topic outline definition of capacitance calculating capacitance combinations of capacitors energy...

Topic Outline

Definition of Capacitance

Calculating Capacitance

Combinations of Capacitors

Energy Stored in a Charged Capacitor

Capacitors with Dielectrics

An Atomic Description of a Dielectric

Capacitance and Dielectric

Capacitors: Device that store electric charge

A capacitor consists of two conductors separated by an insulator.

Capacitance: Depends on its geometry and on the material, called a dielectric, that separates the conductors.


Pictures from Serway & Beichner

A capacitor consists of two conductors (known as plates) carrying charges of equal magnitude but opposite sign.

A potential difference V exists between the conductors due to the presence of the charges.

What is the capacity of the device for storing charge at particular value of V?



Experiments show the quantity of electric charge Q on a capacitor is linearly proportional to the potential difference between the conductors, that is Q ~ V. Or we write Q = C VThe capacitance C of a capacitor is the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between them:

C = Q

VSI Unit: farad (F), 1F = 1 C/V

Typical device have capacitances ranging from microfarad to picofarad.

Parallel - Plate Capacitors


A parallel-plate capacitor consists of two parallel conducting plates, each of area A, separated by a distance d. When the capacitor is charged, the plates carry equal amounts of charge. One plate carries positive charge, and the other carries negative charge. The plates are charged by connection to a

battery.

Parallel-Plate Capacitors


Two parallel metallic plates of equal area A separated by a distance d as shown.

One plate carries a charge Q and the other carries a charge –Q. And surface charge density of each plate is = Q/A.

A

d

If plates are large, then charges can distribute themselves over a substantial area, and the amount of charge that can be stored on a plate for a given potential diff increases as A is increased.

Thus we expect C to be proportional to A

C ~ A

Variation with A



Variation with d

A

d

Potential difference V constant across, E field increases as d decreases.

Imagine d decreases and consider situation before any charges have had a chance to move in response to this change.

Because no charge move E the same but over a shorter distance.

V = Ed means that V decreases.



Variation with d (cont’d)

A

d

The difference between this new capacitor voltage and the terminal voltage of the battery now exists as a potential difference across the wires connecting the battery to the capacitor.

A E field result in the wires that drives more charge onto the plates, increasing the potential diff. V until it matches that of the battery. potential diff. Across wire = 0 flow of charges stop.

More charges has accumulated at the capacitor as a result.

We have d decrease, Q increases. Similarly d increases Q decreases.

Capacitance inversely proportional to d.

C ~ 1/d



(a) The electric field between the plates of a parallel-plate capacitor is uniform near the center but nonuniform near the edges.

(b) Electric field pattern of two oppositely charged conducting parallel plates.



Assume electric field uniform between the plates, we have

(see lecture on Gauss’s Law)

E = Q

o

=

oA

V = E d = Qd

oA

C = Q

V Q

QdoA=

C = oA

d

(As we have argued before)

Example


SB 26 Q8 A 1-megabit computer memory chip contains many 60.0-fF capacitors. Each capacitor has a plate area of 21.0 x 10-12 m2. Determine the plate separation of such a capacitor (assume a parallel-plate configuration). The characteristic atomic diameter is 10-10 m =0.100nm. Express the plate separation in nanometers.

F 100.60 150 d

AC

15

1212

0

100.60100.211085.81

CA

d

d 3.1010 9 m3.10 nm

Cylindrical Capacitors


A solid cylindrical conductor of radius a and charge Q is coaxial with a cylindrical shell of negligible thickness, radius b > a, and charge –Q. Find the capacitance of this cylindrical capacitor if its length is L.

L



Assume that L is >> a and b, neglect the end effects.

E is perpendicular to the long axis of the cylinders and is confined to the region between them.

Potential difference between the two cylinders is given by

L

Vb-Va = - E . ds

b

a

Where E is the E field in the region a < r < b.

Our discussion on Gauss’s Law Er = 2k/r where is the linear charge density of the cylinder.

