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ELECTRICITY AND MAGNETISMELECTRICITY AND MAGNETISMPhy 220Phy 220

Chapter 4: Chapter 4: CapacitorsCapacitors

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a+Q

b-Q

4.1 The definition of capacitance4.1 The definition of capacitance

Capacitor: two conductors (separated by an insulator) usually oppositely charged

The capacitance, C, of a capacitor is defined as a ratio of the magnitude of a charge on either conductor to the magnitude of the potential difference between the conductors Q

CV

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1. A capacitor is basically two parallel conducting plates with insulating material in between. The capacitor doesn’t have to look like metal plates.

2. When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates.

3. Capacitors in circuits symbols:

4. Analysis follow from conservation of energy and conservation of charge

Capacitor for use in high-performance audio systems.

+ -

- -

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Units of capacitanceUnits of capacitance

The unit of C is the farad (F), but most capacitors have values of C ranging from picofarads to microfarads (pF to F).

Recall, micro 10-6, nano 10-9, pico 10-12

If the external potential is disconnected, charges remain on the plates, so capacitors are good for storing charge (and energy).

1 1F C V

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A

A

+Q

-Q

d

4.2 The parallel-plate capacitor4.2 The parallel-plate capacitor

The capacitance of a device The capacitance of a device depends on the geometric depends on the geometric arrangement of the conductorsarrangement of the conductors

where where AA is the area of one of is the area of one of the plates, the plates, dd is the separation, is the separation, 00 is a constant called the is a constant called the

permittivity of free spacepermittivity of free space,,

00= 8.85= 8.851010-12 -12 CC22/N·m/N·m22

0

AC

d

0

1

4ek

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Proof:Proof:According to example 2.4.3,chapter 2, the value of the electric field between the According to example 2.4.3,chapter 2, the value of the electric field between the plates isplates is

Because the field between the plates is uniform, the magnitude of the potential Because the field between the plates is uniform, the magnitude of the potential difference between the plates equals difference between the plates equals Ed, Ed, therefore,

So we find that the capacitance is

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A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine

the capacitancethe charge on each plate

Problem: parallel-plate capacitorProblem: parallel-plate capacitor

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A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine

the capacitancethe charge on each plate

Given:

V=10,000 VA = 2.00 m2

d = 5.00 mm

Find:

C=?Q=?

Solution:

Since we are dealing with the parallel-plate capacitor, the capacitance can be found as

2

12 2 20 3

9

2.008.85 10

5.00 10

3.54 10 3.54

A mC C N m

d m

F nF

9 53.54 10 10000 3.54 10Q C V F V C

Once the capacitance is known, the charge can be found from the definition of a capacitance via charge and potential difference:

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4.3 Combinations of capacitors4.3 Combinations of capacitors

It is very often that more than one capacitor is used in an It is very often that more than one capacitor is used in an electric circuitelectric circuit

We would have to learn how to compute the equivalent We would have to learn how to compute the equivalent capacitance of certain combinations of capacitorscapacitance of certain combinations of capacitors

C1

C2

C3

C5C1

C2

C3

C4

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1 1 1 2 2 2

1 2 1 2 1 2

1 2

1 21 2

1 2

eq

eq

Q CV Q C V

Q Q Q Q Q QQC

V V V V V V

Q Q QC C C

V V V

+Q1

Q1

C1V=Vab

a

b

+Q2

Q2

C2

a. Parallel combinationa. Parallel combination

1 2V V V

Connecting a battery to the parallel combination of capacitors is equivalent to introducing the same potential difference for both capacitors,

1 2Q Q Q

A total charge transferred to the system from the battery is the sum of charges of the two capacitors,

By definition,

Thus, Ceq would be

1 2eqC C C

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Parallel combination: notesParallel combination: notes

Analogous formula is true for any number of capacitors,Analogous formula is true for any number of capacitors,

It follows that the equivalent capacitance of a parallel It follows that the equivalent capacitance of a parallel combination of capacitors is greater than any of the combination of capacitors is greater than any of the individual capacitorsindividual capacitors

1 2 3 ...eqC C C C (parallel combination)

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A 3 F capacitor and a 6 F capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited.

Problem: parallel combination of capacitorsProblem: parallel combination of capacitors

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A 3 F capacitor and a 6 F capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited.

+Q1

Q1

C1V=Vab

a

b

+Q2

Q2

C2Given:

V = 18 VC1= 3 FC2= 6 F

Find:

Ceq=?Q=?

First determine equivalent capacitance of C1 and C2:

12 1 2 9C C C F

Next, determine the charge

6 49 10 18 1.6 10Q C V F V C

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b. Series combinationb. Series combination

1 2V V V

Connecting a battery to the serial combination of capacitors is equivalent to introducing the same charge for both capacitors,

1 2Q Q Q

A voltage induced in the system from the battery is the sum of potential differences across the individual capacitors,

By definition,

Thus, Ceq would be

+Q1

Q1C1

+Q2

Q2C2

V=Vab

a

c

b

1 1 1 2 2 2

1 2 1 2 1 2

1 2

1 2

1 21 2

1

1 1 1

eq

eq

Q CV Q C V

V V V V V VV

C Q Q Q Q Q Q

Q Q Q

V V VC C C

1 2

1 1 1

eqC C C

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Series combination: notesSeries combination: notes

Analogous formula is true for any number of capacitors,Analogous formula is true for any number of capacitors,

It follows that the equivalent capacitance of a series It follows that the equivalent capacitance of a series combination of capacitors is always less than any of the combination of capacitors is always less than any of the individual capacitance in the combinationindividual capacitance in the combination

1 2 3

1 1 1 1...

eqC C C C (series combination)

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A 3 F capacitor and a 6 F capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance.

