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UNIT 4:Capacitors and DielectricUNIT 4:Capacitors and Dielectric

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4.1 Capacitor{ A capacitor is a device that is capable of storing electric charges a device that is capable of storing electric charges

or electric potential energyor electric potential energy.{ It is consist of two conducting plates separated by a small air gap or a

thin insulator (called a dielectric dielectric such as mica, ceramics, paper or even oil).

{ The conducting plates could be in the shape ofz Cylindricalz Sphericalz Parallel plate

{ The electrical symbol for a capacitor is

4.1.1 Uses of Capacitors{ Capacitors are commonly used in many electronic circuits.{ Some examples are :

z Photoflash unit capacitor stores large amount of energy which can be released in a flash when triggered.

or+

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z Giant lasers large amount of energy stored in capacitors could power large lasers.

z On-off switches low energy capacitors can be used as switches in computer motherboards.

z Smoothen d.c. voltages smoothened d.c. voltages are obtained from the rectification of a.c. voltages.

4.1.2 Types of Capacitors{ There are three commercial capacitor designs.{ It is tubular capacitor, high-voltage capacitor and electrolytic capacitor.{ A tubulartubular capacitor whose layers of metallic foil and dielectric are

rolled into a cylinder (figure 4.1a).{{ HighHigh--voltagevoltage capacitor consists of a number of interwoven metallic

plates immersed in silicone oil (figure 4.1b).{ An electrolyticelectrolytic capacitor consists of a metallic foil in contact with an

electrolyte (figure 4.1c).Metal foil

PaperFig. 4.1aFig. 4.1a Fig. 4.1bFig. 4.1b

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{ The capacitance of a capacitor, CDefinition is defined as the ratio of the charge on either plate to the the ratio of the charge on either plate to the

potential difference between thempotential difference between them.Mathematically,

Fig. 4.1cFig. 4.1c

4.2 Capacitance, C

VQC =

where plates theof oneon charge :Qplates two theacross difference potential :V

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{ The unit of capacitance is the farad (F).{{ 1 farad1 farad is defined as the charge of 1 coulomb stored on each of the charge of 1 coulomb stored on each of

the conducting plates as a result of a potential difference of 1the conducting plates as a result of a potential difference of 1volt between the two plates.volt between the two plates.i.e.

{ By rearranging the equation from the definition of capacitance, we get

where the capacitance of a capacitor, C is constant then

{ Note :z The farad is a very large unit. Therefore in many applications the

most convenient units of capacitance are microfaradmicrofarad and the picofaradpicofarad where the unit conversion can be shown below :

F10F1 6 =F10pF1 12 =

voltcoulomb farad 1

11 =

(The charges stored, (The charges stored, QQ is directly proportional to the potential is directly proportional to the potential difference, difference, VV across the conducting plate.)across the conducting plate.)

CVQ =VQ

F10 6=F1 =

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dQ

Q+V

AArea,positive positive terminalterminal

negative negative terminalterminal Fig. 4.3aFig. 4.3a

{ Consider two parallel metallic plate capacitor of equal area A are separated by a distance d and the space between plates is vacuum or air as shown in figure 4.3a.

{ One plate carries a charge +Q and the other carries a charge Q then the potential difference between this two parallel plates is V.

{ Because d is small compared to the dimensions of each plate so that the electric field strength E is uniform between them.

{ The magnitude of the electric field strength is given by

4.3 Parallel-plate Capacitors

Er

0E =

AQ =and

0AQE = (1)(1)

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{ Since Q=CV then equation (1) can be written as

{ Because the field between the plates is uniform, the potential difference between the plates is

Substituting this relation into eq. (2), thus the capacitance of a parallel-plate capacitor is

0ACEdE =

EdV =

or

0ACVE = (2)(2)

dAC 0=

where space free ofty permittivi :0plate theof Area :A

ParallelParallel--plate capacitor plate capacitor separated by a vacuumseparated by a vacuum

dAC = ParallelParallel--plate capacitor plate capacitor separated by a dielectric separated by a dielectric

materialmaterial

plates twoebetween th distance :d

) .( 212120 mNC10x858=

The capacitance of a The capacitance of a parallelparallel--plate capacitor is plate capacitor is proportional to the area proportional to the area of its plates and of its plates and inversely proportional to inversely proportional to the plate separation the plate separation