Note that the charge on outer cylinders does not contribute to E field inside it.



L

Vb-Va = - Er dr

b

a

= - 2k

b

a

drr

= - 2kln( )ba

Using = Q/L , we have

C = = Q

Vln( )

ba

2kQ

L

Q

C =ln( )

ba

2k

L



LC =ln( )

ba

2k

L

What is the capacitance per unit length ?

Example

Co-axial Cable. Read the cable, typically 50 pF/m. Is this sensible ?

Typically a 0.5 mm, b 1.5 mm

50pF/m)3ln(1099.82

1/

9

LC

Cylindrical Capacitors vs Parallel Plate Capacitors


What are the advantages of cylindrical capacitor over that of a parallel plate capacitor?

L

The Spherical Capacitors


A spherical capacitor consists of a spherical conducting shell of radius b and charge –Q concentric with a smaller conducting sphere of radius a and charge Q. Find the capacitance of this device.

The Spherical Capacitors


Gauss’s law E field outside a spherically symmetric charge distribution is radial and given by Er = kQ/r2.

The potential difference between the spheres is

Vb-Va = - Er dr

b

a

= - kQ

b

a

drr2

= kQ1b

1a

C =Vb-Va

Q=

ab

k (b-a)

What happens to the capacitance of this system when the radius of the outer sphere approaches infinity?

Example


SB 26 Q12 A 20.0 F spherical capacitor is composed of two metallic spheres, one having a radius twice as large as the other. If the region between the spheres is a vacuum, determine the volume of this region.

Let the radii be b and a with b = 2a. Put charge Q on the inner conductor and –Q on the outer. Electric field exists only in the volume between them. The potential of the inner sphere is Va=kQ/a; that of the outer is Vb=kQ/b. Then

ababQ

bQk

aQk

VV ba

04==

abab

VVQ

Cba

04==

aa

aC 0

2

0 8=24

=

08=

C

a

The intervening volume is 3

0

33

3

3

43

3

43

3

43

3

4

877

C

aab

3

0

2

3

384

7

C

V = 2.13 x 1016 m3

Example


SB 26 Q16 What is the capacitance of the Earth ?

Think of Earth spherical conductor and the outer conductor of the “spherical capacitor” may be considered as a conducting sphere at infinity where V approaches zero.

m 1037.6mNC 1085.84

46212

0

RC e

C= 7.08 x 10-4 F

A large capacitor !



Parallel Combination

The individual potential differences across capacitors connected in parallel are all the same and are equal to the potential difference applied across the combination.



When the capacitors are first connected, electrons transfer between wires and plates. Leave left plates positively charged and right plates negatively charged.

Energy source for this charge transfer is internal chemical energy stored in the battery.

Flow of charges ceases when the voltage across the capacitors is equal to that across the battery terminals.

Capacitors reach their maximum charge when the flow of charges ceases.




Let the maximum charges on the two capacitors Q1 and Q2.

Total charge Q stored by two capacitors is Q = Q1+Q2.

Voltage across are the same Q1=C1V, Q2=C2V

Define an equivalent capacitor having Ceq s.t. Q = CeqV

We have CeqV = C1V + C2V

And hence Ceq = C1 + C2 (for parallel combination)

In general Ceq = C1 + C2+ C3+ ………….. (for parallel combination)




Series Combination

Start with uncharged situation and follow what happen just after a battery is connected to the circuit.

When a battery is connected, electrons transferred out of the left plate of C1 and into the right plate of C2.

As this charge accumulates on the right plate of C2, an equivalent amount of negative charge is forced off the left plate of C2 and this left plate therefore has an excess positive charge. (cont’d)



Series Combination

The negative charge leaving the left plate of C2 travels through the connecting wire and accumulates on the right plate of C1.

As a result, all right plates end up with a charge –Q and all the left plates end up with a charge +Q.

Thus the charges on capacitors connected in series are the same.



