chapter 26 capacitance 26-1 definition of capacitance 26-2 calculating capacitance 26.3 combinations...

Norah Ali Al moneef 2

Chapter 26 capacitance

26-1 Definition of Capacitance

26-2 Calculating Capacitance

26.3 Combinations of Capacitors

26.4 Energy Stored in a Charged Capacitor

26.5 Capacitors with Dielectrics


Capacitor:Two isolated conductors Equal and opposite charges ±Q Potential difference DV between them.

Units: Coulombs/Volt or Farads

QC

V

C is Always Positive

Experiments show the quantity of electric charge Q on a capacitor is linearly proportional to the potential difference between the conductors, that is Q ~ DV. Or we write Q = C DV

The capacitance C of a capacitor is the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between them:

26.1 definition of capacitanceConsider two conductors carrying charges of equal magnitude and opposite sign, as shown in the Figure. Such a combination of two conductors is called a capacitor.The conductors are called plates. A potential difference Δ V exists between the conductors due to the presence of the charges


Example• A storage capacitor on a random access memory

(RAM) chip has a capacitance of 55 nF. If the capacitor is charged to 5.3 V, how many excess electrons are on the negative plate?

We can find the number of excess electrons on the negative plate if we know the total charge q on the plate. Then, the number of electrons n=q/e, where e is the electron charge in coulomb.

:The charge q of the plate is related to the voltage V to which the capacitor is charged: q=CV.


A 0.75 mF capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor?

V = 16 V, C = 0.75 mF = 0.75 x 10-6 F C = Q/V Q = CVQ = (0.75 x 10-6)(16) = 1.2 x 10-5 C

Example


Capacitance of an isolated sphere

Calculate the capacitance of an isolated spherical conductor of radius R and charge Q by assuming that the second conductor making up the capacitor is a concentric hollow sphere of infinite radius.

Q

Electric potential of the sphere of radius R is

kQ/R and V= 0 at infinity, we have

C = Q

DV=

Q

kQ/R=

R

k= 4peo R

C is proportional to its radius and independent of both the charge on the sphere and the potential difference.

26.2 Calculating Capacitance

The capacitance of a pair of conductors depends on the geometry of the conductors.


Example

(a) If a drop of liquid has capacitance 1.00 pF, what is its radius ? (b) If another drop has radius 2.00 mm, what is its capacitance ? (c) What is the charge on the smaller drop if its potential is 100V ?

C = 4πeo R

R = (9 x 109 N · m2/C2)(1.00 x 10–12 F) = 9 mm

C = 4 π (8.85 x 10-12) x 2.0x10-3 = 0.222 pF

Q =CV = 0.222 pF x 100 V = 2.22 x 10-11 C


ExampleWhat is the capacitance of the Earth ?

Think of Earth spherical conductor and the outer conductor of the “spherical capacitor” may be considered as a conducting sphere at infinity where V approaches zero.

m 1037.6mNC 1085.84

46212

0

RC e

C= 7.08 x 10-4 F

A large capacitor !


: A 0.75 mF capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor?

V = 16 V, C = 0.75 mF = 0.75 x 10-6 F

C = Q/V or Q = CV

Q = (0.75 x 10-6)(16) = 1.2 x 10-5 C

Example


Parallel - Plate CapacitorsA parallel-plate capacitor consists of two

parallel conducting plates, each of area A, separated by a distance d. When the capacitor is charged, the plates carry equal amounts of charge. One plate carries positive charge, and the other carries negative charge. And surface charge density of each plate is s = Q/A

The plates are charged by connection to a battery..


0 0 0 02 2

QE

A

+ + + + + + + + + + + + + +Top Sheet:

02E

02E

- - - - - - - - - - - - - -Bottom Sheet: 02E

02E

0E

0E

Q A

Q A

d?E


Variation with A If plates are large, then charges can distribute themselves over a substantial area, and the amount of charge that can be stored on a plate for a given potential diff increases as A is increased. Thus we expect C to be proportional to A C ~ A

Variation with d Potential difference DV constant across, E field increases as d decreases. Imagine d decreases and consider situation before any charges have had a chance to move in response to this change. Because no charge move E the same but over a shorter distance. DV = Ed means that DV decreases

The difference between this new capacitor voltage and the terminal voltage of the battery now exists as a potential difference across the wires connecting the battery to the capacitor.


A E field result in the wires that drives more charge onto the plates, increasing the potential diff. DV until it matches that of the battery. potential diff. Across wire = 0 flow of charges stop.

More charges has accumulated at the capacitor as a result.

We have d decrease, Q increases. Similarly d increases Q decreases.

Capacitance inversely proportional to d.

C ~ 1/d


Assume electric field uniform between the plates, we have

(see lecture on Gauss’s Law)

Only the geometry of the plates (A and d) affect the capacitance.

The general properties of a parallel-plate capacitor – that the capacitance increases as the plates become larger and decreases as the separation increases – are common to all capacitors.

• Charge density: A

Q

• Electric field:A

QE

00

• Potential diff.: A

QdEdVab

0

1

• Capacitance:d

A

V

QC

ab0

a

b

b

abaab EddEdEVVV


(a) The electric field between the plates of a parallel-plate capacitor is uniform near the center but nonuniform near the edges.

(b) Electric field pattern of two oppositely charged conducting parallel plates.

