Chapter 26A - Chapter 26A - CapacitanceCapacitance

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

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

able to:able to:• Define Define capacitancecapacitance in terms of charge in terms of charge

and voltage, and calculate the and voltage, and calculate the capacitance for a capacitance for a parallel plate parallel plate capacitorcapacitor given separation and area of given separation and area of the plates.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 stored in capacitors.capacitors.

Maximum Charge on a Maximum Charge on a ConductorConductor

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:

; : Q

C Units Coulombs per voltV

; : Q

C Units Coulombs per voltV

Q, V

Capacitance in FaradsCapacitance in Farads

One 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 farad

V volt

(C); (F)

(V)

Q coulombC farad

V volt

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

40 C

8 V

QC

V

C = 5 FC = 5 F

Capacitance of Spherical Capacitance of Spherical ConductorConductor

+Qr

E and V at surface.

At surface of sphere:At surface of sphere:

2;

kQ kQE V

r r 2

; kQ kQ

E Vr r

0

1

4k

Recall:Recall:

04

kQ QV

r r And:And: Capacitance:Capacitance:

QC

V

04

Q QC

V Q r

04C r 04C r

Capacitance, C

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

r = 0.08 m

Capacitance, C

+Qr

Capacitance: C = 4Capacitance: C = 4rr

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 What charge Q is needed to give a potential of 400 V?is 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

; Q

C 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 of a material is that electric intensity that electric intensity EEmm for which the for which the material becomes a conductor. (Charge material becomes a conductor. (Charge leakage.)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 What is the maximum charge that can be placed on a charge that can be placed on a spherical surface one meter in spherical surface one meter in diameter? (R = 0.50 m)diameter? (R = 0.50 m)

rQ

Em = 3 x 106 N/C

Maximum Q

Air

2

2; m

m

E rkQE Q

r k

2

2

6 2NC

9 NmC

(3 x 10 )(0.50 m)

9 x 10Q

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

m

kQE

r 2

mm

kQE

r

Parallel Plate CapacitanceParallel Plate Capacitance

d

Area A+Q

-Q

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

and Q V

C EV d

For these two parallel plates:

0 0

QE

A

0 0

QE

A

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

0

V QE

d A AndAnd

0

Q AC

V d 0

Q AC

V d

Example 3.Example 3. The plates of a parallel The plates of a parallel plate capacitor have an area of plate capacitor have an area of 0.4 m0.4 m22 and are 3 mm apart in air. and are 3 mm apart in air. What is the capacitance?What is the capacitance?

3 mmd

A

0.4 m2

0

Q AC

V d 0

Q AC

V d

2

2

-12 2CNm

(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 Area

0

AC

d 0

AC

d

d

Changing d

Microphone

QV

C

QV

C

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 tuner tuner 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 less probability for electrical discharge.and 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 plates—to be placed on the plates—greater capacitancegreater capacitance C > C C > Coo..

++++++

------

AirAir

CCoo

EEoo

++++++

------

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

C > CC > Coo

E < EE < Eoo

++++++

------

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

DielectricDielectric

reduced reduced EE

• 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 Often it allows for greater mechanical strength.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 Often it allows for greater mechanical strength.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 for a material is the ratio of the capacitance the ratio of the capacitance CC with this with this material as compared with the material as compared with the capacitance capacitance CCoo in a vacuum. in a vacuum.

Dielectric constant: K = 1 for Air

Dielectric constant: K = 1 for Air

0

CK

C

0

CK

C

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 EK

V E

0 0

0

V EK

V E

The Permittivity of a The Permittivity of a MediumMedium

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

0 0 or or A A

C KC C K Cd d

0 0 or or A A

C KC C K Cd d

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

2

2

-12 C0 0 Nm; 8.85 x 10K

2

2

-12 C0 0 Nm; 8.85 x 10K

Example 4: Example 4: Find the capacitance Find the capacitance CC and the charge and the charge QQ if connected to if connected to 200-V200-V battery. Assume the dielectric constant battery. Assume the dielectric constant is 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 10-12-12 C/Nm C/Nm22

2

2

-12 2CNm

(44.25 x 10 )(0.5 m )

0.002 m

AC

d

C = 11.1 nFC = 11.1 nF

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

Q = CV = (11.1 nF)(200 Q = CV = (11.1 nF)(200 V)V)

Q = 2.22 CQ = 2.22 C

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

44.25 x 1044.25 x 10-12-12 C/Nm C/Nm22

' : Q

Gauss law EA

2

2

-6

-12 2

2.22 x 10 C

(44.25 x 10 )(0.5 m )CNm

E

E = 100 N/CE = 100 N/C

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

2 mmd

A

0.5 m2

200 V

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

0 0

0

; V VC

K VC V K

400 V;

5V V = 80.0

VV = 80.0 V

C = KcC = Kcoo = = 5(6 5(6 F)F)

C = 30 FC = 30 F

VVoo = 400 V = 400 V

Mica, K = 5

Air dielectricAir dielectric

Mica Mica dielectricdielectric

Example 5 (Cont.):Example 5 (Cont.): If the If the 400-V400-V battery is battery is reconnected after insertion of the mica, reconnected after insertion of the mica, what what additionaladditional charge will be added to charge will be added to the plates due to the increased the plates due to the increased CC??

QQ00 = C = C00VV00 = = (6 (6 F)(400 F)(400 V)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 F)(400 V)V)

Q = 12,000 Q = 12,000 CC

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

Q = Q = 9600 9600 CC

Energy of Charged Energy of Charged CapacitorCapacitor

The The potential energypotential energy UU of a of a charged capacitor is equal to charged capacitor is equal to the work (the work (qVqV) required to ) required to charge the capacitor.charge the capacitor.If we consider the average If we consider the average potential difference from 0 to Vpotential difference from 0 to Vf f to be to be V/2V/2::

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

221 1

2 2; ; 2

QU QV U CV U

C

221 1

2 2; ; 2

QU QV U CV U

C

Example 6:Example 6: In Ex-4, we found In Ex-4, we found capacitance to be capacitance to be 11.1 nF11.1 nF, the voltage , the voltage 200 V200 V, and the charge , and the charge 2.22 2.22 CC. Find the . Find the potential energy potential energy UU..

212 (11.1 nF)(200 V)U

U = 222 JU = 222 J

212U CV

212U CV

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

2

12 ;

2

QU QV U

C

2

12 ;

2

QU QV U

C

C = 11.1 nF

200 V

Q = 2.22 C

U = ?

Capacitor Capacitor of of Example Example 55..

Energy Density for Energy Density for CapacitorCapacitor

Energy density uEnergy density u is the energy per unit is the energy per unit volume (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 Energy Density Density u u for for

an E-field:an E-field:

AA dd .

U Uu

Vol Ad

2 201 12 2 ( )

AU CV Ed

d

0Recall and :A

C V Edd

21

02 AdEUu

Ad Ad

Energy Density

u: 2102u E

Summary of FormulasSummary of Formulas

(C); (F)

(V)

Q coulombC farad

V volt

(C); (F)

(V)

Q coulombC farad

V volt

04C r 04C r0

Q AC K

V d 0

Q AC K

V d

0 0

0 0

V ECK

C V E

0 0

0 0

V ECK

C V E

221 1

2 2; ; 2

QU QV U CV U

C

221 1

2 2; ; 2

QU QV U CV U

C

2102u E 21

02u E

CONCLUSION: Chapter 25CONCLUSION: Chapter 25CapacitanceCapacitance