capacitance and dielectrics - santa rosa junior collegelwillia2/old42web/42ch26.pdf · chapter 26...

Chapter 26 HW: 7th/6th

4/6, 9/11, 11/13 (on 9 and 11 start from Gauss’s Law!), 17/21, 19/23, 25/29, 28/33, 32/40,39/45,41/49,42/50, 44/53,45/73,49/58, 58/68, 59/67

Capacitance

and

Dielectrics

Recall: The Infinite Charged Plane

02E σ

ε=

Potential Difference in a Uniform Field

Electric field lines always point in the direction of decreasing electric potentialWhen the electric field is directed downward, point B is at a lower potential than point AWhen a positive test charge moves from A to B, the charge-field system loses potential energy

B B

B A A AV V V d E d Ed− = Δ = − ⋅ = − = −∫ ∫E s s

r r

Parallel Plates: In the center we assume the electric field is constant and uniform and that the potential difference between any two points depends on the distance d between those points!

0

E σε

=

V EdΔ = −

Parallel Plate Assumptions

The assumption that the electric field is uniform is valid in the central region, but not at the ends of the platesIf the separation between the plates is small compared with the length of the plates, the effect of the non-uniform field can be ignoredThe charge density on the plates is σ = Q/A

A is the area of each plate, which are equalQ is the charge on each plate, equal with opposite signs

A capacitor consists of two conductorsThese conductors are called platesWhen the conductor is charged, the plates carry charges of equal magnitude and opposite directions

A potential difference exists between the plates due to the chargeCapacitance will always be a positive quantityThe capacitance of a given capacitor is constantThe farad is a large unit, typically you will see microfarads (μF) and picofarads (pF)A capacitor stores electrical energy

Makeup of a Capacitor

Capacitors

Capacitors are devices that store electric charge and energyExamples of where capacitors are used include:

radio receiversfilters in power suppliesenergy-storing devices in electronic flashes

Some Uses of Capacitors

DefibrillatorsWhen fibrillation occurs, the heart produces a rapid, irregular pattern of beatsA fast discharge of electrical energy through the heart can return the organ to its normal beat pattern

In general, capacitors act as energy reservoirs that can be slowly charged and then discharged quickly to provide large amounts of energy in a short pulse

Charging a Parallel Plate Capacitor

Each plate is connected to a terminal of the batteryIf the capacitor is initially uncharged, the battery establishes an electric field in the connecting wires

Active Figure 26.10

AF_2604.html

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

The capacitance is a measure of the capacitor’s ability to store chargeThe SI unit of capacitance is the farad (F)

QCV

=Δ

CapacitanceThe capacitance is proportional to the area of its plates and inversely proportional to the distance between the plates

QCV

=Δ

( )= = = = =

Δ /o

o

o

ε AQ Q Q QC σV Ed Q ε A d ddε

= oε ACd

Circular Plates

(plate radius = 10 cm) ε0= 8.85x10 –12 Farad/m

= oε ACd

QCV

=Δ

Capacitors in SeriesWhen a battery is connected to the circuit, electrons are transferred from the left plate of C1 to the right plate of C2through the batteryAs this negative charge accumulates on the right plate of C2, an equivalent amount of negative charge is removed from the left plate of C2, leaving it with an excess positive chargeAll of the right plates gain charges of –Q and all the left plates have charges of +Q thus:

QCV

=Δ

= =1 2Series: Q Q Q

Capacitors in SeriesThe potential differences add up to the battery voltageΔV = V1 + V2 + …

The equivalent capacitance of a series combination is always less than any individual capacitor in the combination

1 2

1 1 1

eqC C C= + +K

Δ =eq

QVC

Active Figure 26.08Capacitors in Series

Capacitors in ParallelThe total charge is equal to the sum of the charges on the capacitors

Qtotal = Q1 + Q2The potential difference across the capacitors is the same

And each is equal to the voltage of the battery:

ΔV = ΔV1 = ΔV2

The capacitors can be replaced with one capacitor with a capacitance of Ceq

Qtotal = Q1 + Q2

Capacitors in Parallel

Essentially, the areas are combined

= Δtotal eqQ C V

Ceq = C1 + C2 + …

Δ = Δ + Δ1 2eqC V C V C V

Active Figure 26.07Capacitors in Parallel

Capacitor Summary

Δ = Δ + Δ + ⋅ ⋅ ⋅= =

= + +K

1 2

1 2

1 2

Capacitors in Series:

Q1 1 1eq

V V VQ Q

C C C

Δ = Δ = Δ

= +

= + +K

1 2

1 2

1 2

Capacitors in Parallel:

tot

eq

V V VQ Q QC C C

Equivalent CapacitanceDraw the reduced circuit it in stages!

Fig P26-22, p.824

Three capacitors are connected to a battery as shown. Their capacitances are C1 = 3C, C2 = C, and C3 = 5C. (a) What is the equivalent capacitance of this set of capacitors? (b) State the ranking of the capacitors according to the charge they store, from largest to smallest. (c) Rank the capacitors according to the potential differences across them, from largest to smallest.  

