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Advanced Physics

CapacitanCapacitance ce Chapter 25 – Problems 1, 3, 8,

(17), 19, (33), 39, 40 & 49.

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a+Q

b-Q

The definition of capacitanceThe 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 QC

V

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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.

Capacitor for use in high-performance audio systems.

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 circuitssymbolsanalysis follow from

conservation of energy conservation of charge

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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

16.7 The parallel-plate capacitor16.7 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

0ACd

0

14ek

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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

2.008.85 105.00 10

3.54 10 3.54

A mC C N md 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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16.8 Combinations of capacitors16.8 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 circuitWe 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

C1C2

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 VQ Q Q Q Q QQC

V V V V V VQ Q Q

C C CV V V

+Q1Q1

C1V=Vab

a

b

+Q2Q2

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.

+Q1Q1

C1V=Vab

a

b

+Q2Q2

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

+Q1Q1

C1

+Q2Q2

C2V=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 1eq

eq

Q CV Q C VV V V V V VV

C Q Q Q Q Q Q

Q Q QV 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.

+Q1Q1

C1

+Q2Q2

C2V=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

2eqC CC FC C

Next, determine the charge

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

Series / Parallel Combination ExampleSeries / Parallel Combination Example

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Energy stored in a capacitorEnergy stored in a capacitor

C = Q / VC = Q / V Q = V * C Q = V * CV = U / Q U = V * QV = U / Q U = V * QU = V * V * C = CVU = V * V * C = CV2 2 This is not quite right This is not quite right do to averaging, reallydo to averaging, really

U = ½ CVU = ½ CV22 = ½ QV = ½ QV

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DielectricsDielectrics

A non-conducting material used in A non-conducting material used in capacitors to increase capacitance.capacitors to increase capacitance.Dielectric constant K = C / CDielectric constant K = C / C00

Some are polarized Some are polarized

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DielectricsDielectrics

C = KC = K00A / xA / x

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