1

Lecture 5 Capacitance Ch. 25

•Cartoon - Capacitance definition and examples.•Opening Demo - Discharge a capacitor•Warm-up problem•Physlet •Topics

•Capacitance•Parallel Plate Capacitor•Dielectrics and induced dipoles•Coaxial cable, Concentric spheres, Isolated sphere•Two side by side spheres•Energy density•Graphical integration•Combination of capacitance

•Demos•Super VDG•Electrometer•Voltmeter• Circular parallel plate capacitor•Cylindrical capacitor•Concentric spherical capacitor•Dielectric Slab sliding into demo •Show how to calibrate electroscope

2

CapacitanceDefinition of capacitance

A capacitor is a useful device in electrical circuits that allows us to store charge and electrical energy in a controllable way. The simplest to understand consists of two parallel conducting plates of area A separated by a narrow air gap d. If charge +Q is placed on one plate, and -Q on the other, the potential difference between them is V, and then the capacitance is defined as .

The SI unit is , which is called the Farad, named after the famous and creative scientist Michael Faraday from the early 1800’s.

Applications

Radio tuner circuit uses variable capacitor

Blocks DC voltages in ac circuits

Act as switches in computer circuits

Triggers the flash bulb in a camera

Converts AC to DC in a filter circuit

VQC =

VQ

3

Parallel Plate Capacitor

4

Electric Field of Parallel Plate Capacitor

EA =qε0

EAq 0ε=AqE0ε

=

V =Ed =qdε0A

Area of plate =A

+ + + + + + + + + + +

- - - - - - - - - - - E

+ q

- q

C =qV

=q

qdε0A

Coulomb/Volt = Farad

dA

Gaussian surface

C =ε0Ad

Vf −V i =−

rE⋅dr̂

i

f

∫Integrate from - charge to + charge so that

rE ⋅dr̂=−Edr

Vf −V i =+ Edr−

+

∫ =Ed

V =Ed

5

Circular parallel plate capacitor

r

r

d

r = 10 cm = 0.1m

A = r2 = (.1m)2

A = .03 m 2

d = 1 mm = .001 m

C =ε 0Ad

F103C 10−×=

pF300C = p = pico = 10-12

Show Demo Model, calculate its capacitance, and show how to charge it up with a battery.

mmC

NmC

001.03.)10(

211

2

2−= }Farad Volt

Coulomb

6

Demo Continued

Demonstrate

1. As d increases, voltage increases.

2. As d increases, capacitance decreases.

3. As d increases, E0 and q are constant.

7

Dielectrics• A dielectric is any material that is not a conductor, but polarizes well.

Even though they don’t conduct they are electrically active– Examples. Stressed plastic or piezo-electric crystal will produce a

spark. – When you put a dielectric in a uniform electric field (like in between

the plates of a capacitor), a dipole moment is induced on the molecules throughout the volume. This produces a volume polarization that is just the sum of the effects of all the dipole moments. If we put it in between the plates of a capacitor, the surface charge densities due to the dipoles act to reduce the electric field in the capacitor.

8

Permanent dipoles Induced dipoles

++_ _

E0 = the applied field

E’ = the field due to induced dipoles

E = E0 - E’

9

Dielectrics

The amount that the field is reduced defines the dielectric constant from

the formula , where E is the new field and E0 is the old

field without he dielectric.

Since the electric field is reduced and hence the voltage difference is reduced (since ), the capacitance is increased.

where is typically between 2 – 6 with water equal to 80.

Show demo dielectric slab sliding in between plates. Watch how capacitance and voltage change. Also show aluminum slab.

0EE =

VdE =

00

CVQ

VQC

=⎟⎠⎞⎜⎝

⎛==

10

VqC =

V =E0d

00

εσ

=E

Aq=σ

AqE0

0ε

=

V =qdε 0A

C =ε 0Ad

0EE =

V =E0

d

0VV =

VqC =

0VqC

=

0CC =

d

11

Find the capacitance of a ordinary piece of coaxial cable (TV cable)

€

Er = 2kλr

Va −V b =−

rE. dr̂

b

a

∫ = Edrb

a

∫ =2λdrrb

a

∫ =+2λλnrb

a

Va is higher than Vb

LQ=λ

€

k = 14πε 0

airVa −V b =2λλn

ba

rE. dr̂ =Edrcoσ180 =−Edr

r̂ Vf −V i =−

rE⋅dr̂

i

f

∫

Integrate from b to a or - to +

12

capacitance of a coaxial cable cont.

abln

L2QV ,So

0ε=

ab0

lnQL2Q

VQC ε==

ab0

lnL2C ε=

ab

0

ln2

LC ε=

38.1106

4ln106

LC 1111 −− ×=×=

mpF

LC 43=

mpF

LC 86=

ε0 (for air)

For = 2

a = 0.5 mm

b = 2.0 mm

2

Now if a=0.5mm and b=2.0mm, then

And if = 2, then

13

Model of coaxial cable for calculation of capacitance

Signal wire

Outer metal braid

- to +

14

Capacitance of two concentric spherical shells

Integration path

Va−V b =+ Edrb

a

∫ =+qr2b

a

∫ dr=+qdrr2b

a

∫

C =4ε0a

rE. dr̂ =Edσcoσ180 =−Edr

ab)ab(kq)

b1

a1(kq

r1kqV

a

b

−=−==

dr

b

a

+q

-q

E

V =V a−V b =−

rE⋅dr̂

b

a

∫ =+ Edrb

a

∫

as

Let b get very large. Then for an isolated sphere

C =q / V =4ε0

abb−a

15

Spherical capacitor or sphere

Recall our favorite example for E and V is spherical symmetry

The potential of a charged sphere is with V = 0 at r = .