Δ = Δ = Δ= +

= + +K

1 2

1 2

1 2

Capacitors in Parallel:

tot

eq

V V VQ Q QC C C

Δ = Δ + Δ + ⋅ ⋅ ⋅= =

= + +K

1 2

1 2

1 2

Capacitors in Series:

Q1 1 1

eq

V V VQ Q

C C C

Fig P26-27

( )( )

1

1

2

1

1 1 3.33 F5.00 10.02 3.33 2.00 8.66 F

2 10.0 20.0 F

1 1 6.04 F8.66 20.0

s

p

p

eq

C

C

C

C

μ

μ

μ

μ

−

−

⎛ ⎞= + =⎜ ⎟⎝ ⎠

= + =

= =

⎛ ⎞= + =⎜ ⎟⎝ ⎠

Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 μF, C2 = 10.0 μF, and C3 = 2.00 μF.

You Try:

P26.72 Assume a potential difference across a and b, and notice that the potential differenceacross the 8.00 Fμ capacitor must be zero by symmetry. Then the equivalent capacitance can bedetermined from the following circuit:

⇒

⇒

FIG. P26.72

Series? Parallel? Neither Both?

Find the Equivalent Capacitance

P26.75 By symmetry, the potential difference across 3C is zero, so the circuit reduces to

11 1 8 42 4 6 3eqC C CC C

−⎛ ⎞= + = =⎜ ⎟⎝ ⎠

.

You Try:

Energy in a CapacitorConsider the circuit to be a systemBefore the switch is closed, the energy is stored as chemical energy in the batteryWhen the switch is closed, the energy is transformed from chemical to electric potential energy

Energy Stored in a CapacitorAssume the capacitor is being charged and, at some point, has a charge q on itThe work needed to transfer a charge from one plate to the other is

The total work required is

The work done in charging the capacitor appears as electric potential energy U:

qdW Vdq dqC

= Δ =

2

0 2Q q QW dq

C C= =∫

221 1 ( )

2 2 2QU Q V C VC

= = Δ = Δ

This applies to a capacitor of any geometryThe energy stored increases as the charge increases and as the potential difference increasesThe energy can be considered to be stored in the electric field The total work done by a battery charging a capacitor is QV. Half the energy is either dissipated as heat or radiated as electromagnetic waves. For a parallel-plate capacitor, the energy can be expressed in terms of the field as U = ½ (εoAd)E2

It can also be expressed in terms of the energy density (energy per unit volume) uE = ½ εoE2

221 1 ( )

2 2 2QU Q V C VC

= = Δ = Δ

Energy Stored in a Capacitor

Energy ProblemDetermine the energy stored in C2 when C1 = 15 µF, C2 = 10 µF, C3 = 20 µF, and V0 = 18 V.

a. 0.72 mJb. 0.32 mJc. 0.50 mJd. 0.18 mJe. 1.60 mJ

221 1 ( )

2 2 2QU Q V C VC

= = Δ = Δ

If VA – VB = 50 V, how much energy is stored in the 36-µF capacitor?

a. 50 mJb. 28 mJc. 13 mJd. 8.9 mJe. 17 mJ

You Try:2

21 1 ( )2 2 2QU Q V C VC

= = Δ = Δ

Example 26.4: Rewiring Two Charged Capacitors

Fig. 26-12, p. 733

Energy Problem

A 3.0-µF capacitor charged to 40 V and a 5.0-µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 3.0-µF capacitor?a. 11 µCb. 15 µCc. 19 µCd. 26 µCe. 79 µC

221 1 ( )

2 2 2QU Q V C VC

= = Δ = Δ

ProblemWhat is the maximum voltage that can be sustainedbetween 2 parallel plates separated by 2.5 cm of dry air?Dry air supports max field strength of 3x 106 V/m .

V Ed=6(3 10 / )(.025 )x V m m=

47.5 10x V=

75kV=

More than this and the air breaks down and becomes a conductor. LIGHTENING!

The dielectric breakdown strength of dry air, at Standard Temperature and Pressure (STP), between spherical electrodes is approximately 33 kV/cm.

Dielectric Breakdown of Air

DielectricsA dieletric is an insulator that increases the capacitance.

Reduced E field prevents breakdown & discharge between plates.

0 ACd

κε=

The dieletric constant is the ratio of the net electric field magnitude without the dielectric,E0, and with the dielectric, E.

The dielectric reduces the net electric field and potential difference across the plates.

Reduced E field prevents breakdown & discharge between plates.

0EE

κ = 0 /E E κ=~ 1

3.7

80

Air

Paper

Water

κκ

κ

=

=

DielectricsIf the field becomes too great, the dielectric breaks down

and becomes a conductor and the plates discharge.

Reduced E field prevents breakdown & discharge between plates.

