C17 Capacitor and Dielectrics Student1

17.1 Capacitors17.2 Capacitors in series and parallel17.3 Charging and discharging of capacitors17.4 Capacitors with dielectrics

UNIT 17CAPACITOR

& DIELECTRICS

(3 hours)

2

17.1 Capacitors (1/2 hour)SUBTOPIC :

LEARNING OUTCOMES :

a) Define capacitance.

b)Use formulae

b)Calculate the capacitance of parallel plate capacitor.

At the end of this lesson, the students should be able to :

QC

V

17.1 Capacitors

3

•A capacitor , sometimes called a condenser, is a device that can store electric charge.•It is consists of two conducting plates separated by a small air gap or a thin insulator (called a dielectric such as mica, ceramics, paper or even oil).•The electrical symbol for a capacitor is

or

17.1 Capacitors

4

Capacitance, C•The ability of a capacitor to store charge is measured by its capacitance.•Capacitance is defined as the ratio of the charge on either plate to the potential difference between them.

V

QC

charge on one of the plates:Q potential difference across the two plates:V

17.1 Capacitors

5

• The unit of capacitance is the farad (F). 1 farad is the capacitance of a capacitor if the

charge on either of the plates is 1C when the potential difference across the capacitor is 1V.i.e.

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

where the capacitance of a capacitor, C is constant then

1 coulomb1 farad=

1 volt V

QC

CVQ

VQ (The charges stored, Q is directly proportional to the

potential difference, V across the conducting plate.)

17.1 Capacitors

6

• One farad (1F) is a very large unit. • Therefore in many applications the

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

F10μF1 6 F10pF1 12 μF10 6

μμF1

7

Parallel-plate Capacitors

• A parallel–plate capacitor consists of a pair of parallel plates of area A separated by a small distance d. • If a voltage is applied to a capacitor (connected to a battery), it quickly becomes charged.• One plate acquires a negative charge, the other an equal amount of positive charge and the full battery voltage appears across the plates of the capacitor(12 V).

17.1 Capacitors

17.1 Capacitors

8

• The capacitance of a parallel-plate capacitor, C is

d

AC 0

or

d

AC

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

Parallel-plate capacitor Parallel-plate capacitor separated by a dielectric separated by a dielectric materialmaterial

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

0 permittivity of free space :ε

area of the plate:A distance between the two plates :d

0 = 8.85 x 10-12 C2 N-1 m-2

17.1 Capacitors

9

Example 17.1

a)Calculate the capacitance of a capacitor whose plates are 20 cm x 3.0 cm and are separated by a 1.0-mm air gap.

b)What is the charge on each plate if the capacitor is connected to a 12-V battery?

c) What is the electric field between the plates?

17.1 Capacitors

10

Example 17.2

An electric field of 2.80 x 105 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. (Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2)

17.1 Capacitors

11

Exercise

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.

-134.42 x 10 F @ 0.44 pFC -108.84 x 10 CQ

5 -1 -12.50 x 10 N C @V mE

12

17.2 Capacitors in series and parallel(1hour)

SUBTOPIC :

LEARNING OUTCOMES :

a) Deduce and use the effective capacitance of capacitors in series and parallel.

b) Derive and use equation of energy stored in a capacitor.


17.2 Capacitors in series and parallel

V

+Q+Q - Q- Q

CCeqeq,V,V

equivalent to

•Figure above shows 3 capacitors connected in

series to a battery of voltage, V.•When the circuit is completed, the electron from the battery (-Q) flows to one plate of C3 and this plate become negatively charge.