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{ Example 1 :The plates of a parallel-plate capacitor are 8.0 mm apart and each has an area of 4.0 cm2. The plates are in vacuum. If the potential difference across the plates is 2.0 kV, determinea. the capacitance of the capacitor.b. the amount of charge on each plate.c. the electric field strength was produced.d. the surface charge density on each plate.(Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2)Solution: d=8.0x10-3 m, A=4.0x10-4 m2, V=2.0x103 Va. By applying the equation of capacitance for parallel-plate capacitor,

b. From the definition of the capacitance, the amount of charge stored in the capacitor is given by

dAC 0=

pF440F10x424C 13 .@. =

VQC =

C10x848Q 10= .).)(.( 313 10x0210x424Q =

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c. From the relation between uniform E and V , thus

d. By using the equation of surface charge density, then

{ Example 2 :A circular parallel-plate capacitor with radius of 10 cm is connected to a 15 V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. If the separation distance between plates is 35 mm and the medium between plates is air.a. Find the amount of charge on each plate.If their separation is increases to 50 mm after the battery is disconnected, determineb. the amount of charge on each plate.c. the potential difference between plates.d. the capacitance of the capacitor.(Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2)

AQ =

115 mVCN10x502E = @ .

26 m C10x212 = .0E =

0E =

dVE =

or

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Solution: r=10x10-2 m, V=15 V, d=35x10-3 m

a. From the definition of the capacitance, the amount of charge is

The separation, d=50 mm.b. The amount of charge on each plate is not changed because the

electric field between the plates is uniform then

c. From the relation between uniform E and V , thus

CVQ =

C10x191Q 10 . =

V421V .=

VdAQ 0

=

dAC 0=

222 m10x143rA == .and

C10x191Q 10 . =

dVE = EdV = and

AQE0

=d

AQV0

=

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d. The capacitance of the capacitor is given by

{ Example 3 : (exercise)a. A parallel-plate, air-filled capacitor has circular plates separated by

1.80 mm. The charge per unit area on each plate has magnitude 5.60 pC m-2. Find the potential difference between the plates of the capacitor. (Young & Freedman,pg.934.no.24.4) Ans. :1.14 mV

b. An electric field of 2.80x105 V m-1 is desired between two parallel plates each of area 21.0 cm2 and separated by 0.250 cm of air. Find the charge on each plate. (Giancoli,pg.628.no. 14) Ans. :5.20x10-9 C

(Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2){ Example 4 : (exercise)

A 10.0 F parallel-plate capacitor with circular plates is connected to a 12.0 V battery. Calculatea. the charge on each plate.b. the charge on each plate if their separation were twice while the

capacitor remained connected to the battery.c. the charge on each plate if the capacitor were connected to the

12.0 V battery after the radius of each plate was twice without changing their separation (Young & Freedman,pg.934.no.24.5)

Ans. :120 C, 60 C, 480 C

VQC =

F10x565C 12 . = dAC 0=or

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{ Consider a spherical capacitor consists of two thin concentric spherical conducting shells, of radius ra and rb are separated by a vacuum as shown in figure 4.4a.

{ The inner shell carries a uniformly distributed charge +Q on its surface and the outer shell an equal but opposite charge Q.

{ To find the magnitude of the electric field between two spherical shells, a spherical gaussian surface is drawn (figure 4.4a).z The magnitude of the electric field magnitude of the electric field is constantconstant and normalnormal to the

surfacesurface at each point everywhere on the gaussian surface.z By using the Gausss law,

rbr

ar

Q+Q

Fig. 4.4aFig. 4.4a

4.4 Spherical Capacitors

Qqenc =and

Er

Gaussian Gaussian surfacesurface

0

encE

qEdA == 2r4dA =where

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then the magnitude of the electric field between two spherical shells is

{ To obtain the potential difference, Vab between the two conducting shell, we apply the relation between E and V below,