Dielectrics – An Atomic ViewThe molecules that make up the dielectric are modeled as dipolesThe molecules are randomly oriented in the absence of an electric field

Dielectrics – An Atomic ViewAn external electric field is applied which produces a torque on the moleculesThe molecules partially align with the electric fieldThe degree of alignment of the molecules with the field depends inversely upon temperature and directly with the magnitude of the fieldThe degree of alignment of the molecules with the field depends on the polarization of the molecules.Molecules are said to be polarized when a separation exists between the average position of the negative charges and the average position of the positive chargesPolar molecules are those in which this condition is always presentMolecules without a permanent polarization are called nonpolarmolecules

0 0/E σ ε=

Water MoleculesA water molecule is an example of a polar moleculeThe center of the negative charge is near the center of the oxygen atomThe x is the center of the positive charge distribution

The average positions of the positive and negative charges act as point chargesTherefore, polar molecules can be modeled as electric dipoles

Electric DipoleAn electric dipole consists of two charges of equal magnitude and opposite signsThe charges are separated by 2aThe electric dipole moment (p) is directed along the line joining the charges from –q to +qThe electric dipole moment has a magnitude of p = 2aq

3

2y

aqkEx

=

Electric Dipole in an External E FieldEach charge has a force of F = Eq acting on itThe net force on the dipole is zeroThe forces produce a net torque on the dipole that makes it align with the external electric field.

The magnitude of the torque is:τ = 2(Fa sin θ) = 2Eqa sin θ = pE sin θ

The torque can also be expressed as the cross product of the moment and the field: τ = p x E

Induced PolarizationA symmetrical molecule has no permanent polarization (a)Polarization can be induced by placing the molecule in an electric field (b)Induced polarization is the effect that predominates in most materials used as dielectrics in capacitorsInduced polar molecules can also be modeled as electric dipoles

An external field can polarize the dielectric whether the molecules are polar or nonpolarThe charged edges of the dielectric act as a second pair of plates producing an induced electric field in the direction opposite the original electric field, thus reducing the net electric field in the dielectric:

Dielectrics – An Atomic View

0/ind indE σ ε=

0 0/E σ ε=

0 indE E E= − 0 /E E κ=

Dielectrics – An Atomic View

0/ind indE σ ε=

0 0/E σ ε=0 indE E E= −

0 0 0

indσσ σκε ε ε

= −

0 /E E κ=

1ind

κσ σκ−

=

1 indκ σ σ> → <

Induced Charge and Field

The electric field due to the plates is directed to the right and it polarizes the dielectricThe net effect on the dielectric is an induced surface charge that results in an induced electric fieldIf the dielectric were replaced with a conductor, the net field between the plates would be zero

0 indE E E= −

ProblemDetermine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 0 mm.

( ) ( )12 4 2110

5

2.10 8.85 10 F m 1.75 10 m8.13 10 F 81.3 pF

4.00 10 m

AC

d

κ− −

−−

× ×∈= = = × =

×

( ) ( )6 5m ax m ax 60.0 10 V m 4.00 10 m 2.40 kVV E d −Δ = = × × =

(a)

(b)

Dielectric Problem

Capacitance of a Cylindrical Capacitor

From Gauss’s Law, the field between the cylinders isE = 2keλ / rΔV = -2keλ ln (b/a)The capacitance becomes

( )= =

Δl

00 2 ln /e

QCV k b a ( )

≡≡ ≡= = =Δ

l0 2 ln /e

QC CV k b a

Coaxial Cable

Radio-grade flexible coaxial cable.A: outer plastic sheathB: copper screenC: inner dielectric insulatorD: copper core

Coaxial cables are often used as a transmission line for radio frequency signals. In a hypothetical ideal coaxial cable the electromagnetic field carrying the signal exists only in the space between the inner and outer conductors. Practical cables achieve this objective to a high degree. A coaxial cable provides protection of signals from external electromagnetic interference, and effectively guides signals with low emission along the length of the cable.

Types of Capacitors – Tubular Metallic foil may be interlaced with thin sheets of paraffin-impregnated paper or MylarThe layers are rolled into a cylinder to form a small package for the capacitor

Types of Capacitors – Oil Filled

Common for high-voltage capacitorsA number of interwoven metallic plates are immersed in silicon oil

Types of Capacitors –Electrolytic

Used to store large amounts of charge at relatively low voltagesThe electrolyte is a solution that conducts electricity by virtue of motion of ions contained in the solution

Variable CapacitorsVariable capacitors consist of two interwoven sets of metallic platesOne plate is fixed and the other is movableThese capacitors generally vary between 10 and 500 pFUsed in radio tuning circuits

Next Week’s Lab

0C Cκ=

0 ACd

κε=

0 /V V κΔ = Δ

Circular Plates

(plate radius = 10 cm) ε0= 8.85x10 –12 Farad/m

Example 26.7: The Effect of a Metallic Slab

Fig. 26-24, p. 742