Ch 17 student_201516

Topic --- Capacitor & Dielectrics

17.1 Capacitance & Capacitors In Series &

Parallel (1 Hour)

17.2 Charging & Discharging Of Capacitors

(1 Hour)

17.3 Capacitors With Dielectrics (1 Hour)


Ca

pa

cito

r &

Die

lectr

ics

Capacitance

Capacitors

In series

In parallel

Charging & discharging of capacitor

Dielectric


17.1 CAPACITANCE & CAPACITORS IN SERIES & PARALLEL

(a) Define and use capacitance,

(b) Derive and determine the effective

capacitance of capacitors in series and

parallel

(c) Derive and use energy stored in a capacitor,

V

QC

C

QQVCVU

22

2

1

2

1

2

1


Capacitor (Condenser)

A device that is capable of storing electric charges orelectric potential energy

Consist of two conducting plates separated by a small air gap or a thin insulator(dielectric) such as mica, ceramics or even oil

Uses: Photo-flash, on-off switches, smoothen direct current (d.c) voltages

OR


MAKE presents- The Capacitor.mp4

MAKE presents- The Capacitor.mp4


CVQ V

QC

VQ



V

+Q1QIC1,V1

+Q2

Q2C2,V

2+Q3Q3

C3,V3

+Q

QCeff,

V

V321

111

CCCQ

V

eff

1

CQ

V

capacitors in series

Where and

11

11

C

Q

C

QV

22

22

C

Q

C

QV

The magnitude of the charge Q on each plate is the same

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

The total potential difference V is given by

321 QQQQ

33

33

C

Q

C

QV

321 VVVV

321 C

Q

C

Q

C

Q

nC...

CCCC

11111

321eff


V +Q1QI

C1,V1

+Q2

Q2C2,V2

+Q3Q3

C3,V3

+Q

QCeff,

V

V

The potential difference across each capacitoris the same as the supply voltage (V)

321 VVVV

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

The total charge Q on the effective capacitor is given by

where

;1111 VCVCQ ;2222 VCVCQ VCVCQ 3333

321 QQQQ VCVCVC 321

and321 CCC

V

Q effC

V

Q

nC...CCCC 321eff

capacitors in parallel


Calculate the total capacitance in (a), (b) and (c).



Determine the potential difference across and charges on capacitors X, Y and Z.



Determine the effective capacitance of the configuration shown in Figure 17.1. All the capacitors are identical and each has a capacitance of 2 F.

Figure 17.1

μF 11

30eff C OR μF 73.2



In Figure 17.2, C1= 100 F, C2

= 200 F and C3 = 300 F. The applied potential difference between points A and B is VAB = 8.0 V. Calculate

(a) the charge on each capacitor

(b) the potential difference across each capacitor

(c) the potential difference between points A and D.

A

B

D

1C

2C

3C

Figure 17.2

μC 1200123 QQQ

V 0.4321 VVV

V 0.412AD VV



• When a switch in circuit closed, charges begin to accumulate on the plates

• 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

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

dQ

VandVdQdW C

QV

dQC

QdW

QW

dQC

QdW

00

C

QWU

2

2

1

2

2

1CVU

QVU2

1



Figure 17.3 shows a combination of three capacitors where C1= 100 F, C2 = 22 F and C3 = 47 F. A 20 Vsupply is connected to the combination.Determine(a) the effective capacitance in the circuit,(b) the charge stored in the capacitor C1,(c) the potential difference across the

capacitor C2,(d) the energy stored in the capacitor C3,(e) the area of the each plate in capacitor C1 if the distance

between two plates is 0.02 m and the region between plates is vacuum.

(Permittivity of free space, 0 = 8.85 1012 C2 N1 m2)

1C

2C 3C

V 02

Figure 17.3



14

Consider the circuit shown in Figure 17.4, where C1= 50 F, C2 = 25 F and V = 25.0 V.

Capacitor C1 is first charged by

closing a switch S1. Switch S1 is

then opened, and then the

Charged capacitor is connected

to the uncharged capacitor C2 by

closing a switch S2.

Calculate the initial charge acquired by C1 and the final charge on each capacitor.