Capacitance configurations

sphere individualan of ecapacitanc

- /,With

)(

)11

(2

e

e

e

b

a

e

kaCb

abk

abC

baQk

r

drQkV

Cylindrical capacitor

)ln(2

)ln(22

ab

k

lC

a

b

l

Qk

r

drkV

e

e

b

a

e

Spherical Capacitance


Example• A huge parallel plate capacitor consists of two

square metal plates of side 50 cm, separated by an air gap of 1 mm

• What is the capacitance?

C = e0A/d

= (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m)= 2.21 x 10–9 F


Example

• We have a parallel plate capacitor constructed of two parallel plates, each with area 625 cm2 separated by a distance of 1.00 mm.

• What is the capacitance of this parallel plate capacitor?

Result: A parallel plate capacitor constructed out of square conducting plates 25 cm x 25 cm separated by 1 mm produces a capacitor with a capacitance of about 0.5 nF.

F 1053.5

m 0.001

m 0.0625 F/m108.85

10

2-120

dA

C

C = 0.553 nF


Example• We have a parallel plate capacitor constructed of two

parallel plates separated by a distance of 1.00 mm.• What area is required to produce a capacitance of 1.00 F?

Result: A parallel plate capacitor constructed out of square conducting plates 10.6 km x 10.6 km (6 miles x 6 miles) separated by 1 mm produces a capacitor with a capacitance of 1 F.

28

12-0

m 1013.1

F/m 108.85

m 0.001 F 1

Cd

A


• What is the AREA of a 1 F capacitor that has a plate separation of 1 mm?

d = 1 mm = 0.001 m, eo = 8.85 x 10-12 C2/(Nm2)

Sides

A

Ax

D

AC o

001.01085.81 12

1.13 x 108 m2

10629 m

Example.


: Changing Dimensions

1. V increases, Q increases

2. V decreases, Q increases3. V is the same, Q increases4. V increases, Q is the same5. V decreases, Q is the

same6. V is the same, Q is the same7. V increases, Q

decreases8. V decreases, Q

decreases

9. V is the same, Q decreases

A parallel-plate capacitor has plates with equal and opposite charges ±Q, separated by a distance d, and is connected to a battery. The plates are pulled apart to a distance D > d. What happens?

example


With a battery connected to the plates the potential V between them is held constant

In this situation, sinceV = E d

As d increases, E must decrease.

Since the electric field is proportional to the charge on the plates, Q must decrease as well.

Answer: 9. V is the same, Q decreases


A parallel plate capacitor is constructed with plate of an area of 0.028 m2 and a separation of 0.55 mm. Find the magnitude of the charge of this capacitor when the potential difference between the plate is 20.1 V.

A = 0.028 m2 d = 0.55 mm = 0.00055 m, V = 20.1 VQ = CVC = eo A/d C = (8.85 x 10 -12)(0.028) /(0.00055)= 4.51 x 10-10 FQ = (4.51 x 10-10 )(20.1) = 9.06 x 10-9 C

Example.


Example

Assume that charge builds up on the cloud and on the ground until a uniform electric field of 3.00 x 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold?

C= eoA/d = 8.85x10-12 x (1000)2/800 = 11.1 nFPotential between ground and cloud is DV = Ed = 3.0 x106 x 800 = 2.4 x 109 VQ = C(DV) = 26.6 C

Regarding the Earth and a cloud layer 800 m above the Earth as the “plates” of a capacitor, calculate the capacitance if the cloud layer has an area of 1.00 x 1.00 km2. Assume that the air between the cloud and the ground is pure and dry.


26.3 Combinations of Capacitors


Parallel Combination

For parallel capacitors

n

jjeq CC

1

21

21

VVV

QQQ

VCVCQ

VCVCQ

2222

1111

VCQ eq21 CCCeq

• When a potential difference V is applied across several capacitors connected in parallel, that potential difference V is applied across each capacitor.

• The total charge q stored on the capacitors is the sum of the charges stored on all the capacitors.

• Capacitors connected in parallel can be replaced with an equivalent capacitor that has the same total charge q and the same potential difference V as the actual capacitors.


•When a potential difference V is applied across several capacitors connected in series, the capacitors have identical charge q. •The sum of the potential differences across all the capacitors is equal to the applied potential difference V. •Capacitors that are connected in series can be replaced with an equivalent capacitor that has the same charge q and the same total potential difference V as the actual series capacitors.

n

j jeq CC 1

11For series capacitors

21

21

QQQ

VVV

22

22

11

11

C

Q

C

QV

C

Q

C

QV

eqC

QV

21

111

CCCeq

Capacitors in Series

SERIES: • Q is same for all capacitors• Total potential difference = sum of V


Example: Equivalent Capacitance

In parallel use C=C1+C2

In series use 1/C=1/C1+1/C2

6.00 mF

20.00 mF

2.50 mF

8.50 mF

20.00 mF


5.965 mF



In series use 1/CB=1/C+1/C+1/C

In series use 1/CA=1/C+1/C C

C/2

C/3

In parallel use Ceq=C+C/2+C/3

Example:Step 1:

Cp=C1+C2

Cp=0.10 mF+0.20 mF

Cp =0.30 mF

Step 1

Step 2

Step 2:

1/Cs=1/C3+1/Cp

F20.0FF60.0

FF60.0

p3s

CC

CCC p3


Find the Charge stored in the capacitor.

12)(1020( 6

Q

CVQV

QC

12V 20μF

CQ

VV

CQ

CVQV

QC

4

6

104.2

)12)(1020(

Example


Example Find the total charge stored in each the capacitor.