==0

2

Q)r4E(dAE

EdrdV =

20r4

QE =

drdVE =

= ab

a

b

r

r

V

VEdrdV

= ab

r

r 20

ba drr1

4QVV abba VVV =

=

ba0ab r

1r1

4QV

and

=

ba

ab

0ab rr

rr4

QV

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{ From the definition of the capacitance,

then the capacitance of a spherical capacitor is given byabV

QC =

= abba

0 rrrr4C

=

ba

ab

0 rrrr

4Q

QC

where space free ofty permittivi :0shellinner theof radius :ar

shellouter theof radius :br

spherical capacitor spherical capacitor separated by a vacuumseparated by a vacuum

or

= abba

rrrr4C spherical capacitor spherical capacitor separated by a dielectric separated by a dielectric

material.material.

material dielectric ofty permittivi :

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{ Example 5 :A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF.a. Find the radius of the outer sphere.b. If the potential difference between the two spheres is 220 V,

calculate the magnitude of charge on each sphere.(Young & Freedman,pg.935.no.24.12)

(Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2)Solution: ra=15.0x10-2 m, C=116x10-12 F a. By using the equation of capacitance for spherical capacitor,

then the radius of the outer shell is given by

= abba

0 rrrr4C

m180rb .=a0

ab r4C

Crr =

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b. Given V=220 VFrom the definition of the capacitance, thus the charge on each sphere is

{ Example 6 :A 20.0 F spherical capacitor is composed of two concentric metal spheres, one having a radius twice as large as the other. The region between the spheres is a vacuum. Determine the volume of this region.(Serway & Jewett,pg.823,no.12)

(Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2)Solution: ra=r, rb=2ra=2r, C=20.0x10-6 F

CVQ =C10x552Q 8 . =

= rr2

r2r4C 0))((

By applying the equation of capacitance for spherical capacitor,

= abba

0 rrrr4C

m10x009r 4 .=

Q

ar

brQ+

vacuumvacuum

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The volume of the vacuum region, V is given by

4.5.1 Capacitors connected in series{ Figure 4.5a shows 3 capacitors connected in series to a battery of

voltage, V.

[ ]3a3b rr34V =3 . m10x132V 16=

3r3

28V =

V = V = Volume of the Volume of the outer sphere, outer sphere, VVbb --Volume of the Volume of the inner sphere, inner sphere, VVaa

4.5 Combinations of Capacitors

V

+Q+Q11 --QQ11 +Q+Q22 --QQ22 +Q+Q33 --QQ33

CC11,V,V11 CC22,V,V22 CC33,V,V33

Fig. 4.5aFig. 4.5a

V

+Q+Q --QQ

CCeqeq,V,Vequivalent toequivalent to

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{ When the circuit is complete, electrons are transferred onto the plates such that the magnitude of the charge Q on each plate is the same.

{ Thus the total charge (Q) on the equivalent capacitor is

{ The potential difference across each capacitor C1,C2 and C3 are V1,V2 and V3 respectively. Hence

{ Since the total potential difference V is given by

Therefore the equivalent (effective) capacitance Ceq for n capacitors connected in series is given by

321 QQQQ ===

;11

11 C

QCQV ==

and

n321eq C1

C1

C1

C1

C1 ++++= ...

321 VVVV ++=

;22

22 C

QCQV ==

33

33 C

QCQV ==

321 CQ

CQ

CQV ++=

321 C1

C1

C1

QV ++=

eqC1

QV =

capacitors connected capacitors connected in seriesin series

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4.5.2 Capacitors connected in parallel{ Figure 4.5b shows 3 capacitors connected in parallel to a battery of

voltage, V.

{ The potential difference across each capacitor is the same as the supply voltage (V).

{ Thus the total potential difference (V) on the equivalent capacitor is

{ The charges stored by each capacitor C1,C2 and C3 are Q1,Q2 and Q3 respectively. Hence

;VCVCQ 1111 ==

321 VVVV ===

equivalent toequivalent to

V

+Q+Q11 --QQ11

+Q+Q22 --QQ22

+Q+Q33 --QQ33

CC11,V,V11

CC22,V,V22

CC33,V,V33

Fig. 4.5bFig. 4.5b

V

+Q+Q --QQ

CCeqeq,V,V

;VCVCQ 2222 == VCVCQ 3333 ==

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{ Since the total charge Q on the equivalent capacitor is given by

Therefore the equivalent (effective) capacitance Ceq for n capacitors connected in parallel is given by

{ Example 7 :Determine the equivalent capacitance of the configuration shown in figure below. All the capacitors are identical and each has capacitance of 1 F.

and

n321eq CCCCC ++++= ...