1C 2CV

1S 2S

Figure 17.4



Given 0 = 8.85 1012 C2 N1 m2

1. Four capacitors are connected as shown in Figure 17.5.

Figure 17.5

Calculate

(a) the equivalent capacitance between points a and b,

(b) the charge on each capacitor if Vab=15.0 V.

(Physics for scientists and engineers,6th

edition,Serway&Jewett, Q21, p.823)

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



2. Determine the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure 17.6.

Figure 17.6

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

(Physics for scientists and engineers,6th

edition,Serway&Jewett, Q27, p.824)

ANS: 6.04 F



17

3. An electronic flash unit for a camera contains a capacitor of capacitance 850 F. When the unit is fully charged and ready for operation, the potential difference across the plates is 330 V.

(a) What is the magnitude of the charge on each plate of the fully charged capacitor?

(b) Calculate the energy stored in the “charged-up” flash unit.(Physics,3rd edition, J.S.Walker, Q59, p.692)

ANS: 0.28 C; 46 J4. A parallel-plate capacitor has plates with an area of 405 cm2 and an

air-filled gap between the plates that is 2.25 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.

(a) How much energy is stored in the capacitor?(b) The separation between the plates is now increased to 4.50 mm.

How much energy is stored in the capacitor now?(c) How much work is required to increase the separation of the

plates from 2.25 mm to 4.50 mm?(Physics,3rd edition, J.S.Walker, Q60, p.692)

ANS: 2.63 105 J; 5.27 105 J; 2.63 105 J



17.2 CHARGING & DISCHARGING OF CAPACITORS

18

(a) Define and use time constant, = RC

(b) Sketch and explain the characteristics of Q-t and I-t

graph for charging and discharging of a capacitor

(c) Use

for discharging and

for charging

RC

t

eQQ

0

RC

t

eQQ 10


Charging Discharging



20

Time constant,

• Scalar quantity

• Unit: s

• A measurementof how quickly the capacitor charges or discharges

RC



• The voltage V across the capacitor, increase from zero at t = 0 to maximum values V0 after a very long time

• is defined as the time required for the capacitor to reach (1 e1) = 0.63 or 63% of its maximum voltage

Charging

0 t,time

V

0V

063.0 V

RCτ

RC

t

eVV 10

00 V

C

Q

RC

t

eC

Q

C

QV 10

and



• The charge Q across the capacitor, increase from zero at t = 0 to maximum values Q0 after a very long time

• is defined as the time required for the capacitor to reach (1 e1) = 0.63 or 63% of its maximum charge

Charging

0 t,time

Q

0Q

063.0 Q

RCτ

RC

t

eQQ 10

00 CVQ and



• the current drops exponentially in time constant

• is defined as the time required for the current drops to 1/e = 0.37 or 37% of its initial value(I0)

Q0: maximum chargeV0: maximum (supply) voltage

I0: maximum currentR: resistance of the resistor

C: capacitance of the capacitor

23

Charging

23

0I

0 t,time

I

037.0 I

RCτ

and

RC

t

eII

0

R

VI 0

0



24

0V

0 t,time

V

037.0 V

RCτ

RC

t

eVV

0

• the charge Q, the voltage Vand the current I is seen to decrease exponentially in time constant

• is defined as the time required for the charge on the capacitor (or voltage across it or current in the resistor) decreases to 1/e = 0.37 or 37% of its initial value

Discharging: V-t, Q-t & I-t graph


Discharging


25

Charge on capacitor Current through resistor

0I

0 t,time

I

RCτ

037.0 I

The charge on the capacitor decreases exponentially with time

The current through the resistor decreases exponentially with time

0Q

0 t,time

Q

037.0 Q

RCτ

RC

t

eQQ

0

RC

t

eII

0


Discharging


In the RC circuit shown in Figure 17.15, the battery has fully charged the capacitor. At time t = 0 s, a switch S is thrown from position a to b. The battery voltage V0 is 12.0 V and the capacitance C = 3.00 F. The current I is observed to decrease to 0.45 of its initial value in 60 s.Determine(a) the value of R.(b) the time constant, (c) the value of Q, the charge on the capacitor at t = 0.(d) the value of Q at t = 100 s

C

R

0V

S

b

a

Figure 17.7


17.3 CAPACITORS WITH DIELECTRICS

(a) Calculate capacitance of air-filled parallel plate

capacitor,

(b) Define and use dielectric constant

(c) Describe the effect of dielectric on a parallel plate

capacitor.