12V 10μF

20μF

30μF

First calculate the total capacitance.

Then find the total charge

In a series circuit the charge stored on each capacitor is the same and the voltage is split.

CVFCVQ

V

QC

56 1054.6)12)(1045.5(

FCC

CC

ss

i is

45.530

1

10

1

20

11

11


Example Find the total charge stored in all the capacitors.

12V10μF20μF 30μF

First find the total capacitance.

calculate the total charge.

In a parallel circuit the charge stored on each capacitor can be different and the voltage must be the same.

CVFCVQ

V

QC

46 102.7)12)(1060(

FC

CC

p

iip

60301020


Example Find the total capacitance of all the capacitors.

12V

12μF

8μF4μF

First find the total capacitance of the parallel capacitors

FC p 1248 combine this value in series with

12μF capacitor.

FCC total

total

6 12

1

12

11

Energy Stored in a Capacitor

2622

121062

1

2

1

22

1 CV

C

QQVUC

Units: Joules (J)


½ C 2 C1/3 C

3 C

C2/5 C

Example.

Example

C3

C1 C2

2112 CCC 312123

111

CCC

312

312123 CC

CCC

V

C3

C12

Step 1

V C123

Step 2

V

parallel

series

C1 = 12.0 mF, C2 = 5.3 mF, C3 = 4.5 mF

C123 = (12 + 5.3)4.5/(12+5.3+4.5) mF = 3.57 mF

Example

C3

C1

C2

V

C3

C4

C5

C6

C3

C6

series

54

5445 CC

CCC

C45

parallel

64511456 CCCC

parallel

3223 CCC

Example

C1456

C23

Vseries

231456

231456123456 CC

CCC

54

5445 CC

CCC

64511456 CCCC

3223 CCC

231456

231456123456 CC

CCC

Complete solution

32654

541

32654

541

123456

)(

CCCCC

CCC

CCCCC

CCC

C


26.4 Energy Stored in Capacitor

Suppose that, at a given instant, a charge q has ′been transferred from one plate of a capacitor to the other. The potential difference DV between the plates at that instant will be q / C.

If an extra increment of charge dq is then transferred, the increment of work required will be,

dW = DV dq - +


Energy To Charge Capacitor

1. Capacitor starts uncharged.2. Carry +dq from bottom to top.

Now top has charge q = +dq, bottom -dq

3. Repeat4. Finish when top has charge q =

+Q, bottom -Q

+q

-q

At some point top plate has +q, bottom has –q

Potential difference is DV = q / C

Work done lifting another dq is dW = dq DV


So work done to move dq is:

Total energy to charge to q = Q:

dW dq V

0

1Q

W dW q dqC

1qdq q dq

C C

21

2

Q

C

A plot of potential difference versus charge for a capacitor is a straight line having a slope 1/C. The work required to move charge dq through the potential difference DV across the capacitor plates is given by the area of the shaded rectangle. The total work required to charge the capacitor to a final charge Q is the triangular area under the straight line, W = QDV/2.

1V = 1 J/C hence the unit for the area is joule J.


Energy Stored in Capacitor

QC

V

Since

22

2

1

2

1

2VCVQ

C

QU

Where is the energy stored???

Work done in charging the capacitor = electric potential energy U stored in the capacitor.


Energy Stored in Capacitor

2

field energy density2

oE

Eu E

21

2U CV

Parallel-plate capacitor: ando AC V Ed

d

Energy stored in the E field!

2

21( )

2 2o oA E

Ed Add

( )Eu volume

Ad = Volume occupied by the E field. This lead to a new quantity known as Energy Density uE = U/Volume = U/Ad

The energy stored in a capacitor can be put to a number of uses: a camera flash; a cardiac defibrillator; and others. In addition, capacitors form an essential part of most electrical devices used today.


Example

d

A

- - - - -+ + + +

• Suppose the capacitor shown here is charge to Q and then the battery is disconnected.

• Now suppose I pull the plates further apart so that the final separation is d1.

• How do the quantities Q, C, E, V, U change?

• How much do these quantities change?

• Q:• C:• E:• V:• U:

remains the same.. no way for charge to leave.

increases.. add energy to system by separating

decreases.. since capacitance depends on geometry

increases.. since C ¯, but Q remains same (or d but E the same)remains the same... depends only on charge density

Answers: C1 d

d1

C V1 d1

dV

U

1

d1

dU


• Suppose the battery (V) is kept attached to the capacitor.

• Again pull the plates apart from d to d1.

• Now what changes?

• C:• V:• Q:• E:• U:

decreases (capacitance depends only on geometry)

must stay the same - the battery forces it to be V

must decrease, Q=CV charge flows off the plate

d

A

- - - - -+ + + +V

• How much do these quantities change?.. exercise for student!!

Answers: U

1

d

d1

U

C

1

d

d1

C E

1

d

d1

E

must decrease ( , ) E

E

0 E

V

Dmust decrease ( )

U 1

2CV 2

Example.


A parallel-plate capacitor, disconnected from a battery, has plates with equal and opposite charges, separated by a distance d. Suppose the plates are pulled apart until separated by a distance D > d. How does the final electrostatic energy stored in the capacitor compare to the initial energy?

1. The final stored energy is smaller2. The final stored energy is larger3. Stored energy does not change.

example


As you pull apart the capacitor plates you increase the amount of space in which the E field is non-zero and hence increase the stored energy. Where does the extra energy come from? From the work you do pulling the plates apart.

Answer: 2. The stored energy increases


Example

A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled ?