321 QQQQ ++=VCVCVCQ 321 ++=

321 CCCVQ ++= eqCV

Q =

capacitors connected capacitors connected in parallelin parallel

1 F

1 F

1 F

1 F

1 F1 F

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Solution:z Label all the capacitors in the circuit.

z To find the equivalent capacitance for circuit above, it is easier to solve it from the end of the circuit (left) to the terminal (right) shown by an arrow in figure above.

{ Capacitors C1, C2 and C3 connected in series, then

321x C1

C1

C1

C1 ++=

F31Cx =

CC22 CC55

CC11CC33 CC44

CC66

CC55

CCxxCC44

CC66

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{ Capacitors Cx, C4 and C5 connected in series, then

{ Capacitors Cy and C6 connected in parallel, then the equivalent capacitance Ceq is given by

54xy C1

C1

C1

C1 ++=

F51Cy =

CCyy CC66

6yeq CCC +=F1F

51Ceq +

=

F56Ceq =

CCeqeq

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{ Example 8 :

In the circuit shown in figure above, C1= 2.00 F, C2 = 4.00 F and C3 = 9.00 F. The applied potential difference between points a and b is Vab = 61.5 V. Calculatea. the charge on each capacitor.b. the potential difference across each capacitor.c. the potential difference between points a and d.similar to (Young & Freedman,pg.936.no.24.14)Solution:

CC22aa

CC11

CC33bb

dd

CC2 2 ,V,V22aa

CC1 1 ,V,V11

CC3 3 ,V,V33bb

dd

aaCCxx ,,VVxx

CC3 3 ,V,V33bb

dd

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a. From the figure, Capacitors C1 and C2 connected in parallel then the equivalent capacitance Cx is given by

The equivalent capacitance, Ceq in the circuit is given by

The total charge Q stored in the equivalent capacitance Ceq is

Since the capacitors Cx and C3 connected in series then the charge stored in each capacitor is the same as the total charge.

The potential difference across the capacitor C3 is

F4F2CCC 21x +=+=F6Cx =

3xeq C1

C1

C1 +=

( )( )56110x63VCQ 6abeq .. ==F63Ceq .=

C221QQQ x3 ===

C10x212Q 4 221 @ C . =

( )( )9

221CQV

3

33

==V624V3 .=

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then the potential difference across the equivalent capacitance Cx is given by

Since the capacitors C1 and C2 connected in parallel then the potential difference across each capacitor is the same as Vx.

Therefore

b. The potential difference across the capacitors C1 and C2 is

and the potential difference, V3 is given by

c. The potential difference between points a and d is given by

3abx VVV =V936Vx .=

V936VVV x21 .===( )( ) C8739362VCQ 111 .. ===

C147Q2 =1x2 QQQ = 222 VCQ =

V624V3 .=

or

V936VV 21 .==

V936VV xad .==

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{ Example 9 : (exercise)Four capacitors are connected as shown in figure below.

{ Example 10 : (exercise)Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in figure below.

Calculatea. the equivalent capacitance between

points a and b.b. the charge on each capacitor if

Vab=15.0 V. (Serway & Jewett,pg.823,no.21)Ans. : 5.96 F, 89.5 C on 20 F, 63.2 C on 6 F, 26.3 C on 15 F and on 3 F

Take C1 = 5.00 F, C2 = 10.0 F and C3 = 2.00 F. (Serway & Jewett,pg.824,no.27)Ans. : 6.04 F

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{ When the switch is closed in figure 4.6a, charges begin accumulate on the plates.

{ The total work W required to increase the accumulated charge from zero to Q is given by

{ Note : No charges will accumulate on each plate if the capacitor is not charged.

4.6 Energy stored in a Charged capacitor, U

and

Fig. 4.6aFig. 4.6a

dQ

V

+++

{ A small amount of work (dW ) is done in bringing a small amount of charge (dQ) from the battery to the capacitor. This is given by

VdQdW =CQV =

dQCQdW =

CQ

21WU

2

== = Q0 dQCQdW

also QV21U =2CV

21U = and

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{ Example 11 :Two capacitors, C1= 3.00 F and C2 = 6.00 F are connected in series and charged with a 4.00 V battery as shown in figure below.