V1 V2 V3

Q1 Q2 Q3

Capacitors connected in series


• This negative charge induces a charge +Q on the other plate of C3 because electrons on one plate of C3 are repelled to the plate of C2. Hence this plate is charged –Q, which induces a charge +Q on the other plate of C2.• This in turn produces a charge –Q on one plate of C1 and a charge of +Q on the other plate of of capacitor C1.•Hence the charges on all the three capacitors are the same, Q. • The potential difference across capacitor C1,C2 and C3 are

;11

11 C

Q

C

QV ;

22

22 C

Q

C

QV

33

33 C

Q

C

QV


• The total potential difference V is given by

• If Ceq is the equivalent capacitance, then

321 VVVV

321 C

Q

C

Q

C

QV

321 C

1

C

1

C

1

Q

V

1 2 3

1 1 1 1 1...

eq nC C C C C

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

capacitors connected in series


Capacitors connected in parallel

V

+Q+Q -Q-Q

CCeqeq,V,Vequivalent to

• Figure above shows 3 capacitors connected in parallel to a battery of voltage V. • When three capacitors are connected in parallel to a battery, the capacitors are all charged until the potential differences across the capacitors are the same.


• If not, the charge will flow from the capacitor of higher potential difference to the other capacitors until they all have the same potential difference, 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 charge on each capacitor is

321 VVVV

17

1 1 1 1Q C V C V

2 2 2 2Q C V C V VCVCQ 3333


• The total charge is

and

321 QQQQ VCVCVCQ 321

321 CCCV

Q eqC

V

Q

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

1 2 3 ...eq nC C C C C capacitors connected in parallel


Example 17.3

50 V C1 = 1µF C2 = 2µF

In the circuit shown above, calculate thea)charge on each capacitorb)equivalent capacitance


Example 17.4

In the circuit shown below, calculate thea)equivalent capacitance b) charge on each capacitorc) the pd across each capacitor

1 2

1 2 50

V V V

V V

50 V

C1 = 1µF C2 = 2µF

V1 V2


Example 17.5

In the circuit shown below, calculate thea)equivalent capacitance b)charge on each capacitor c) the pd across each capacitor

12 V

C1 = 6.0µF

C2 = 4.0µF

V1 V2 =V3

C3 = 8.0µF

12 V

C1 = 6.0µF C23 = 12.0µF

V1 V2


Example 17.6

Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in figure below.

Take C1 = 5.00 F, C2 = 10.0 F C3 = 2.00 F.


Solution 17.6

C1 = 5.00 F, C2 = 10.0 F and C3 = 2.00 F.

Series bSeries a

parallel

C12 C12

C22


Solution 17.6

C1 = 5.00 F, C2 = 10.0 F and C3 = 2.00 F.

• b

• a

C12 C12C3

C22

Parallel

Ca


Solution 17.6

• b

• a

Ca

C22

series

Ceq

17.2 Capacitors in series and parallelExample 17.7

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

1 F

1 F

1 F

1 F

1 F

1 F


Solution 17.7

1 F

1 F

1 F

1 F

1 F

1 F

Ca

series

CCaa

1 F

1 F

1 Fseries

CCbb


Solution 17.7

parallel

CeqCeqCCbb 1 F


Exercise

CC22

aa

CC11

CC33

bbdd

1. 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.

3 221 Q μCμC147Q2

1 73 8 .Q μC

V624V3 .V936VV 21 .

36 9 .adV V


2. Four capacitors are 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. 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.


3. A 3.00-µF and a 4.00-µF capacitor are connected in series and this combination is connected in parallel with a 2.00-µF capacitor.

a) What is the net capacitance? b) If 26.0 V is applied across the whole

network, calculate the voltage across each capacitor.

3.71-µF, 26.0 V, 14.9 V, 11.1 V


Energy stored in a capacitor, U

• A charged capacitor stores electrical energy.• The energy stored in a capacitor will be equal to the work done to charge it.• A capacitor does not become charged instantly. It takes time.• Initially, when the capacitor is uncharged , it requires no work to move the first bit of charge over.• When some charge is on each plate, it requires work to add more charge of the same sign because of the electric repulsion.• The work needed to add a small amount of charge dq, when a potential difference V is across the plates is,

dW Vdq


• Since V=q/C at any moment , where C is the capacitance, the work needed to store a total charge Q is

0

0

2

0

2

1

1

2

1

2

Q

Q

Q

W Vdq

W qdqC

qW

C

QW

C

• Thus the energy stored in a capacitor is 21

2

QW U

C

2CV2

1U

QV2

1U

or

or


Example 17.8

A camera flash unit stores energy in a 150 µF capacitor at 200 V. How much energy can be stored?