U = Q2/2C and C = oA/d and d2 = 2 d1 then C1= C2/2 and the energy stored doubles.


In a typical defibrillator, a 175 mF, is charged until the potential difference between the plates is 2240 V. A.) What is the charge on each plate?

V = 2240 V, C = 175 mF = 175 x 10-6 F Q = CV = (175 x 10-6)(2240) = 0.392 C

B.) Find the energy stored in the charged up defibrillator.U = ½ CV2 U = Q2/(2C) U = ½ QV = ½ (0.392)(2240)U = 439 J

example


Consider the circuit as shown, where C1 = 6.00mF and C2= 3.00 mF and DV =20.0V. Capacitor C1 is first charged by 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.

Example

S1 close, S2 open C = Q/V Q = 120 mCAfter S1 open, S2 close Q1 + Q2 = 120 mCSame potential Q1 /C1 = Q2 / C2 (120-Q2)/C1= Q2/C2

(120 - Q2)/6 = Q2/ 3 Q2 = 40 mC

Q 1= 80 mC

52

--Find the charge on (Q) and potential difference (V) across each capacitor. What is the total energy stored in the system?

example

Cequ C2 C8 (parallel)

= 2F +8F

=10F

1

Cequ

1

C5

1

C10

1

C2

(series)

1

5F

1

10F

1

2F

8

10F

Cequ 5F

4

53

Q CV

Qtot (5

4F)(12V ) 15C

Q5 Q10 Q2 Qtot 15C (in series)

V5 Q5

C5

15C

5F3V

V10 Q10

C10

15C

10F1.5V

V2 Q2

C2

15C

2F7.5V

VC QC

CC

(always)

V2 = V8 = V10 =1.5V (in parallel)

QC CCVC (always)

Q2 C2V2 (2F )(1.5V ) 3C

Q8 C8V8 (8F )(1.5V ) 12C

UC Q2

2C

1

2C(V )2

(15F)2

2(5 /4F)90J


26.5 Capacitors with Dielectrics

A dielectric is a nonconducting material, such as rubber, glass, or waxed paper. Dielectric constant is a property of a material and varies from one material to another.

A charged capacitor (a) before and (b) after insertion of a dielectric between the plates. The charge on the plates remains unchanged, but the potential difference decreases from DVo to DV = DVo/k.

Thus the capacitance increases from Co to k Co.

Note no battery is involved in this example.


When a dielectric is inserted between the plates of a capacitor, the capacitance increases.

If the dielectric completely fills the space between the plates, the capacitance increases by a dimensionless factor k , which is called the dielectric constant.

Capacitance increases by the factor k when dielectric completely fills the region between the plates.

If the dielectric is introduced while the potential difference is being maintained constant by a battery, the charge increases to a value

Q = k Qo . The additional charge is supplied by the battery and the

capacitance again increases by the factor k.

For parallel plate capacitor: C = keoA d


Dielectrics

• Definition:The dielectric constant of a material is the ratio of the capacitance when filled with the dielectric to that without it:

k values are always > 1 (e.g., glass = 5.6; water = 78)They INCREASE the capacitance of a capacitor (generally good, since

it is hard to make “big” capacitors)They permit more energy to be stored on a given capacitor than

otherwise with vacuum (i.e., air):

C

C0

U

CV 2

2C

0V 2

2U

V

V0

E

E0


Dielectric Strength

. If magnitude of the electric field in the dielectric exceeds the dielectric strength, then the insulating properties break down and the dielectric begins to conduct.

If the electric field in a dielectric becomes too large, it can tear the electrons off the atoms, thereby enabling the material to conduct. This is called dielectric breakdown; the field at which this happens is called the dielectric strength. For any given separation d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the dielectric strength (maximum electric field) of the dielectric.


the advantages of dielectric material in a capacitor?

•Increase the capacitance

• Increase the maximum operating voltage

• Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C.


Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected it is filled with a dielectric.

Compare the voltages of the two capacitors.

a) V1 > V2 b) V1 = V2 c) V1 < V2

Compare the electric fields between the plates of both capacitors.

a) E1 > E2 b) E1 = E2 c) E1 < E2

Example.


When we insert the dielectric into the capacitor C2 we do:

a) positive work

Recall the meaning of negative work is the energy of the system is reduced. The dielectric is sucked into the capacitor. When the charge is constant, the total energy of the capacitor decreases because the presence of the dielectric increases the capacitance . It turns out that the dielectric is pulled in even if the voltage is held constant, e.g., via a battery. On a microscopic scale the force on the dielectric arises due to “fringing fields” at the edges of the capacitor.

b) negative work c) no work

Example.


Examples: (a) A parallel-plate capacitor is connected to a power supply that maintains a constant potential difference V0 between the two plates. If the gap between the plates is first filled with air (≈ vacuum) and then a dielectric with dielectric constant κ=3 is inserted, how do the following quantities change:•The capacitance of the capacitor?•The charge on the positive plate of the capacitor?•The electric field between the capacitor plates?•The energy stored in the capacitor?

If V doesn’t change then: if C↑3 Q↑3

d

AC 0

If A and d don’t change then: if κ ↑3 C↑3

CVQ

Two identical parallel plate capacitors are connected in series to a battery as shown below. If a dielectric is inserted in the lower capacitor, which of the following increase for that capacitor?