Problem: series combination of capacitorsProblem: series combination of capacitors

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A 3 F capacitor and a 6 F capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance and total charge deposited.

+Q1

Q1C1

+Q2

Q2C2

V=Vab

a

c

b

Given:

V = 18 VC1= 3 FC2= 6 F

Find:

Ceq=?Q=?

First determine equivalent capacitance of C1 and C2:

1 2

1 2

2eq

C CC F

C C

Next, determine the charge

6 52 10 18 3.6 10Q C V F V C

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4.4 Energy stored in a charged capacitor4.4 Energy stored in a charged capacitor

The energy (U) stored on a capacitor is given by:The energy (U) stored on a capacitor is given by:

1

2U QV

221

2 2

QCV

C

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Find electric field energy density (energy per unit volume) in a parallel-plate capacitor

2

0

20

20

1

2

/ energy density

1= ( ) /( )

21

2

volume A

U CV

AC V Ed

du U volume

AEd d

E

d

Ad

u

Example: electric field energy in parallel-plate Example: electric field energy in parallel-plate capacitorcapacitor

Recall

Thus,

and so, the energy density is

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C1

C2

V

C3

In the circuit shown V = 48V, C1 = 9F, C2 = 4F and C3 = 8F.(a) determine the equivalent capacitance of the circuit,(b) determine the energy stored in the combination by

calculating the energy stored in the equivalent capacitance.

Example: stored energyExample: stored energy

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C1

C2

V

C3

In the circuit shown V = 48V, C1 = 9F, C2 = 4F and C3 = 8F.(a) determine the equivalent capacitance of the circuit,(b) determine the energy stored in the combination by calculating the energy stored in the equivalent capacitance,

First determine equivalent capacitance of C2 and C3:

23 2 3 12C C C F

22 6 31 15.14 10 48 5.9 10

2 2U CV F V J

The energy stored in the capacitor C123 is then

Next, determine equivalent capacitance of the circuit by noting that C1 and C23 are connected in series

1 23123

1 23

5.14eq

C CC C F

C C

Given:

V = 48 VC1= 9 FC2= 4 FC3= 8 F

Find:

Ceq=?U=?

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Q QQ Q

V0 V

4.5 Capacitors with dielectrics4.5 Capacitors with dielectrics

A dielectrics is an insulating material (rubber, glass, etc.)Consider an insolated, charged capacitor

Notice that the potential difference decreases (k = V0/V)Since charge stayed the same (Q=Q0) → capacitance increases

dielectric constant: k = C/C0

Dielectric constant is a material property

0 0 00

0 0

Q Q QC C

V V V

Insert a dielectric

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Capacitors with dielectrics - notesCapacitors with dielectrics - notes

Capacitance is multiplied by a factor Capacitance is multiplied by a factor kk when the when the dielectric fills the region between the plates dielectric fills the region between the plates completelycompletely

E.g., for a parallel-plate capacitorE.g., for a parallel-plate capacitor

There exists a maximum of the electric field, sometimes There exists a maximum of the electric field, sometimes called called dielectric strengthdielectric strength, that can be produced in the , that can be produced in the dielectric before it breaks downdielectric before it breaks down

0

AC

d

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Dielectric constants and dielectric Dielectric constants and dielectric strengths of various materials at room strengths of various materials at room temperaturetemperatureMaterialMaterial Dielectric Dielectric

constant, kconstant, kDielectric Dielectric strength (V/m)strength (V/m)

VacuumVacuum 1.001.00 -- --

AirAir 1.000591.00059 3 3 101066

WaterWater 8080 ----

Fused quartzFused quartz 3.783.78 9 9 101066

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Take a parallel plate capacitor whose plates have an area of 2000 cm2 and are separated by a distance of 1cm. The capacitor is charged to an initial voltage of 3 kV and then disconnected from the charging source. An insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1 kV. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material.

ExampleExample

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Take a parallel plate capacitor whose plates have an area of 2 m2 and are separated by a distance of 1cm. The capacitor is charged to an initial voltage of 3 kV and then disconnected from the charging source. An insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1 kV. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material.

Given:

V1=3,000 VV2=1,000 VA = 2.00 m2

d = 0.01 m

Find:

C=?C0=?Q=?k=?

Since we are dealing with the parallel-plate capacitor, the original capacitance can be found as

2

12 2 20 0 3

2.008.85 10 18

1.00 10

A mC C N m nF

d m

9 50 1 18 10 3000 5.4 10Q C V F V C

The charge on the capacitor can be found to be

The dielectric constant and the new capacitance are

10 0

2

3 18 54V

C C C nF nFV