lesson 9 capacitance, electrostatic energysdyang/courses/em/lesson09... · 2011. 4. 15. · lesson...

Lesson 9Capacitance, Electrostatic Energy

楊尚達 Shang-Da YangInstitute of Photonics TechnologiesDepartment of Electrical EngineeringNational Tsing Hua University, Taiwan

Outline

CapacitanceElectrostatic energy

Sec. 9-1Capacitance

1. Single-conductor capacitors2. Two-conductor capacitors3. Methods to evaluate capacitance

Single-conductor capacitor-1

0=ρ0=E

v

sec10 19−≈t

0=t


,1

∑=

=n

kkqQ ∑

=

=n

k k

k

RqV

1041πε

On the surface Inside the conductor

Spatial distribution of maintains


,2QQ =′ VRqV

n

k k

k 224

110

==′⇒ ∑=πε

)(rsvρ


The constant ratio of the deposited charge to the resulting potential is defined as the capacitance of a single conductor.

VQC ≡

Example 9-1: Conducting sphere (1)

Find the capacitance of a conducting sphere of radius b

b

Q Assume charge Q is deposited

Spherical symmetry, ⇒

is uniformly distributed, ⇒

)(rsvρ

⎪⎩

⎪⎨

⎧

<<

≥=

bR

bRR

QaE R

0 if ,0

if ,4 2

0πεv

v


The surface potential is:

b

∞

( )

Qb

Q

dRaR

Qa

bRVb

RR

∝=

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛−=

=

∫∞

0

20

4

4

)(

πε

πεvv

bVQC 04πε=≡⇒

Two-conductor capacitor-1

∫ =⋅−→→±→2

1 1212 VldEEQVvvv

Ev2 1

any path

Two-conductor capacitor-2

ErrQrVVv

→±→=′ 1212

The constant ratio of the deposited charge to the voltage difference is defined as the capacitance of the conducting pair:

12VQC ≡

Comment

∞→R

Evaluation of capacitance (Method 1)

1. Assume charges are depositedQ±

2. Find by Gauss’s law orEv

∫+

′′′

=21

20 ),(

)(4

1)(SS

sR sd

rrRrarE vv

vvvv ρ

πε

3. Find by( )QV ∝12 ∫ ⋅−=2

1 12 ldEVvv

4. Find by , independent of QC12V

QC =

Evaluation of capacitance-reference figure

Ev2 1

)(2 rsvρ )(1 rs

vρ

3. Find by


1. Assume between the conductors

2. Find by solving with BC

Ev

6. Find C by , independent of12V

QC =

12V

)(rV v02 =∇ V

)(rVE vv−∇=

4. Find of either conductor by)(rsvρ

0ερ s

nE =

5. Find deposited Q by ( )12 )( VdsrQ

S s ∝= ∫vρ

12V


1. Assume or V for the two conductorsQ±

2. Find and by M1, M2Ev

3. Find the stored energy

22

or iswhich ,

21 VQdvEDW

Ve ∝⎟⎠⎞

⎜⎝⎛ ⋅= ∫ ′

vv

4. Find by orC

Dv

CQWe 2

2

=2

2CVWe =

Example 9-2: Parallel-plate capacitor (1)


2. By Gauss’s law (planar sym.):S

QaE y εvv

−=


3.

4. dS

VQC ε==12

( )S

QddyaEVd

y ε=⋅−= ∫

0 12vv

Example 9-3: Coaxial cable capacitor (1)


2. By Gauss’s law (cylindrical sym.):rL

QaE r πε2vv

=

( )abL

VQC

ln2

12

πε==


3.

4.

( ) ⎟⎠⎞

⎜⎝⎛=⋅−= ∫ a

bL

QdraEVa

b r ln2

12 πεvv

Now you know how to evaluate the capacitance per unit length of coaxial TX lines!