Calculatea. the total capacitance for the circuit above.b. the charge on each capacitor.c. the potential difference across each capacitor.d. the energy stored in each capacitor.e. the area of the each plate in capacitor C1 if the distance between two

plates is 0.01 mm and the region between plates is vacuum.(Given permittivity of free space, 0 = 8.85 x 10-12 F m-1)Solution: V=4.00 Va. Since capacitors C1 and C2 connected in series, thus the total

capacitance, C is given by

1C

V004 .

2C

F0061

F0031

C1

C1

C1

21 . . +=+=F002C .=

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b. Since capacitors C1 and C2 connected in series, thus the charge on

capacitors C1 and C2 is equal to the total charge Q in the circuit.

Therefore

c. The potential difference across:Capacitor C1,

Capacitor C2,

d. The energy stored in:Capacitor C1,

Capacitor C2,

( )( )004002CVQ . . ==C008Q .=

21 QQQ ==

J10x071U 51 .=

V672V1 .=1

11 C

QV =

V331V2 .=12 VVV = or

2

22 C

QV =

( )( )262111 67210x00321VC21U .. ==

J10x315U 62 .=

2222 VC2

1U =

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e. Given C1=3.00 x 10-6 F, d = 0.01 x 10-3 m and 0 = 8.85 x 10-12 F m-1By applying the formula of capacitance for parallel plate capacitor,thus the area of the each plate in capacitor C1 is given by

{ Example 12 :Consider the circuit shown in figure below, where C1= 6.00 F, C2 = 3.00 F and V = 20.0 V.

Solution:After the switch S1 is closed. The capacitor C1 is fully charged and the charge has been placed on it is given by

2m393A .=dAC 01

=

Capacitor C1 is first charged by the closing of switch S1. Switch S1 is then opened, and the charged capacitor is connected to the uncharged capacitor by the closing of S2. Calculate the initial charge acquired by C1 and the final charge on each capacitor.(Serway & Jewett,pg.824,no.23)

C120Q1 =( )( )020006VCQ 11 . .==

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After the switch S2 is closed and S1 is opened. The capacitors C1and C2 (uncharged) are connected in parallel and the equivalent capacitance is

By using the principle of conservation of charge, the total charge Q on the circuit is given by

The potential difference V across each capacitor is the same (parallel) and given by

Therefore the final charge accumulates oncapacitor C1 :

capacitor C2 :

F009CCC 21eq .=+=

C879Q1 .' =

C1200120QQQ 21 =+=+=

V313V .'=eqC

QV ='

''12 QQQ =

( )( )313006VCQ 11 . .'' ==or

C240Q2 .' =

'' VCQ 22 =

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4.7.1 Charging a capacitor through a resistor{ Figure 4.7a shows a simple circuit for charging a capacitor.

{ As charge accumulates on the capacitor, the potential difference across it increases and the current is reduced until eventually the maximum voltage across the capacitor equals the voltage supplied by the battery, V0.

{ At this time, no further current flows (I = 0) through the resistor R and the charge Q on the capacitor thus increases gradually and reaches a maximum value Q0.

4.7 Charging and Discharging of a Capacitor

Fig. 4.7aFig. 4.7a

0V

R

++ +

S,switch

CAB

++ +

e

e

{ When the switch S is closed, current I0 immediately begins to flow through the circuit.

{ Electrons will flow out from the negative terminal of the battery, through the resistor R and accumulate on the plate B of the capacitor.

{ Then electrons will flow into the positive terminal of the battery, leaving a positive charge on the plate A.

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0I

t,time

I

0I370.

RC =

{ The charging process can be shown through the graphs in figure 4.7b, 4.7c and 4.7d.

Fig. 4.7b : the potential Fig. 4.7b : the potential difference across capacitor difference across capacitor increases with time.increases with time.

0

0V

t,time

V

0V630.

RC =Fig. 4.7c : the charge on the Fig. 4.7c : the charge on the capacitor increases with capacitor increases with time.time.

0

0Q

t,time

Q

0Q630.

RC =

Fig. 4.7d : the current through the Fig. 4.7d : the current through the resistor decreases exponentially resistor decreases exponentially with time.with time.