Example 17.9

A 2 µF capacitor is charged to 200V using a battery. Calculate thea)charge delivered by the batteryb)energy supplied by the battery.c)energy stored in the capacitor.

17.2 Capacitors in series and parallelExercise

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.

1C

4.00 V

2C

2.00 µF

8.00 µCV1 = 2.67 V, V2 = 1.33 V

U1 = 1.07 x 10 -5 J, U2 = 5.31 x 10-6 J

3.39 m 2

37

SUBTOPIC :

LEARNING OUTCOMES :

a) Define and use time constant, τ = RC.b)Sketch and explain the characteristics of Q-t and I-t

graph for charging and discharging of a capacitor.c)Use formula for discharging and for charging.


17.3 Charging and discharging of capacitors (1 hour)

/t RCoQ Q e

1 /( )t RCoQ Q e

17.3 Charging and discharging of capacitorsCharging a capacitor through a resistor

0V

R

S,switch

CAB

e

e

•Figure below shows a simple circuit for charging a capacitor.•When the switch S is closed, current Io 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.

17.3 Charging and discharging of capacitors

• 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, Vo.

• 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 Qo.

0

0Q

time s, ( )t

Charge C, ( )Q

0Q630.

RCτ 0

0I

time s, ( )t

Current A, ( )I

0I370.

RCτ

The charge on the capacitor increases exponentially with time

The current through the resistor decreases exponentially with time

Charge on charging Charge on charging capacitor :capacitor :

RC

t

0 e1QQ

Current in resistor :Current in resistor :

RC

t

0eII

where

0 maximum charge :Q

maximum current:oI resistance of the resistor :R capacitance of the capacitor :C


Discharging a capacitor through a resistor

R

S,switch

CAB

e

e

• Figure below shows a simple circuit for discharging a capacitor.

0V

• When a capacitor is already charged to a voltage Vo and it is allowed to discharge through the resistor R as shown in figure below.

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


• 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.


0

0Q

time s, ( )t

Charge C, ( )Q

0Q370.

RCτ

0

0I

time s, ( )t

Current A, ( )I

0I370.

RCτ

The charge on the capacitor decreases exponentially with time.

The current through the resistor decreases exponentially with time.

Charge on discharging Charge on discharging capacitor :capacitor :

Current in resistor :Current in resistor :

RC

t

0eQQ

RC

t

0eII

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

For calculation of current in discharging process, ignore the negative sign in the formula.


Time constant, • It is a measure of how quickly the capacitor

charges or discharges.• Its formula, .• Its unit is second (s).

RC

Charging process• The time constant is defined as the time required for the capacitor to reach 0.63 or 63% of its maximum charge (Qo).• The time constant is defined as the time required for the current to drop to 0.37 or 37% of its initial value(I0).

RC

t

0 e1QQ

when t=RC

0 1

1 0 37

0 63

.

.

RC

RC

o

o

Q Q e

Q Q

Q Q

RC

t

0eII

0

0 37.

RC

RC

o

I I e

I I

when t=RC


RC

t

0eII

0

0 37.

RC

RC

o

I I e

I I

when t=RC

Discharging process• The time constant is defined as the time required for the charge on the capacitor/current in the resistor decrease to 0.37 or 37% of its initial value.

0

t

RCQ Q e

0

0 37.

RC

RC

o

Q Q e

Q Q

when t=RC


Example 17.10Consider the circuit shown in figure below, where C1= 6.00 F, C2 = 3.00 F and V = 20.0 V.

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.