A. I and III.B. I, II and IV.C. I, II and III.D. All except II.E. All increase.

I. Capacitance of capacitor

II. Voltage across capacitor

III. Charge on capacitor

IV. Energy stored on capacitor

CVq d

AC 0 2

21

2

2CV

C

qU

V

C

C

k

Example.

A Closer Look

V

C

C

k

• Insert dielectric

kC

q

qq’

q’V

V

Capacitance goes up by k

Charge increases

Charge on upper plate comes from upper capacitor, so its charge also increases.

Since q’ = CV1 increases on upper capacitor, V1 must increase on upper capacitor.

Since total V = V1 + V2 = constant, V2 must decrease.

You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant.

What effect does adding the dielectric have on the energy stored in the capacitor?

A. The stored energy increases.

B. The stored energy remains the same.

C. The stored energy decreases.

D. not enough information given to decide

Example.

You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant.

What effect does adding the dielectric have on the amount of charge on each of the capacitor plates?

A. The amount of charge increases.

B. The amount of charge remains the same.

C. The amount of charge decreases.


Example.

You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant.

What effect does adding the dielectric have on the energy stored in the capacitor?

A. The stored energy increases.

B. The stored energy remains the same.

C. The stored energy decreases.


Example.


A parallel-plate capacitor is attached to a battery that maintains a constant potential difference V between the plates. While the battery is still connected, a glass slab is inserted so as to just fill the space between the plates. The stored energy

1. increases.2. decreases.3. remains the same.

Example.


1. Increases 2. Decreases3. Stays the Same

A parallel plate capacitor is charged to a total charge Q and the battery removed. A slab of material with dielectric constant k in inserted between the plates. The charge stored in the capacitor

+ + + + + + +

- - - - - - -

k

Since the capacitor is disconnected from a battery there is no way for the amount of charge on it to change.

Answer: 3. Charge stays the same

Example.


1. Increases 2. Decreases3. Stays the Same

A parallel plate capacitor is charged to a total charge Q and the battery removed. A slab of material with dielectric constant k in inserted between the plates. The energy stored in the capacitor

+ + + + + + + +

- - - - - - - -

k

The dielectric reduces the electric field and hence reduces the amount of energy stored in the field. The easiest way to think about this is that the capacitance is increased while the charge remains the same so U = Q2/2CAlso from energy density:

Answer: 2. Energy stored decreases

2

2,0 0 0 ,0

1 1

2 2E E

Eu E u

Example.

Calculating Capacitance Units

1 F = 1 C2/N m (Note [e]=C2/N m2)

e0 = 8.85 x 10-12 F/m

1 mF = 10-6 F, 1 pF = 10-12 F

Example : Size of a 1-F capacitor ,calculate its area

F 0.1 , mm 1 Cd

2812

3

0

m 101.1F/m 1085.8

m) 100.1F)(0.1(

Cd

A

Calculate the Capacitance ,the charge on the plates and the electric field

Example

kV 10.0 V 000,10 , m 00.2 , mm 00.5

in vacuumcapacitor palte-parallelA 2 VAd

F 0.00354F 1054.3

m1000.5

)m F/m)(2.00 1085.8(

5

3

212

0

d

AC

C 35.4C1054.3

V) 10C/V)(1.00 1054.3(5

49

abCVQ

N/C 1000.2

)m 00.2)(mN/C 1085.8(

C 1054.3

6

22212

5

00

A

QE

128.85 10 F/m r AC

d

Find the capacitance of a 4.0 cm diameter sensor immersed in oil if the plates are separated by 0.25 mm.

The plate area is

The distance between the plates is

4.0 for oilr

3 2

123

4.0 1.26 10 m 8.85 10 F/m

0.25 10 mC

30.25 10 m

178 pF

2 2 3 2π 0.02 m 1.26 10 mA r

Example.

Capacitors in Series and Parallel

example

(a) Express the capacitance C0 in terms of the potential difference V0

between the plates and the charge Q if air is between the plates.(b) Express the dielectric constant k in terms of the capacitance C0 (air gap) and the capacitance C with material of the dielectric constant k). (c) Using the results of (a) and (b), express the ratio of the potential difference V/V0 if Q is the same, where V is the potential difference between the plates and a dielectric material dielectric constant is k fills the space between them.(d) A voltmeter reads 45.0 V when placed across the capacitor. When dielectric material is inserted completely filling the space, the voltmeter reads 11.5 V. Find the dielectric constant of this material.

In this problem you try to measure dielectric constant of a material. Firsta parallel-plate capacitor with only air between the plates is charged byconnecting it to a battery. The capacitor is then disconnected from thebattery without any of the charge leaving the plates.

(a)00 /VQC

(b) 0/ CC(c) /1/// 000 VVCCVV

(d) From (c) 91.35.11/0.45/0 VV

If a 22 mF capacitor is connected to a 10 V source, the charge is

220 mC

Example.

If a 0.001 mF capacitor is connected in series with an 800 pF capacitor, the total capacitance is 444 pF

0 .001 µ F 800 pF

C 1 C 2

Example.

1800 pF

If a 0.001 mF capacitor is connected in parallel with an 800 pF capacitor, the total capacitance is

0 .001 µ F 800 pF

C 1 C 2

Example.

If a 0.015 mF capacitor is in series with a 6800 pF capacitor, the total capacitance is

a. 1568 pF

b. 4678 pF

c. 6815 pF

d. 0.022 mF

Example.

Two capacitors that are initially uncharged are connected in series with a dc source. Compared to the larger capacitor, the smaller capacitor will have

a. the same charge

b. more charge

c. less voltage

d. the same voltage

Example.

