q = 12,000 CC –– 2400 2400 CC

Q = Q = 9600 9600 CC

Energy of Charged CapacitorEnergy of Charged CapacitorThe The potential energypotential energy UU of a charged of a charged capacitor is equal to the work (capacitor is equal to the work (qVqV) ) required to charge the capacitor.required to charge the capacitor.

If we consider the average potential If we consider the average potential difference from 0 to difference from 0 to VVff to be to be V/2V/2::

Work = Q(V/2) = ½QVWork = Q(V/2) = ½QV

221 1

2 2; ; 2QU QV U CV UC

Example 6:Example 6: In ExIn Ex--4, we found capacitance to 4, we found capacitance to be be 11.1 11.1 nFnF, the voltage , the voltage 200 V200 V, and the , and the charge charge 2.22 2.22 CC. Find the potential energy . Find the potential energy UU..

212 (11.1 nF)(200 V)U

U = 222 JU = 222 J

212U CV

Verify your answer from the Verify your answer from the other formulas for P.E.other formulas for P.E.

21

2 ; 2QU QV UC

C = 11.1 nF

200 V

Q = 2.22 C

U = ?

Capacitor of Capacitor of Example 5Example 5..

Energy Density for CapacitorEnergy Density for CapacitorEnergy density uEnergy density u is the energy per unit volume is the energy per unit volume ((J/mJ/m33). For a capacitor of area ). For a capacitor of area AA and separation and separation dd, the energy density , the energy density u u is found as follows:is found as follows:

Energy Density Energy Density u u for an Efor an E--field:field: AA dd .

U UuVol Ad

2 201 12 2 ( )AU CV Ed

d

0Recall and :AC V Edd

2102 AdEUu

Ad Ad

Energy Density u:21

02u E

Summary of FormulasSummary of Formulas(C); (F)

(V)Q coulombC faradV volt

04C r0Q AC KV d

0 0

0 0

V ECKC V E

221 1

2 2; ; 2QU QV U CV UC

2102u E

CONCLUSION: Chapter 25CONCLUSION: Chapter 25 CapacitanceCapacitance