capacitance and dielectrics - srjcsrjcstaff.santarosa.edu/~lwillia2/42/42ch26.pdf · parallel...

45
Capacitance and Dielectrics

Upload: duongtram

Post on 04-Sep-2018

219 views

Category:

Documents


0 download

TRANSCRIPT

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 potential

When the electric field is directed downward, point B is at a lower potential than point A

When 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− = ∆ = − ⋅ = − = −∫ ∫E s s

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 Ed∆ = −

Parallel Plate Assumptions

The assumption that the electric field is uniform is valid in the central region, but not at the ends of the plates

If the separation between the plates is small compared with the length of the plates, the effect of the non-uniform field can be ignored

The charge density on the plates is σ = Q/A A is the area of each plate, which are equal Q is the charge on each plate, equal with opposite

signs

A capacitor consists of two conductors These conductors are called plates When the conductor is charged, the plates

carry charges of equal magnitude and opposite directions

A potential difference exists between the plates due to the charge

Capacitance will always be a positive quantity The capacitance of a given capacitor is constant The 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 energy

Examples of where capacitors are used include: radio receivers filters in power supplies energy-storing devices in electronic flashes

Some Uses of Capacitors Defibrillators

When fibrillation occurs, the heart produces a rapid, irregular pattern of beats

A 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 battery

If the capacitor is initially uncharged, the battery establishes an electric field in the connecting wires

Capacitance The 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 charge

The SI unit of capacitance is the farad (F) F = C/V

QCV

=∆

Capacitance The capacitance is proportional

to the area of its plates and inversely proportional to the distance between the plates

QCV

=∆

( )= = = = =

∆ /o

o

o

ε AQ Q Q QC σV Ed Q ε A d ddε

= oε ACd

Capacitors in Series When a battery is connected

to the circuit, electrons are transferred from the left plate of C1 to the right plate of C2 through the battery

As 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 charge

All of the right plates gain charges of –Q and all the left plates have charges of +Q thus:

QCV

=∆

= =1 2Series: Q Q Q

Capacitors in Series The potential differences

add up to the battery voltage

ΔV = 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= + +

∆ =eq

QVC

Capacitors in Parallel The total charge is

equal to the sum of the charges on the capacitors Qtotal = Q1 + Q2

The 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

= ∆total eqQ C V

Ceq = C1 + C2 + …

∆ = ∆ + ∆1 2eqC V C V C V

Capacitor Summary

∆ = ∆ + ∆ + ⋅ ⋅ ⋅= =

= + +

1 2

1 2

1 2

Capacitors in Series:

Q1 1 1eq

V V VQ Q

C C C

∆ = ∆ = ∆= +

= + +

1 2

1 2

1 2

Capacitors in Parallel:

tot

eq

V V VQ Q QC C C

Equivalent Capacitance Draw the reduced circuit 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.

∆ = ∆ = ∆= +

= + +

1 2

1 2

1 2

Capacitors in Parallel:

tot

eq

V V VQ Q QC C C

∆ = ∆ + ∆ + ⋅ ⋅ ⋅= =

= + +

1 2

1 2

1 2

Capacitors in Series:

Q1 1 1eq

V V VQ Q

C C C

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

∆ = ∆ = ∆= +

= + +

1 2

1 2

1 2

Capacitors in Parallel:

tot

eq

V V VQ Q QC C C

∆ = ∆ + ∆ + ⋅ ⋅ ⋅= =

= + +

1 2

1 2

1 2

Capacitors in Series:

Q1 1 1eq

V V VQ Q

C C C

Recall from Chapter 25 Problem

What is the maximum voltage that can be sustained between 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.

Dielectrics A 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

3.780

Air

Paper

Water

κκ

κ

=

=

Dielectrics If 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.

Dielectric Breakdown of Air

Problem Determine (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.