(d) Use capacitance with dielectric, 0CC r

0

rε

εε

d

AεC 0


• Consider a two parallel metallic plates capacitor of equal area A are separated by a distance d and the space between plates is vacuum or air as shown in Figure 17.8.

• When the capacitor is charged, its plates have charges of equal magnitudes but opposite signs: + Q and Q then the potential difference Vacross the plates is produced.

dQ

QV

A

positive

terminal

negative

terminal

E

Figure 17.8



• Since d << A so that the electric field strength E is uniformbetween the plates.

• The magnitude of the electric field strength within the plates is given by

where, : surface charge density on either side (Unit: C m-2 and a scalar quantity)

• Since Q=CV and V=Ed (Uniform E) then equation (17.1) can be written as

0ε

σE

A

Qσ and

0Aε

QE (17.1)

0Aε

CVE

0Aε

CEdE



0 = permittivity of free space = 8.85 x 10-12 C2 N-1 m-2

= permittivity of dielectric materiald = distance between the two plates

A = area of each plate

• From equations (17.2) and (17.3), The capacitance, C of a

parallel-plate capacitor is proportional to the area, A of its plates

and inversely proportional to the plate separation, d.

(17.2)

d

AεC 0

and

(17.3)

d

εAC

Parallel-plate capacitor

separated by a vacuum

Parallel-plate capacitor

separated by a dielectric

material



A parallel-plate capacitor has plates of area 280 cm2 are separated by

a distance 0.550 mm. The plates are in vacuum. If a potential

difference of 20.1 V is supplied to the capacitor, determine

a. 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.



A circular parallel-plate capacitor with radius of 1.2 cm is connected to

a 6.0 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 between plates is 2.5 mm and the medium between plates

is air.

a. Calculate the amount of charge on each plate.

If their separation is increase to 8.0 mm after the battery is

disconnected, determine

b. the amount of charge on each plate.

c. the potential difference between the plates.

d. the capacitance of the capacitor.

(Given permittivity of free space, 0 = 8.85 1012 C2 N1 m2)



1. 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 of 5.60 pC m2. Calculate the potential difference between the plates of the capacitor.

(University physics,11th edition, Young&Freedman, Q24.4, p.934)

b. An electric field of 2.80 105 V m1 is desired between two parallel plates each of area 21.0 cm2 and separated by 0.250 cm of air. Determine the charge on each plate. (Physics for scientist & engineers ,3rd

edition, Giancoli, Q14, p.628) ANS: 1.14 mV; 5.20 109 C

2. When the potential difference between the plates of a capacitor is increased by 3.25 V, the magnitude of the charge on each plate increases by 13.5 C. What is the capacitance of this capacitor?

(Physics,3rd edition, J.S.Walker, Q86, p.694) ANS: 4.15 F



3. A 10.0 F parallel-plate, air-filled 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.

(University physics,11th edition, Young&Freedman, Q24.5, p.934) ANS: 120 C; 60 C; 480 C

4. A capacitor stores 100 pC of charge when it is connected across a potential difference of 20 V. Calculatea. the capacitance of the capacitor,b. the amount of charge to be removed from the capacitor to reduce its potential difference to 15 V.

ANS: 5.0 pF; 25 pC



• Consider a parallel-plate capacitor as shown in Figure 17.19

Figure 17.9

Q Qd

0E



• Initially the plates are separated by a vacuum and connected to a battery, giving the charge on the plates +Q and –Q

• The battery is now removed and the charge on the plates remains constant

• The electric field between the plates is uniform and has a magnitude of E0

• Meanwhile the separation between plates is d

• When a dielectric is placed in the electric field between the plates, the molecules of the dielectric tend to become orientedwith their positive ends pointing toward the negatively charged plate and their negative ends pointing toward the positively charged plated as shown in Figure 17.10

Q Q

Figure 17.10

0E

E



• The result is a buildup of positive charge on one surface of the dielectric and of negative charge on the other as shown in Figure 17.11 Q Q

Figure 17.11

E



38

• From Figure 17.11, the number of field lines within the dielectric is reduced thus the applied electric field E0 is partially canceled.