(a) Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm and are separated by a 1.0-mm air gap. (b) What is the charge on each plate if a 12-V battery is connected across the two plates? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1 F, given the same air gap d.

Solution: a. C = 53 pF.b. Q = CV = 6.4 x 10-10 C.c. E = V/d = 1.2 x 104 V/m.d. A = C d /ε0 = 108 m2.

Example.

Determine the capacitance of a single capacitor that will have the same effect as the combination shown.

Solution: First, find the equivalent capacitance of the two capacitors in parallel (2C); then the equivalent of that capacitor in series with the third (2/3 C).

Example.

Example

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

Solution: a. U = ½ CV2 = 3.0 J.

The plates of a parallel-plate capacitor have area A, separation x, and are connected to a battery with voltage V. While connected to the battery, the plates are pulled apart until they are separated by 3x. (a) What are the initial and final energies stored in the capacitor? (b) How much work is required to pull the plates apart (assume constant speed)? (c) How much energy is exchanged with the battery?

a. U = ½ CV2, and increasing the plate separation decreases the capacitance, so the initial and final energies can be calculated from this.b. The work is equal to the negative of change in energy.c. The energy of the battery is increased by the work done minus the change in potential energy of the capacitor (which is negative).

Example.

. (b) The dielectric is carefully removed, without changing the plate separation nor does any charge leave the capacitor. Find the new values of capacitance, electric field strength, voltage between the plates, and the energy stored in the capacitor.

ExampleA parallel-plate capacitor, filled with a dielectric with K = 3.4, is connected to a 100-V battery. After the capacitor is fully charged, the battery is disconnected. The plates have area A = 4.0 m2 and are separated by d = 4.0 mm

(a) Find the capacitance, the charge on the capacitor, the electric field strength, and the energy stored in the capacitor.

Solution: a. C = Kε0A/d = 3.0 x 10-8 F. Q = CV = 3.0 x 10-6 C. E = V/d = 25 kV/m. U = ½ CV2 = 1.5 x 10-4 J.B. Now C = 8.8 x 10-9 F, Q = 3.0 x 10-6 C (no change), V = 340 V, E = 85 kV/m, U = 5.1 x 10-4 J. The increase in energy comes from the work it takes to remove the dielectric.

Since Q = CV and the two capacitors are

identical, the one that is connected to

the greater voltage has more charge,

which is C2 in this case.

Capacitor C1 is connected across a

battery of 5 V. An identical

capacitor C2 is connected across a

battery of 10 V. Which one has

more charge?

1) C1

2) C2

3) both have the same charge

4) it depends on other factors

Example

Since Q = CV, in order to increase the charge that a

capacitor can hold at constant voltage, one has to

increase its capacitance. Since the capacitance is

given by , that can be done by either

increasing A or decreasing d.

1) increase the area of the plates

2) decrease separation between the plates

3) decrease the area of the plates

4) either (1) or (2)

5) either (2) or (3)

dAC 0

What must be done to a

capacitor in order to

increase the amount of

charge it can hold (for a

constant voltage)?

+Q –Q

Example

Since the battery stays connected, the voltage

must remain constant! Since

, when the spacing d is doubled, the

capacitance C is halved. And since Q = CV, that

means the charge must decrease.

+Q –Q

dAC 0

A parallel-plate capacitor

initially has a voltage of 400 V

and stays connected to the

battery. If the plate spacing is

now doubled, what happens?

1) the voltage decreases

2) the voltage increases

3) the charge decreases

4) the charge increases

5) both voltage and charge change

Example

Once the battery is disconnected, Q has to remain

constant, since no charge can flow either to or from

the battery. Since , when the

spacing d is doubled, the capacitance C is halved.

And since Q = CV, that means the voltage must

double.

A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage?

1) 100 V

2) 200 V

3) 400 V

4) 800 V

5) 1600 V

+Q –Q

dAC 0

Example

The 2 equal capacitors in series add up as

inverses, giving 1/2C. These are parallel to

the first one, which add up directly. Thus,

the total equivalent capacitance is 3/2C.

o

o

C CCCeq

1) Ceq = 3/2C

2) Ceq = 2/3C

3) Ceq = 3C

4) Ceq = 1/3C

5) Ceq = 1/2C

What is the equivalent capacitance, Ceq ,

of the combination below?

Example

1) V1 = V2

2) V1 > V2

3) V1 < V2

4) all voltages are zero

C1 = 1.0 mF C3 = 1.0 mF

C2 = 1.0 mF

10 V

The voltage across C1 is 10 V. The

combined capacitors C2 + C3 are

parallel to C1. The voltage across C2

+ C3 is also 10 V. Since C2 and C3 are

in series, their voltages add. Thus the voltage across C2 and C3 each

has to be 5 V, which is less than V1.

How does the voltage V1 across the

first capacitor (C1) compare to the

voltage V2 across the second capacitor

(C2)?

Follow-up: What is the current in this circuit?

Example

C1 = 1.0 mF C3 = 1.0 mF

C2 = 1.0 mF

10 V

We already know that the voltage

across C1 is 10 V and the voltage

across both C2 and C3 is 5 V each.

Since Q = CV and C is the same for

all the capacitors, we have V1 > V2

and therefore Q1 > Q2.

1) Q1 = Q2

2) Q1 > Q2

3) Q1 < Q2

4) all charges are zero

How does the charge Q1 on the first

capacitor (C1) compare to the charge

Q2 on the second capacitor (C2)?