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{ The equations involved in the charging process :

= RC

t

0 e1QQ

= RC

t

0 e1VV

Charge on charging Charge on charging capacitor :capacitor :

Potential difference (voltage) Potential difference (voltage) across charging capacitor :across charging capacitor : 0

0 VCQ =

== RC

t0 e1

CQ

CQV and

Current in resistor :Current in resistor :RCt

0eII=

where charge maximum :0Q

current maximum :0Iresistor theof resistance :R

tagesupply vol voltagemaximum : =0V

capacitor theof ecapacitanc :C

RVI 00 =and

(4.7(4.7--1)1)

(4.7(4.7--2)2)

(4.7(4.7--3)3)

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4.7.2 Discharging a capacitor through a resistor{ Figure 4.7e shows a simple circuit for discharging a capacitor.

{ Initially, the potential difference (voltage) across the capacitor is maximum, V0 and then a maximum current I0 flows through the resistor R.

{ When part of the positive charges on plate A is neutralized by the electrons, the voltage across the capacitor is reduced.

{ The process continues until the current through the resistor is zero.{ At this moment, all the charges at plate A is fully neutralized and the

voltage across the capacitor becomes zero.{ The discharging process can be shown through the graphs in figure 4.7f,

4.7g and 4.7h.

{ When a capacitor is already charged to a voltage V0 and it is allowed to discharge through the resistor R as shown in figure 4.7e.

{ When the switch S is closed, electrons from plate B begin to flow through the resistor R and neutralizes positive charges at plate A.

Fig. 4.7eFig. 4.7e

0V

R

++ +

S,switch

CAB

++ +

e

e

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0

0I

t,time

I

0I370.

RC =

Fig. 4.7f : the potential Fig. 4.7f : the potential difference across capacitor difference across capacitor decreases exponentially decreases exponentially with time.with time.

Fig. 4.7g : the charge on the Fig. 4.7g : the charge on the capacitor decreases capacitor decreases exponentially with time.exponentially with time.

Fig. 4.7h : the current through Fig. 4.7h : the current through the resistor decreases the resistor decreases exponentially with time.exponentially with time.

0

0V

t,time

V

0V370.

RC = 0

0Q

t,time

Q

0Q370.

RC =

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{ The equations involved in the discharging process :

{ Note : For calculation of current in discharging processcurrent in discharging process, ignore the ignore the negative sign in the formulanegative sign in the formula.

RCt

0eQQ=

RCt

0eVV=

Charge on discharging Charge on discharging capacitor :capacitor :

Potential difference (voltage) Potential difference (voltage) across discharging capacitor :across discharging capacitor :

Current in resistor :Current in resistor : RCt

0eII=

The negative sign indicates that as the capacitor discharges, the current direction opposite its direction when the capacitor was being charged.

RCQI 00 =and

(4.7(4.7--4)4)

(4.7(4.7--5)5)

(4.7(4.7--6)6)

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4.7.3 Time constant, { The quantity RCRC that appears in the exponent for all equation is called

time constanttime constant or relaxation timerelaxation time of the circuit or mathematically

{ Its dimension is the dimension of timedimension of time, then the unit is second (s).second (s).{ It is a measure of how quickly the capacitor charges or discharges.{ Charging process

z From eqs. 4.7-1 and 4.7-2, the charge Q on the capacitor and the voltage V across it, increase from zero at t = 0 to maximum values Q0 and V0 after a very long time. The time constant is defined as defined as the time required for the capacitor to reach (1the time required for the capacitor to reach (1--ee--11)=0.63 or 63% )=0.63 or 63% of its maximum charge/voltage.of its maximum charge/voltage.

z From eq. 4.7-3, the current drops exponentially in time constant equal to . The time constant is defined as the time required for defined as the time required for the current to drop to 1/e = 0.37 or 37% of its initial value(the current to drop to 1/e = 0.37 or 37% of its initial value(II00).).

{ Discharging processz From eqs. 4.7-4, 4.7-5 and 4.7-6, the charge Q, the voltage V and

the current I is seen to decrease exponentially in time with the same time constant .The time constant is defined asdefined as the time required the time required for the charge on the capacitor/voltage across it/current in tfor the charge on the capacitor/voltage across it/current in the he resistor decrease to resistor decrease to 1/e = 0.37 or 37% of its initial value.1/e = 0.37 or 37% of its initial value.