( )abLC

ln2πε

=

Sec. 9-2Electrostatic Energy

1. Energy of charges2. Energy of fields

Energy of two charges-1

RQRV

0

1

4)(

πε=

The work done to move Q2 from ∞ to P2 against V(R) due to Q1 is:

120

21222 4 R

QQVQWπε

==


RQRV

0

2

4)(

πε=′

The work done to move Q1 from ∞ to P1 against V'(R) due to Q2 is:

120

21112 4 R

QQVQWπε

==


The electrostatic energy stored by a system of two charges Q1-Q2 is:

( )221122112 21 VQVQVQVQW +===

Potential of Q1

at P1 due to Q2

Potential of Q2

at P2 due to Q1

Energy of three charges-1

)()( 23133 RVRVV ′+=The extra work done to move Q3 from ∞ to P3

against V(R), V'(R) due to Q1, Q2 is:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

=Δ

230

2

130

13

33

44 RQ

RQQ

VQW

πεπε

⎟⎟⎠

⎞⎜⎜⎝

⎛++=Δ+=

23

32

13

31

12

21

023 4

1R

QQRQQ

RQQWWW

πε


The electrostatic energy stored by a system of three charges Q1-Q2-Q3 is:


⇒ ( )3322113 21 VQVQVQW ++=

,44 130

3

120

21 R

QR

QVπεπε

+=

Vk denotes the potential of charge Qk at position Pk (k =1, 2, 3) due to the remaining charges, i.e.,

,44 230

3

120

12 R

QR

QVπεπε

+=

230

2

130

13 44 R

QR

QVπεπε

+=

Energy of N charges

The electrostatic energy stored by a system of Ndiscrete charges Q1-Q2-…-QN is:

∑=

=N

kkke VQW

121

∑≠

=kj jk

jk R

QV

041πε

Energy of continuous charge distributions

The electrostatic energy stored by a system of continuous charge distribution over V' is:)(rvρ

∫ ′=

Ve dvrVrW

)()(21 vvρ

Potential at source point due to the total charge distribution

rv

Example 9-4: Sphere of uniform charge density (1)

Find the energy stored in a sphere of radius

with uniform volume charge density

b

bρ


ρ

)(REaE Rvv

=

By Gauss’s law, ⇒

if ,

3

0 if ,3

20

30

⎪⎪⎩

⎪⎪⎨

⎧

≥

<<

=bR

Rba

bRRaE

R

R

ερ

ερ

v

v

v


For any point P inside the sphere

P

Rρ

( )bR <<0

( )

( )22

0

23

0

2

3

0

0

20

3

3623

3

33

)()(

RbRRb

RdRRdRb

RdRRdRb

RdaRERV

R

b

b

R

b

b

R

b

b

R

R

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡ ′+

′−−=

⎥⎦

⎤⎢⎣

⎡′′+′

′−=

⎥⎦

⎤⎢⎣

⎡′

′+′

′−=

′⋅′−=

∞

∞

∞

∞

∫∫

∫∫

∫

ερ

ερ

ερ

ερ

ερ

vv∞


5

0

52

0

2

154

)4)((21

bb

dRRRVWb

e

∝=

= ∫

επρ

πρ

∫ ′=

Ve dvrVrW

)()(21 vvρByρ

bQ Total charge is:3

4 3ρπbQ =

bbQWe

120

3

0

2

∝=πε

(if ρ = constant)

(if Q = constant)

b

Energy of electric fields-1

In real applications (especially electromagnetic waves), sources are usually far away from the region of interest, only the fields are given.

)(rvρ ∞→RDEvv

,

ROI


)(rvρ

( )∫∫ ′′⋅∇==

VVe VdvDdvrVrW

21)()(

21 vvvρ

Dr

⋅∇=ρ

V ′

Av

f

( ) ( ) fAAfAf ∇⋅+⋅∇=⋅∇vvv

By vector identity:

( )∫∫ ′′∇⋅−⋅∇=

VVe dvVDdvDVW 2

1)(21 vv

(1)

(2)

contain all the source charges


∫∫ ′′⋅=⋅∇⇒

SVsdDVdvDV

)( vvv( )∫∫ ⋅∇=⋅

VSdvAsdA

,

vvvQ

)(rvρ

V ′ S ′

,VE −∇=v

Q ( ) ( )∫∫ ′′⋅−=∇⋅⇒

VVdvEDdvVD

vvv

( )∫∫ ′′⋅+⋅=

VSe dvEDsdDVW 2

1 21 vvvv

1I 2I

(3)


∞→RS ′

S ′

21

RD ∝v

RV 1

∝

Obs. pt.