Series Combination

Voltage V across battery terminals is split between two capacitors.

V = V1 + V2

Where V1 and V2 are potential diff across capacitors C1 and C2.

Suppose we have equivalent capacitor Ceq = Q/V

For each capacitor, we have V1 = Q/C1 and V2=Q/C2

Q/Ceq = Q/C1 + Q/C2

1/Ceq = 1/C1 + 1/C2 (series combination)

In general 1/Ceq = 1/C1 + 1/C2 + 1/C3 + ….. (series combination)

Example: Equivalent Capacitance


In parallel use C=C1+C2

In series use 1/C=1/C1+1/C2

6.00 F

20.00 F

2.50 F

8.50 F

20.00 F


5.965 F

Example: Equivalent Capacitance





Example: Equivalent Capacitance 26.22


In series use 1/CB=1/C+1/C+1/C

In series use 1/CA=1/C+1/C C

C/2

C/3

In parallel use Ceq=C+C/2+C/3

Energy Stored in a Charged CapacitorSuppose q is the charge on the capacitor at some instant during the charging process.

At the same instance, the potential difference across the capacitor is V=q/C.

The work necessary to transfer an increment of charge dq from the plate carrying charge –q to the plate carrying charge q (which is at the higher electric potential) is

dW = V dq = dqq

C

The total work required to charge the capacitor from q = 0 to some final charge q = Q is

W =

Qdq0

q

C=

1

C

Q

0

q dq =

Q2

2C


Work done in charging the capacitor = electric potential energy U stored in the capacitor.

U =

Q2

2C=

1

2QV =

1

2C (V)2

This result applies to any capacitors, regardless of its geometry.



A plot of potential difference versus charge for a capacitor is a straight line having a slope 1/C. The work required to move charge dq through the potential difference V across the capacitor plates is given by the area of the shaded rectangle. The total work required to charge the capacitor to a final charge Q is the triangular area under the straight line, W = QV/2.

1V = 1 J/C hence the unit for the area is joule J.

Energy Density Stored in a Charged CapacitorFor parallel plate capacitor, V = E d, C = oA/d we have

U =

1

2

oA

d(E d)2 =

1

2(oA d) E 2

Ad = Volume occupied by the E field. This lead to a new quantity known as Energy Density

u = U/Volume = U/Ad

u =

1

2o E 2

Although above equation was derived for parallel-plate capacitor, the expression is generally valid.

Energy Density in any electric field is proportional to the square of the magnitude of the electric field at a given point.

Example


26.33: A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled ?

U = Q2/2C and C = oA/d and d2 = 2 d1 then C2= C1/2 and the energy stored doubles.

Example


26.35: A parallel-plate capacitor has a charge Q and plates of area A. Show that the force exerted on

each plate by the other is F = Q2/2oA. Is this force attractive or repulsive. Does this equation follow from common sense?

F =

W U F dx

AxQ

dxd

CQ

dxd

dxdU

0

22

22

F = Q2/2oA



A dielectric is a nonconducting material, such as rubber, glass, or waxed paper.

When a dielectric is inserted between the plates of a capacitor, the capacitance increases.

If the dielectric completely fills the space between the plates, the capacitance increases by a dimensionless factor , which is called the dielectric constant.

Dielectric constant is a property of a material and varies from one material to another.



A charged capacitor (a) before and (b) after insertion of a dielectric between the plates. The charge on the plates remains unchanged, but the potential difference decreases from Vo to V = Vo/. Thus the capacitance

increases from Co to Co.

Note no battery is involved in this example.



Capacitance increases by the factor when dielectric completely fills the region between the plates.

If the dielectric is introduced while the potential difference is being maintained constant by a battery, the charge increases to a value Q = Qo . The additional charge is supplied by the battery and the capacitance again increases by the factor .

For parallel plate capacitor:

C = A

d

Dielectric Strength


For any given separation d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the dielectric strength (maximum electric field) of the dielectric.