Example

How much energy is stored in a 2.0 mF capacitor that has been charged to 5000 V? What is the average power dissipation if the capacitor is discharged in 10 ms?

The spacing between the plates of a 1.0 mF capacitor is 1 mm.(a)What is the surface area A of the plates?(b)How much charge is on the plates if this capacitor is attached to a 1.5 V battery?

Two flat parallel plates are d = 0.40 cm apart. The potential difference between the plates is 360 V. The

electric field at the point P at the center is approximately

A.90 kN/C. B.180 N/C. C.0.9 kN/C. D.Zero. E.3.6 ´ 105 N/C

Two large metallic plates are parallel to each other and charged. The distance between the plates is d. The

potential difference between the plates is V. The magnitude of the electric field E in the region between the

plates and away from the edges is given by

A. d/V. B. V2/d. C. d V. ∙ D. V/d2. E. V/d .

A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V. If the charge Q on the plates is doubled to

2Q, A. the capacitance becomes (1/2)V. B. the capacitance becomes 2C. C. the potential changes to (1/2)V. D.the potential changes to 2V. E. the potential does not change.

If a capacitor of capacitance 2.0 µF is given a charge of 1.0 mC, the potential difference across

the capacitor is

A.0.50 kV. B.2.0 V. C.2.0 µV. D.0.50 V. E.None of these is correct.

If the area of the plates of a parallel-plate capacitor is doubled, the capacitance is

A.not changed. B.doubled. C.halved. D.increased by a factor of 4. E.decreased by a factor of 1/4.

An 80-nF capacitor is charged to a potential of 500 V. How much charge accumulates on each

plate of the capacitor?

A.4.0 ´ 10–4 C B.4.0 ´ 10–5 C C.4.0 ´ 10–10 C D.1.6 ´ 10–10 C E.1.6 ´ 10–7 C

As the voltage in the circuit is increased (but not to the breakdown voltage), the capacitance

A.increases. B.decreases. C.does not change.

Doubling the potential difference across a capacitor

A.doubles its capacitance. B.halves its capacitance. C.quadruples the charge stored on the capacitor. D.halves the charge stored on the capacitor. E.does not change the capacitance of the

capacitor.

If the area of the plates of a parallel plate capacitor is halved and the separation between the plates tripled, then by what factor does the

capacitance change? A.It increases by a factor of 6. B.It decreases by a factor of 2/3. C.It decreases by a factor of 1/6. D.It increases by a factor of 3/2. E. It decreases by a factor of ½.

Which of the following statements is false?

A. In the process of charging a capacitor, an electric field is produced between its plates.

B. The work required to charge a capacitor can be thought of as the work required to create the electric field between its plates.

C. The energy density in the space between the plates of a capacitor is directly proportional to the first power of the electric field.

D. The potential difference between the plates of a capacitor is directly proportional to the electric field.

E. None of these is false.

Which of the following statements about a parallel plate capacitor is false?

A.The two plates have equal charges of the same sign.B. The capacitor stores charges on the plates.C. The capacitance is proportional to the area of the

plates.D. The capacitance is inversely proportional to the

separation between the plates.E. A charged capacitor stores energy.

If you increase the charge on a parallel-plate capacitor from 3 µC to 9 µC and increase the plate separation

from 1 mm to 3 mm, but keep all other properties the same, the energy stored in the capacitor changes by a

factor of A.27. B.9. C.3. D.8. E.1/3.

The energy stored in a capacitor is directly proportional to

A. the voltage across the capacitor. B. the charge on the capacitor. C. the reciprocal of the charge on the capacitor. D.the square of the voltage across the capacitor. E. None of these is correct.

A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates

are separated by a distance d = 2 mm and a voltage applied between the plates. The electric field strength within the plates is E = 4000 V/m. The energy stored in

the capacitor is

A.0.71 nJ. B.1.42 nJ. C.2.83 nJ. D.3.67 nJ. E.Zero.

A circuit consists of a capacitor, a battery, and a switch, all connected in series. Initially, the switch is open and the capacitor is uncharged. The switch is then closed

and the capacitor charges. While the capacitor is charging, how does the net charge within the battery

change?

A.It increases. B.It decreases. C.It stays the same

Several different capacitors are hooked across a DC battery in parallel.

The charge on each capacitor is A.directly proportional to its capacitance. B.inversely proportional to its capacitance. C.independent of its capacitance.

Several different capacitors are hooked across a DC battery in parallel. The voltage across each capacitor is

A.directly proportional to its capacitance. B.inversely proportional to its capacitance. C.independent of its capacitance.

Several different capacitors are hooked across a DC battery in series. The charge on

each capacitor is

A.directly proportional to its capacitance. B.inversely proportional to its

capacitance. C.independent of its capacitance.

Several different capacitors are hooked across a DC battery in series. The voltage across each capacitor is

A.directly proportional to its capacitance. B.inversely proportional to its

capacitance. C.independent of its capacitance.

If C1 < C2 < C3 < C4 for the combination of capacitors shown, the equivalent capacitance is

A.less than C1.

B.more than C4.

C.between C1 and C4.

The equivalent capacitance of two capacitors in series is

A.the sum of their capacitances. B.the sum of the reciprocals of their

capacitances. C.always greater than the larger of their

capacitances. D.always less than the smaller of the

capacitances. E.described by none of the above.