The capacitance is

RkR

RkQQ

VQC 04ε====

Where is the other plate (conducting shell)?

It’s at infinity where it belongs, since that’s where the electric lines of flux terminate.k = 1010 and R in meters we have

€

C = R1010 =10−10R(m) =10−12R(cm)

pFcmRC )(=

Earth: C = (6x108 cm)pF = 600 F

Marble: 1 pF

Basketball: 15 pF

You: 30 pFDemo: Show how you measured capacitance of electroscope

RkQV =

Q

R

16

Capacitance of one charged conducting sphere of radius a relative to another oppositely charged sphere of radius a

d

C = 4ε0a (1+m+m2+m3+m4+…..)

m= a/d

d >>a

a a

d

d =20 cma =10 cmm =0.5C=10-10(.1) (1+.5 +.25 +.125….)C=10-10(.1)(1/(1-m))C= 0.2 x 10-10 FC= 0.02 nF =20 pF

If d gets very large, then C= 10 pF

17

Electric Potential Energy of CapacitorAs we begin charging a capacitor, there is initially no potential difference

between the plates. As we remove charge from one plate and put it on the other, there is almost no energy cost. As it charges up, this changes.

At some point during the charging, we have a charge q on the positive plate.

The potential difference between the plates is

As we transfer an amount dq of positive charge from the negative plate to the positive one, its potential energy increases by an amount dU.

.dqCqVdqdU ==

The total potential energy increase is

CQ

Cqdq

CqU

Q

22

22

0

===∫

Also CQCVQVU

22

21

21

21

=== using VQC =

CqV =

18

Graphical interpretation of integration

∫=Q

VdqU0

where

Qdq

q/c

q

V

V = q/c

∑=

Δ=N

i

ii qqC

U1

1= Area under the triangle

Area under the triangle is the value of the integral dqCqQ∫0

Area of the triangle is also =

€

dU =Vdq

€

Area = 12

(b)(h) = 12

(Q)(QC

) = 12Q2

C

CqV =

hb⋅21

qCdq =

q2

2 0

Q

0

Q

∫ =Q2

2C

19

Where is the energy stored in a capacitor?

• Find energy density for parallel plate capacitor. When we charge a capacitor we are creating an electric field. We can think of the work done as the energy needed to create that electric field. For the parallel plate capacitor the field is constant throughout, so we can evaluate it in terms of electric field E easily.

Use U = (1/2)QV

AQEεε

σεσ

===0

and We are now including dielectric effects: ε

Solve for Q = εAE, V = ES and substitute in

)(21))((

21

21 2 ASEESAEQVU εε ===

volume occupied by Eηε == 2

21 E

ASU

2

21 Eεη =

Electrostatic energy density general result for all geometries.

To get total energy you need to integrate over volume.

ESV =

20

How much energy is stored in the Earth’s atmospheric electric field?(Order of magnitude estimate)

atmosphere

h

R

Earth

20 km

R = 6x106 m

210100 == VE

VolumeEU ⋅= 20

21 ε

hRVolume 24=318426 106.8)102()106(4 mVolume ×=××=

)106.8)(10)(10(21 318211

2

2 mU mV

NmC ×= −

JU 11103.4 ×=This energy is renewed daily by the sun. Is this a lot?

The total solar influx is 200 Watts/m2

dayJsJUsun211626 102102)106(14.3200 ×=×≅×⋅=

10102 −×≅σunUUOnly an infinitesimal fraction gets converted to electricity.

World consumes about 1018 J/day. This is 1/2000 of the solar flux.

21

Parallel Combination of Capacitors

Typical electric circuits have several capacitors in them. How do they combine for simple arrangements? Let us consider two in parallel.

We wish to find one equivalent capacitor to replace C1 and C2. Let’s call it C.

The important thing to note is that the voltage across each is the same and equivalent to V. Also note what is the total charge stored by the capacitors? Q.

VCCVCVCQQQ )( 212121 +=+=+=

2121 CCCCCVQ

+=⇒+=

VQC =

Q1Q2

22

Series Combination of Capacitors

What is the equivalent capacitor C?

Voltage across each capacitor does not have to be the same.

The charges on each plate have to be equal and opposite in sign by charge conservation.

The total voltage across each pair is:

)1()11(2121

21C

QCC

QCQ

CQVVV =+=+=+=

So 211

111CCC

+= Therefore, 21

21

CCCCC+

=

Q QV1 V2

CQVVQC

=

=

23

Sample problem

C1 = 10 F

C2 = 5.0 F

C3 = 4.0 F

a) Find the equivalent capacitance of the entire combination.

C1 and C2 are in series.

21

2112

2112

111CCCCC

CCC +=⇒+=

FC 3.31550

510510

12 ==+×

=

C12 and C3 are in parallel.

FCCCeq 3.70.43.3312 =+=+=

24

Sample problem (continued)

C1 = 10 F

C2 = 5.0 F

C3 = 4.0 F

b) If V = 100 volts, what is the charge Q3 on C3?

C = Q/V

100100.4 633 ⋅×== −VCQ

c) What is the total energy stored in the circuit?

CoulombsQ 43 100.4 −×=

JVFVCU eq22462 106.310103.7

21

21 −− ×=×××==

JU 2106.3 −×=