Lecture 8Lecture 8

Capacitance and capacitors

CapacitorsCapacitors

Capacitors are devices that store energy in an electric Capacitors are devices that store energy in an electric field.field.Capacitors are used in many every-day applicationsCapacitors are used in many every-day applications Heart defibrillatorsHeart defibrillators Camera flash unitsCamera flash units

Capacitors are an essential part of electronics.Capacitors are an essential part of electronics. Capacitors can be micro-sized on computer chips or Capacitors can be micro-sized on computer chips or

super-sized for high power circuits such as FM radio super-sized for high power circuits such as FM radio transmitters.transmitters.

Definition of CapacitanceDefinition of Capacitance

The definition of capacitance isThe definition of capacitance is

The units of capacitance are coulombs per volt.The units of capacitance are coulombs per volt.

The unit of capacitance has been given the name farad The unit of capacitance has been given the name farad (abbreviated F) named after British physicist Michael (abbreviated F) named after British physicist Michael Faraday (1791 - 1867)Faraday (1791 - 1867)

A farad is a very large capacitanceA farad is a very large capacitance Typically we deal with Typically we deal with F (10F (10-6-6 F), nF (10 F), nF (10-9-9 F), F),

or pF (10or pF (10-12-12 F) F)

Vq

CVq

C

V 1

C 1 F 1

The proportionality constant between the charge q and the electric potential difference V is the capacitance C.

CVq

A

A

+Q

-Q

d

The parallel-plate capacitorThe parallel-plate capacitor

TheThe capacitance capacitance of a device of a device depends on the area of the depends on the area of the plates and the distance plates and the distance between the platesbetween the plates

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

constant (constant (permittivity of free permittivity of free space)space),,

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

0

AC

d

Example: 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:

Combinations of capacitorsCombinations 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

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

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)

Example: 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

Series combinationSeries 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

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)

Example: 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

1

2U QV

Energy stored in a charged capacitorEnergy stored in a charged capacitor

Consider a battery connected to a capacitor

A battery must do work to move electrons from one plate to the other. The work done to move a small charge q across a voltage V is W = V q.

As the charge increases, V increases so the work to bring q increases. Using calculus we find that the energy (U) stored on a capacitor is given by:

V

V

q

221

2 2

QCV

C

Q

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

C1

C2

V

C3

Example: 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=?

Q QQ Q

V0 V

Capacitors with dielectricsCapacitors 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

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

The capacitance is limited from above by the electric The capacitance is limited from above by the electric discharge that can occur through the dielectric material discharge that can occur through the dielectric material separating the platesseparating the plates

In other words, there exists a maximum of the electric In other words, there exists a maximum of the electric field, sometimes called field, sometimes called dielectric strengthdielectric strength, that can be , that can be produced in the dielectric before it breaks downproduced in the dielectric before it breaks down

0

AC

d

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

For a more complete list, see Table 16.1

Example: 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 3

2.008.85 10 18

1.00 10

A mC C N m nF

d m

9 518 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

0.33 18 6V

C C C nF nFV

How does an insulating dielectric material reduce electric fields by producing effective surface charge densities?

Reorientation of polar molecules

Induced polarization of non-polar molecules

Dielectric Breakdown: breaking of molecular bonds/ionization of molecules.

(1) Two capacitors, C1=2mF and C2=16mF, are connected in parallel.  What is the value of the equivalent capacitance of the combination?

(2) Calculate the equivalent capacitance of the two capacitors in the previous exercise if they are connected in series.

(3) A 100pF capacitor is charged to a potential difference of 50V, the charging battery then being disconnected.  The capacitor is then connected in parallel with a second (initially uncharged) capacitor.  If the measured potential difference drops to 35V, what is the capacitance of this second capacitor?

(4) A parallel-plate capacitor has circular plates of 8.0cm radius and 1.0mm separation.  What charge will appear on the plates if a potential difference of 100V is applied?

(5) In figure 6.18 the battery supplies 12V.  (a) Find the charge on each capacitor when switch S1 is closed, and (b) when later switch S2 is also

closed.  Assume C1=1F, C2=2F, C3=3F, and C4=4F.

(14) A 16pF parallel-plate capacitor is charged by a 10V battery.  If each plate of the capacitor has an area of 5cm2, what is the energy stored in the capacitor?  What is the energy density (energy per unit volume) in the electric field of the capacitor if the plates are separated by air?

(15) The energy density in a parallel-plate capacitor is given as 2.1 ´l0-9J/m3.  What is the value of the electric field in the region between the plates?

(15) The energy density in a parallel-plate capacitor is given as 2.1 ´l0-9J/m3.  What is the value of the electric field in the region between the plates?(a) Determine the equivalent capacitance for the capacitors shown in figure 6.20.  (b) If they are connected to 12V battery, calculate the potential difference across each capacitor and the charge on each capacitor                                                                           

(7) A 6.0mF capacitor is connected in series with a 4.0mF capacitor and a potential difference of 200 V is applied across the pair. (a) What is the charge on each capacitor? (b) What is the potential difference across each capacitor?

(8) Repeat the previous problem for the same two capacitors connected in parallel.