RC=

either oneeither one

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{ Example 13 : In the RC circuit shown in figure below, the battery has fully charged the capacitor.

Then at t = 0 s the switch S is thrown from position a to b. The battery voltage is 20.0 V and the capacitance C = 1.02 F. The current I is observed to decrease to 0.50 of its initial value in 40 s. Determinea. the value of R.b. the time constant, b. the value of Q, the charge on the capacitor at t = 0.c. the value of Q at t = 60 sSolution: V0=20.0 V, C=1.02x10-6 F, I=0.50I0 , t =40x10-6 sa. By applying the equation of current for discharging process (ignore

the negative sign) :

then taking natural logs on both sides, thus the value of R is

C

R

0V

Sb

a

RCt

0eII=

( )66

10x021R10x40

00 eII500

= ..

57R =

SF027 40

b. The time constant is given by

c. By using the equation of charge for discharging process and the time, t = 0 hence

d. By using the equation of charge for discharging process and the time, t = 60x10-6 s hence

RC =

RCt

0eQQ=

s10x85 5= .

00 CVQ =0QQ = and0CVQ =

( ) 5610x85 10x605 e10x042Q = ..

C10x042Q 5= .

RCt

0eQQ=

C10x37Q 6= .

SF017

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{ Definition is defined as the nonthe non--conducting (insulating) material conducting (insulating) material placed between the plates of a capacitor.placed between the plates of a capacitor.

{ When a dielectric (such as rubber, glass or waxed paper) is inserted between the plates of a capacitor, the capacitance increases.

{ This capacitance increases by a factor or r which is called the dielectric constant (relative permittivity) dielectric constant (relative permittivity) of the material.

{ The advantages of inserting the dielectric between the plates of the capacitor arez Increase in capacitancez Increase in maximum operating voltage.z Possible mechanical support between the plates, which allows the

plates to be close together without touching, thereby decreasing d and increasing C.

4.8.1 Dielectric constant, (r){ Definition is defined as the ratio between the capacitance of given the ratio between the capacitance of given

capacitor with space between plates filled with capacitor with space between plates filled with dielectric, dielectric, CC with the capacitance of same capacitor with the capacitance of same capacitor with plates in a vacuum, with plates in a vacuum, CC00..

Mathematically,

4.8 Dielectric

0CC = (4.8(4.8--1)1)

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{ It is dimensionless constant (no unit).{ For parallel-plates capacitor:

then the equation 4.8-1 can be written as

{ From the definition of the capacitance,

hence the equation 4.8-1 can be written as

00

== or

dAC =

dAC 00

=and

=

dA

dA

0

VQC =

(4.8(4.8--2)2)

where material dielectric ofty permittivi :

00 V

QC =and Q is constantQ is constant

VV 0=

wherein vacuumcapacitor across difference potential :0V

dielectricwith capacitor across difference potential :V

(4.8(4.8--3)3)

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{ From the relationship between E and V for uniform electric field,

thus the equation 4.8-3 can be written as

{ The dielectric constant depends on the insulating material used. Table below shows the value of dielectric constant and the dielectric strength fro several materials.

EdV = dEV 00 =and

wherein vacuumcapacitor theofstrength field electric :0E

EE 0= (4.8(4.8--4)4)

EddE 0=

dielectricwith capacitor theofstrength field electric :E

602.1Teflon-80Water

152.5Silicone oil163.7Paper73.2Mylar31.00059Air

Dielectric Strength (106 V m-1)Dielectric constant, Material

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{ The dielectric strength is defined as the electric field strength at the electric field strength at which dielectric breakdown occurs and the material becomes a which dielectric breakdown occurs and the material becomes a conductor.conductor.

{ Since V=Ed for a uniform electric field, the dielectric strength determines the maximum potential difference that can be applied across a capacitor per meter of plate spacing.

{ Summary :

{ Example 14 : A parallel-plate capacitor has plates of area A = 2x10-10 m2 and separation d = 1 cm. The capacitor is charged to a potential difference V0 = 3000 V. Then the battery is disconnected and a dielectric sheet of the same area A is placed between the plates as shown in figure below.

dielectricdielectric

d

In the presence of the dielectric, the potential difference across the plates is reduced to 1000 V. Determine

a. the initial capacitance of the air-filled capacitor.

b. the charge on each plate before the dielectric is

inserted.

c. the capacitance after the dielectric is in place.