If magnitude of the electric field in the dielectric exceeds the dielectric strength, then the insulating properties break down and the dielectric begins to conduct.

Dielectric Constant and Dielectric Strength of Various Materials at Room Temperature


Material Dielectric Constant Dielectric Strength (V/m)

Air (dry) 1.00059 3 x 106

Bakelite 4.9 24 x 106

Fused quartz 3.78 8 x 106

Neoprene rubber 6.7 12 x 106

Nylon 3.4 14 x 106

Paper 3.7 16 x 106

Polystyrene 2.56 24 x 106

Polyvinyl Chloride 3.4 40 x 106

Porcelain 6 12 x 106

Pyrex Glass 5.6 14 x 106

Silicone Oil 2.5 15 x 106

Strontium Titanate 233 8 x 106

Teflon 2.1 60 x 106

Vacuum 1.00000 -

Water 80 -

Capacitors with Dielectric Material


What are the advantages of dielectric material in a capacitor?

• Increase the capacitance

• Increase the maximum operating voltage

• Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C.

Types of Capacitors


(a) A tubular capacitor, whose plates are separated by paper and then rolled into a cylinder. (b) A high-voltage capacitor consisting of many parallel plates separated by insulating oil. (c) An electrolytic capacitor.

Energy Stored Before and After


A parallel-plate capacitor is charged with a battery to a charge Qo as shown (a). The battery is then removed

and a slab of material that has a dielectric constant is inserted between the plates as shown (b). Find the energy stored in the capacitor before and after the dielectric is inserted.

Energy Stored Before and After


Before

(a)

Uo = Qo

2

2 Co

Charge on the capacitor the same before and after. (Why?)

U = = = Qo

2

2 C

Qo2

2 Co

Uo

Energy reduced, where does the “missing” energy go to?

Dielectric, when inserted, gets pulled into the device. External agent do negative work to keep dielectric from accelerating.

Work = U-Uo

Capacitors with Dielectric


The nonuniform electric field near the edges of a parallel-plate capacitor causes a dielectric to be pulled into the capacitor. Note that the field acts on the induced surface charges on the dielectric, which are nonuniformly distributed.

Induced polarization


This molecule is non-polar since the center of negative charges coincide with the center of positive charges.

The presence of an external E field causes the charges to change their positions and such that the center of negative charges does not coincide with the center of positive charges. We say that a dipole moment has been induced due to the presence of the E field.

This is called induced polarization.

An Atomic Description of Dielectrics


Potential difference Vo between the plates of a capacitor is reduced to Vo/ when a dielectric is introduced.

How is this possible?

Think about the E field, if Eo is the E field without the dielectric, then the field in the presence of E field is E = Eo/. I.e. The field is reduced. How is this possible?



(a) Polar molecules are randomly oriented in the absence of an external electric field.

(b) When an external field is applied (to the right as shown), the molecules partially align with the field. (dielectric is polarized!)

What is the effective E field inside the dielectric?



(a) When a dielectric is polarized, the dipole moments of the molecules in the dielectric are partially aligned with the external field Eo. (b) This polarization causes an induced charge on the opposite side. This separation of charge results in a reduction in the net electric field within the dielectric.

The net effect on the dielectric is the formation of an induced positive surface charge density ind on the right face and an equal negative surface charge density –ind on the left face.



The induced surface charge give rise to an induced electric field Eind in the direction opposite the external field Eo.

Therefore, the net electric field E in the dielectric has a magnitude

E = Eo-Eind

What is the magnitude of the induced charge density?


Induced charge on a dielectric placed between the plates of a charged capacitor. Note that the induced charge density on the dielectric is less than the charge density on the plates.

For parallel plate capacitor,

External field Eo = /o

Induced Field Eind = ind/o

And E = Eo/ = /o

Substitute into E = Eo-Eind gives

o

=o o

ind

And ind =

topic outline definition of capacitance calculating capacitance combinations of capacitors energy...

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