• Because the new electric field strength (E < E0) is less then the potential difference, V across the plates is less as well.

• Since V is smaller while Q remains the same the capacitance,

is increased by the dielectric.

EdV

V

QC



• is defined as a ratio between the capacitance of a given capacitor with space between plates filled with dielectric, Cand the capacitance of the same capacitor with plates in a vacuum, C0

Mathematically,

No unit for dielectric constant.• For parallel-plates capacitor:

0

rC

Cε (17.10)

d

εAC

d

AεC 0

0

and

d

Aε

d

εA

ε0

r



then the equation (17.10) can be written as

: permittivity of dielectric material• From the definition of the capacitance,

hence the equation (17.10) can be written as

V: potential difference across capacitor with dielectricV0: potential difference across capacitor in vacuum

0

rε

εε 0rεεε OR (17.11)

V

QC

0

0V

QC and where Q is constant

V

Vε 0

r (17.12)



• From the relationship between E and V for uniform electric field,thus the equation (17.12) can be written as

E0: electric field strength of the capacitor in vacuumE0: electric field strength of the capacitor with dielectric

• The dielectric constant depends on the insulating material used.

• Table 17.1 shows the value of dielectric constant and the dielectric strength for several materials.

EdV dEV 00 and

E

Eε 0

r (17.13)

Ed

dE

V

Vε 00

r



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

Material Dielectric constant, r

Dielectric Strength

(106 V m1)

Air 1.00059 3

Mylar 3.2 7

Paper 3.7 16

Silicone oil 2.5 15

Water 80 -

Teflon 2.1 60

Table 17.1

Note:

E

E

V

V

ε

ε

C

Cε 00

00

r



A vacuum parallel-plate capacitor has plates of area A = 150 cm2 and

separation d = 2 mm. The capacitor is charged to a potential difference

V0 = 2000 V. Then the battery is disconnected and a dielectric sheet of

the same area A is placed between the plates as shown in Figure

17.12. In the presence of the dielectric, the potential difference across

the plates is reduced to 500 V. Determine

(a) the initial capacitance of the 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 material,

(f) the initial electric field,

(g) the electric field after the dielectric is inserted.

(0 = 8.85 1012 C2 N1 m2)

Figure 17.12

dielectric

d



A parallel-plate capacitor has the space between the plates filled with two

slabs of dielectric constants 1 and 2 as shown in Figure 17.13.

Each slab has thickness d/2, where d is the plate separation. Show that

the capacitance of the capacitor is

Figure 17.13

21

2102

d

AC

d/2

d/22

1



Given 0 = 8.85 1012 C2 N1 m2

1. What are the maximum and minimum equivalent capacitances that

can be obtained by combinations of three capacitors of 1.5 F, 2.0

F and 3.0 F?

(College Physics,6th edition, Wilson, Buffa & Lou, Q97, p.566)

ANS: 6.5 F; 0.67 F

2. The dielectric of a parallel-plate capacitor is to be constructed from

teflon that completely fills the volume between the plates. The area

of each plate is 0.50 m2.

a. What is the thickness of the teflon if the capacitance is to

be 0.10 F?

b. Calculate the charge on the capacitor if it is connected to a

12 V battery.

(Dielectric constant for teflon is 2.1)

ANS: 92.9 m; 1.2 C



3. Explain clearly why the electric field between two parallel plates of

a capacitor decreases when a dielectric is inserted if the capacitor

is not connected to a power supply, but remains the same when it

is connected to a power supply.

(College Physics,6th edition, Wilson, Buffa & Lou, Q79, p.566)

4. An air-filled parallel-plate capacitor has rectangular plates with

dimensions of 6.0 cm 8.0 cm. It is connected to a 12 V battery.

While the battery remains connected, a sheet of 1.5 mm thick

paper is inserted and completely fills the space between the

plates.

a. Explain briefly what is happen to the charge on the plates

of the capacitor while the dielectric was being inserted.

b. Determine the change in the charge storage of the

capacitor because of the dielectric insertion.

(Dielectric constant for paper is 3.7)

ANS: 0.92 nC



47

Next Chapter…CHAPTER 18 :

Electric current

&

direct-current circuits

Ch 17 student_201516

Education

Transcript of Ch 17 student_201516