EE

VV

CC 00

00

====

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SF027 45

d. the relative permittivity.e. the permittivity of dielectric sheet.f. the initial electric field.g. the electric field after the dielectric is inserted.(Given permittivity of free space, 0 = 8.85 x 10-12 F m-1)Solution: d=1x10-2 m, A=2x10-10 m2, V0=3000 V , V=1000 Va. Before the dielectric is in place, the capacitance is given by

b. The charge on each plate is

c. In the presence of the dielectric, the charge on each plate is the same as before the dielectric was inserted. Therefore the capacitance is

dAC 00

=

000 VCQ =F10x771C 190 .

=

VQC 0=

C10x315Q 160 .=

F10x315C 19 . =

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d. From the definition of dielectric constant, thus relative permittivity is

e. The permittivity of dielectric sheet is

f. By applying the relationship between E and V for uniform electric field, the magnitude of the initial electric field is

g. The magnitude of the final electric field is

0CC =

0 =3 =

dVE 00=

2121-11 mNCmF10x662 = @ .

1-1-50 @V mCN10x3E =

EE 0=

1-5 CN10x1E =( )

310x3E

5

=

SF017

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4.8.2 Dielectric effect on the parallel-plate capacitor(a) Polar dielectrics{ The molecules of some dielectrics like water have permanent electric

dipole moments where the concentration of positive and negative charges are separated.

{ When no electric fields is present the polar molecules are oriented randomly as shown in figure 4.8a . The electric dipoles tend to line up when the external electric field is applied to them as in figure 4.8b.

{ The alignment of the electric dipoles produces an electric field that is directed opposite the applied field and smaller in magnitude.

+ _ + _

+ _

+ _

+ _+ _

Fig. 4.8bFig. 4.8b

+ _

+ _

+ _ + _ + _+ _

+ _+ _

+ _

+ _+ _ + _

Fig. 4.8aFig. 4.8a

+ _

+ _

+_

+_

+_

+ _ + _

+_

+ _

+_+_

+_

+_ + _

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(b) Non-polar dielectrics{ Non-polar molecules such as glass or paraffin oil have their positive and

negative charge centres at the same point in the absence of an external electric field as shown in figure 4.8c.

{ When the non-polar molecules are placed in an external electric field, these centres become separated slightly and the molecules acquire induced dipole moments. These induced dipole moments tend to align with the electric field and the dielectric is polarized as shown in figure 4.8d.

Fig. 4.8cFig. 4.8c

+++ +++ +++ +++ +++ +++Fig. 4.8dFig. 4.8d

+ + + + + + + + + + + + + + + + + +

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(c) Dielectric in a parallel-plate capacitor{ Consider a capacitor whose plates are separated by a dielectric

material (either polar or non-polar). This capacitor has a charge +Q on one plate and Q on the other, so that the electric field E0 is produced between the plates. Because of the electric field, all the dielectric molecules tend to become oriented as shown in figure 4.8e.

{ The net effect in either case polar or non-polar is as if there were a net negative charge on the outer edge of the dielectric facing the positive plates and a net positive charge on the opposite side as shown in figure 4.8f.

Q+

0Er

Fig. 4.8eFig. 4.8e

+ _ + _

+ _

+ _

+ _+ _

+ _

+ _

+ _ + _ + _+ _

+ _+ _

+ _

+ _+ _ + _

+++++++

Q

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{ The electric field lines do not pass through the dielectric but instead end on charges induced on the surface of the dielectric as shown in figure 4.8f. Therefore the electric field within the dielectric is less than in air.

0Er

Fig. 4.8fFig. 4.8f

+++++++

+++

Fig. 4.8gFig. 4.8g

+++++++

+++

0Er

indEr

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SF027 51

{ According to figure 4.8g, the electric field within the dielectric E is given by

{ since

ind0 EEErrr +=

ind0 EEE =EE 0=

EEE 0ind0 =

then

or

= 11EE 0ind

whereplates on the charges the todue field electric :0Eon the charge induced the todue field electric :indE dielectric theof surfaces