0111

4)()(21

21

22

2

1

→∝⋅⋅∝

⋅≈⋅= ∫ ′

RR

RR

RRDRVsdDVIS

πvvv

∫ ′==⇒

V ee dvrwIW 2 )(v

)mJ( 21 3ED

vv⋅ …energy density


∞→RV ′

∫ ′⎟⎠⎞

⎜⎝⎛ ⋅=

Ve dvEDW 2

1 vv

dvEDdWe ⎟⎠⎞

⎜⎝⎛ ⋅=

vv

21


Planar symmetry, ⇒ uniform E-field:

Find the stored electrostatic energy of:

EDvv

ε=,dVaE y

vv−=


Total stored energy:

Energy density:2

2

21

21

21

⎟⎠⎞

⎜⎝⎛==⋅=

dVEEDwe εε

vvv

22

21

21 V

dSSd

dVWe ⎟

⎠⎞

⎜⎝⎛=⋅⎟

⎠⎞

⎜⎝⎛= εε

VQC =

QVC

QCVWe 21

221 2

2 ===⇒



2. By Gauss’s law (cylindrical sym.):rL

QaE r πε2vv

=

Find the capacitance of:


3.

4.

Instead of evaluating V by line integral of ,Ev

2

21

21 EEDwe

vvvε=⋅=

( ) ( )

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛==

∫

∫∫

ab

LQ

rdr

LQ

rdrLLr

QdvEW

b

a

b

ae

ln44

222

121

2

2

22

πεπε

ππε

εεv

( )abL

WQC

e ln2

2

2 πε==5.


Find the energy stored in a sphere of radius

with uniform volume charge density

b

bρ


ρ

)(REaE Rvv

=

By Gauss’s law, ⇒

if ,

3

0 if ,3

20

30

⎪⎪⎩

⎪⎪⎨

⎧

≥

<<

=bR

Rba

bRRaE

R

R

ερ

ερ

v

v

v


Instead of evaluating V by line integral of ,

and arriving at

Ev

∫ ′=

Ve dvrVrW

)()(21 vvρ

bρ

:0For bR <<

0

52

0

4

0

2

0

22

001

452

92

)4(32

1

επρ

επρ

πε

ρε

bdRR

dRRRW

b

b

e

=⎟⎠⎞⎜

⎝⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

∫

∫dV

2Ev


bρ

:For bR >

dV

0

52

20

62

22

2

3

02

921

92

)4(32

1

επρ

επρ

περε

bdRR

b

dRRRbW

b

be

=⎟⎠⎞

⎜⎝⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

∫

∫∞

∞

2Ev

bQbWWW eee

0

2

0

52

21 203

154

πεεπρ

==+=(1:5)


The corresponding “capacitance” is:

bρ

310

20

2 bWQC

e

πε==


Find the stored energy and the capacitance of a conducting sphere of total charge Q

b

Q Spherical symmetry, ⇒

is uniformly distributed, ⇒

)(rsvρ

⎪⎩

⎪⎨

⎧ ≥=

otherwise ,0

if ,4 2

0

bRR

QaE R πε

vv

bQ

dRRR

QWbe

0

2

22

20

0

8

)4(42

1

πε

ππε

ε

=

⎟⎟⎠

⎞⎜⎜⎝

⎛= ∫

∞


Instead of evaluating V by line integral of ,

and arriving at

Ev

∫ ′=

Ve dvrVrW

)()(21 vvρ

:For bR >dV

2Ev

bQ

bWQC

e0

2

42

πε==⇒

lesson 9 capacitance, electrostatic energysdyang/courses/em/lesson09... · 2011. 4. 15. · lesson...

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