Solution 17.1017.3 Charging and discharging of capacitors

Solution 17.10


Example 17.11 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 s

C

R

0V

Sb

a


Solution 17.11


51

SUBTOPIC :

LEARNING OUTCOMES :

a) Define dielectric constant.

b)Describe the effect of dielectric on a parallel plate capacitor.

c)Use formula


17.4 Capacitors with dielectrics

r oC C


• A dielectric is an insulating material. Hence no free electrons are available in it.• When a dielectric (such as rubber, plastics, ceramics, glass or waxed paper) is inserted between the plates of a capacitor, the capacitance increases.• The capacitance increases by a factor or

r which is called the dielectric constant

(relative permittivity) of the material.


• Two types of dielectric : i) non-polar dielectricFor an atom of non-polar dielectric, the center of the negative charge of the electrons ‘coincides’ with the center of the positive charge of the nucleus.* It does not become a permanent dipole.

ii) polar dielectric* Consider the molecule of waters. * Its two positively charge hydrogen ions are ‘attracted’ to a negatively charged oxygen ion.* Such an arrangement of ions causes the center of the negative charge to be permanently separated slightly away from the center of the positive charge, thus forming a permanent dipole.

++ - -+

• Dielectric constant, (r) is defined as the ratio between the capacitance of given capacitor with space between plates filled with dielectric, C with the capacitance of same capacitor with plates in a vacuum, C0.

0r

C

C

d

εAC

d

AC 0

0

0r

εAd

ε A

d

00

or r r

εε ε

ε

permittivity of dielectric material:


0r

C

C

• From the definition of the capacitance,

V

QC

00 V

QC and Q is constant

0r

V

V

where

0 potential difference across capacitor in vacuum:V potential difference across capacitor with dielectric:V

• From the relationship between E and V for uniform electric field, EdV dEV 00 and

0r

E d

Ed

0r

E

E 0 electric field strength of the capacitor in vacuum:E

where

electric field strength of the capacitor with dielectric:E


Material Dielectric constant, εr Dielectric Strength

(106 V m-1)

Air 1.00059 3

Mylar 3.2 7

Paper 3.7 16

Silicone oil 2.5 15

Water 80 -

Teflon 2.1 60

• The dielectric strength is the maximum electric field before dielectric breakdown (charge flow) occurs and the material becomes a conductor.

0 0

0 0r

V EC

C V E


Example17.12

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


Example 17.12

In the presence of the dielectric, the potential difference across the plates is reduced to 1000 V. Determinea) 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.d)the relative permittivity.e)the permittivity of dielectric sheet.the initial electric field.the electric field after the dielectric is inserted. (Given permittivity of free space, 0 = 8.85 x 10-12 F m-1)


Solution 17.12


Dielectric effect on the parallel-plate capacitor

In part a, the region between the charged plates is empty. The field lines point from the positive toward the negative plate

In part b, a dielectric has been inserted between the plates. Because of the electric field between the plates, the molecules of the dielectric (whether polar or non-polar) will tend to become oriented as shown in the figure, the negative ends are attracted to the positive plate and the positive ends are attracted to the negative plate. Because of the end-to-end orientation, the left surface of the dielectric become negatively charged, and the right surface become positively charged.


• Because of the surface charges on the dielectric, not all the electric field lines generated by the charges on the plates pass through the dielectric. • As figure c shows, some of the field lines end on the negative surface charges and begin again on the positive surface charges.

0r

E

E

• Thus, the electric field inside the dielectric is less strong than the electric field inside the empty capacitor, assuming the charge on the plates remains constant. • This reduction in the electric field is described by the dielectric constant εr which is the ratio of the field magnitude Eo without the dielectric to the field magnitude E inside the dielectric:


Quantity Capacitor without

dielectric

Capacitor with

dielectric

Relationship

Electric field

Eo E E < Eo

Potential difference

Vo V V < Vo

Charge Qo Q Q = Qo

Capacitance Co C C > Co

0 0

0 0r

V EC

C V E


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C17 Capacitor and Dielectrics Student1