Dielectric Problem

This Week’s Lab

Energy in a Capacitor Consider the circuit to

be a system Before the switch is

closed, the energy is stored as chemical energy in the battery

When the switch is closed, the energy is transformed from chemical to electric potential energy

Energy Stored in a Capacitor Assume the capacitor is being charged and, at some

point, has a charge q on it The magnitude of the 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:

∆∆ = → = ∆ =

0

U qV dW Vdq dqq C

= =∫2

0 2Q q QW dq

C C

221 1 ( )

2 2 2QU Q V C VC

= = ∆ = ∆

This applies to a capacitor of any geometry The energy stored increases as the charge increases and

as the potential difference increases The 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

= = ∆ = ∆

Energy Stored in a Capacitor

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

a. 0.72 mJ b. 0.32 mJ c. 0.50 mJ d. 0.18 mJ e. 1.60 mJ

221 1 ( )

2 2 2QU Q V C VC

= = ∆ = ∆

∆ = ∆ = ∆= +

= + +

1 2

1 2

1 2

Capacitors in Parallel:

tot

eq

V V VQ Q QC C C

∆ = ∆ + ∆ + ⋅ ⋅ ⋅= =

= + +

1 2

1 2

1 2

Capacitors in Series:

Q1 1 1eq

V V VQ Q

C C C

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

a. 50 mJ b. 28 mJ c. 13 mJ d. 8.9 mJ e. 17 mJ

221 1 ( )

2 2 2QU Q V C VC

= = ∆ = ∆You Try:

∆ = ∆ = ∆= +

= + +

1 2

1 2

1 2

Capacitors in Parallel:

tot

eq

V V VQ Q QC C C

∆ = ∆ + ∆ + ⋅ ⋅ ⋅= =

= + +

1 2

1 2

1 2

Capacitors in Series:

Q1 1 1eq

V V VQ Q

C C C

Recall: The Dipole Electric Field Bisector

In practice, we almost always observe the electric field of a dipole only for distance much greater than the charge separation and the E ~1/r3 holds everywhere.

2 2 3/2

2( )y

kQaEx a

=+

E

Near: Far: 3 3

2y

aqk pkEx x

= =

Electric Dipole in an External E Field Each charge has a force of

F = Eq acting on it The net force on the dipole is zero The forces produce a net torque

on the dipole that makes it align with the external electric field.

Define the electric dipole moment p = 2aq

The magnitude of the net torque is:

τ = 2(Fa sin θ) = 2Eqa sin θ = pE sin θ

=

2qp a τ = ×p E

Electric Dipole Energy The potential energy can be expressed as a function of the orientation of the dipole with the field: Uf – Ui = pE(cos θi – cos θf) Set Ui =0, then, U= - pE cos θ

= − ⋅

U p E

sinU d pE dτ θ θ θ∆ = =∫ ∫

8. A small rigid object carries positive and negative 3.50-nC charges. It is oriented so that the positive charge has coordinates (–1.20 mm, 1.10 mm) and the negative charge is at the point (1.40 mm, –1.30 mm). (a) Find the electric dipole moment of the object. The object is placed in an electric field E = (7 800î – 4 900ĵ) N/C. (b) Find the torque acting on the object. (c) Find the potential energy of the object–field system when the object is in this orientation.

τ = ×p E

= − ⋅

U p E

=

2qp a

( ) ( ) ( )ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B× = − − − + −A B i j k

x x y y z zA B A B A B A B⋅ = + +

Dielectrics – An Atomic View The molecules that

make up the dielectric are modeled as dipoles

The molecules are randomly oriented in the absence of an electric field

Dielectrics – An Atomic View An external electric field is applied which

produces a torque on the molecules The molecules partially align with the

electric field The degree of alignment of the molecules

with the field depends inversely upon temperature and directly with the magnitude of the field

The 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 charges

Polar molecules are those in which this condition is always present

Molecules without a permanent polarization are called nonpolar molecules

0 0/E σ ε=

Induced Polarization A 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 capacitors

Induced polar molecules can also be modeled as electric dipoles

An external field can polarize the dielectric whether the molecules are polar or nonpolar

The 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 dielectric

The net effect on the dielectric is an induced surface charge that results in an induced electric field

If the dielectric were replaced with a conductor, the net field between the plates would be zero

0 indE E E= −

Water Molecules A water molecule is an

example of a polar molecule

The center of the negative charge is near the center of the oxygen atom

The x is the center of the positive charge distribution

The average positions of the positive and negative charges act as point charges

Therefore, polar molecules can be modeled as electric dipoles