The equivalent capacitance of three capacitors in series is

A.the sum of their capacitances. B.the sum of the reciprocals of their

capacitances. C.always greater than the larger of their

capacitances. D.always less than the smaller of the

capacitances. E.described by none of the above.

The equivalent capacitance of two capacitors in parallel is

A.the sum of the reciprocals of their capacitances.

B.the reciprocal of the sum of the reciprocals of their capacitances.

C.always greater than the larger of their capacitances.

D.always less than the smaller of the two capacitances.

E.described by none of the above.

The capacitance of a parallel-plate capacitor

A.is defined as the amount of work required to move a charge from one plate to the other.

B.decreases if a dielectric is placed between its plates.

C.is independent of the distance between the plates.

D.has units of J/C. E.is independent of the charge on the capacitor.

A capacitor is connected to a battery as shown. When

a dielectric is inserted between the plates of the

capacitor, A.only the capacitance changes. B.only the voltage across the capacitor

changes. C.only the charge on the capacitor changes. D.both the capacitance and the voltage change. E.both the capacitance and the charge change.

Two identical capacitors A and B are

connected across a battery, as shown. If

mica (k = 5.4) is inserted in B,

A. both capacitors will retain the same charge. B. B will have the larger charge. C. A will have the larger charge. D. the potential difference across B will increase. E. the potential difference across A will increase.

If a dielectric is inserted between the plates of a parallel-plate capacitor that is connected to a

100-V battery, the A.voltage across the capacitor decreases. B.electric field between the plates decreases. C.electric field between the plates increases. D.charge on the capacitor plates decreases. E.charge on the capacitor plates increases.

A charged capacitor has an initial electric field E0 and potential difference V0 across its plates. Without connecting any source of emf, you insert a dielectric (k > 1) slab between the plates to produce an

electric field Ed and a potential difference Vd across the capacitor. The pair of statements that best represents the relationships between the

magnitude of the electric fields and potential differences is

A. Ed > E0; Vd > V0. D. Ed < E0; Vd > V0.

B. Ed = E0; Vd > V0. E. Ed < E0; Vd < V0.

C. Ed > E0; Vd = V0.

Does the capacitance always increase when a dielectric is inserted into the gap of

a capacitor?

A.Yes, it always increases.B.No, it always decreases. C.No, it may increase or decrease depending on

the dielectric constant of the material.

An external electric field, E, is applied to a region which contains a dielectric. Which of the following statements

is true?

A. The electric field within the dielectric is less than E.B. The dielectric produces an electric field in the opposite

direction to E.C. The molecules in the dielectric become polarized.D. The electric field will produce a torque on molecules in

the dielectric that have permanent dipoles.E. All the above statements are true.

In real life we want to store more charge at lower voltage, hence large capacitances are needed

Increased area, decreased separations, “stronger” insulators

Electronic circuits – like a shock absorber in the car, capacitor smoothes power fluctuations

Response on a particular frequency – radio and TV broadcast and receiving

Undesirable properties – they limit high-frequency operation

Summary• Capacitance says how much charge is on an

arrangement of conductors for a given potential. • Capacitance depends only on geometry

– Parallel Plate Capacitor

– Cylindrical Capacitor

– Spherical Capacitor

– Isolated Sphere• Units, F (farad) = C2/Nm or C/V (note e0 = 8.85 pF/m)• Energy and energy density stored by capacitor.

CVq

d

AC 0

)/ln(2 0 ab

LC

ab

abC

04

RC 04

221 CVU 2

021 Eu

Phys 133 -- Chapter 30 131

Capacitors: series equivalent

Q1 Q2 Qequivalent

V1 V2 Vequivalent

Q1

C1

Q2

C2

Qequivalent

Cequivalent

1

C1

1

C2

1

Cequivalent

132

Capacitors: parallel equivalent

Q1 Q2 Qequivalent

V1 V2 Vequivalent

C1V C2V2 CequivalentVequivalent

C1 C2 Cequivalent

133

series parallel

V V1+V2= Veq V1=V2

Q Q1=Q2 Q1+Q2= Qeq

1

Ceq

1

C1

1

C2

Ceq C1 C2

Capacitors

n

jjeq CC

1

n

j jeq CC 1

11

Dielectrics change the potential difference

• The potential between to parallel plates of a capacitor changes when the material between the plates changes.

K C

C0

V V0

K

E E0

K

k is a unitless number

Field lines as dielectrics change

• Moving from part (a) to part (b) of Figure 24.15 shows the change induced by the dielectric.

K0

C K0

A

d A

d

u 1

2E 2

In dielectric

In vacuum, energy density is

u 1

20E 2


• With battery attached, V=const, so more charge flows to the capacitor

With battery disconnected, q=const, so voltage (for given q) drops.

CVq

CVq

C

qV

C

qV


, capacitors with and without dielectrics

• If capacitor is disconnected from circuit, inserting a dielectric changes decreases electric field, potential and increases capacitance, but the amount of charge on the capacitor is unchanged.

• If the capacitor is hooked up to a power supply with constant voltage, the voltage must remain the same, but capacitance and charge increase

Applications of Capacitors – Camera Flash• The flash attachment on a camera uses a capacitor

– A battery is used to charge the capacitor– The energy stored in the capacitor is released when the button

is pushed to take a picture– The charge is delivered very quickly, illuminating the subject

when more light is needed• 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 slowly charged and then discharged quickly to provide large amounts of energy in a short pulse

chapter 26 capacitance 26-1 definition of capacitance 26-2 calculating capacitance 26.3 combinations...

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