25. Capacitance25. Capacitance

• Capacitor : storing charges, e.g. Photoflashconsists two conductors, which are charged oppositely.

Opposite Charges.

• Capacitance : Capacitor 의용량 , C

∆V

VqC

∆≡

VCF 11 =

C

Unit : farad (F)

∆V : 1 Coulomb per Voltq

VCq ∆⋅=

1F is too large. => Most capacitor units: µF (10-6 F) or pF (10-12 F)

: Typical Capacitor

• Charging a Capacitor :

25-3. Calculating the Capacitance• Spherical Conductor

+q Rq

RqkV e

041πε

==∆

RkR

VqC

e04πε==

∆=

Assume another conductor at ∞.

RC ∝

• Calculating the capacitance

(i) Assume stored charges +q and –q in the two conductors

(ii) Use Gauss’s Law to find E-field and ∆V between two conductorsqAdE =⋅∫rr

0ε

VqC

∆=(iii) Determine the capacitance

• Parallel Plate Capacitor Supposed stored charges +q and –q

Electric field

qAdE =⋅∫rr

0ε

AqE

AAE

00

0

εεσ

σε

==

=

Potential difference ∆V

dA

qV ⋅=∆0ε

Capacitance C

dA

VqC 0ε

=∆

= d,A 1∝

mpFmFmNC /85.8/1085.8/1085.8 1222120 =×=⋅×= −−εPermittivity :

For A = 0.012 m2 = 10-4 m2, d = 1 mm= 10-3 m C = 0.885 pF

• A Cylindrical Capacitor

λπεεε ⋅=⋅⋅=⋅=Φ ∫ LELrAdEc 2000

rr

rr

E ˆ2

1

0

λπε

=r

drr

drEV b

a

a

b

λπε

=⋅−=∆ ∫∫02

1

Lq

=λ ⎟⎠⎞

⎜⎝⎛

πελ

=abln

02⎟⎠⎞

⎜⎝⎛=

abLq ln

2 0πε

Linear charge density λ =q/L

L

Capacitance C( )ab

LVqC

/ln2 0πε

=∆

=

Capacitance per unit length

( ) ( )abkabLC

e /ln21

/ln2 0 ==πε

• A Spherocal Capacitor

rrqE ˆ

41

20πε

=r

drrqdrEV

b

a

a

b 202

1∫∫ =⋅−=∆

πε

ababq

baq −

=⎟⎠⎞

⎜⎝⎛ −=

00 411

4 πεπε

Capacitance C

abab

VqC

−=

∆= 04πε

25-4. Capacitors in Parallel and in Series• Symbols in Circuits

Capacitor

+− Battery

Switch

• Capacitors in Parallel (병렬)

33 qVC =11 qVC = 22 qVC =

( )VCCCqqqq 321321 ++=++=

321 CCCCeq ++=

⎟⎟⎠

⎞⎜⎜⎝

⎛=+++= ∑

jjeq CCCCC L321

• Capacitors in Series (직렬)

333

222

111

VCqVCqVCq

===

qqqq === 321

Charge Conserved

33 C

qV =1

1 CqV =

22 C

qV =

eqCQ

CCCqVVVV =⎟⎟

⎠

⎞⎜⎜⎝

⎛++=++=

321321

111

321

1111CCCCeq

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛=+++= ∑

j jeq CCCCC11111

321

L

• Example. Capacitors in Parallel and Series

2112 CCC +=321312123

11111CCCCCC

++

=+=

( )( )

321

321123

321

321

123

1

CCCCCCC

CCCCCC

C

+++

=

+++

=

25-5. Energy Stored in an Electric Field+++++++

−−−−−−−

CVq = CdVdq =⇒

CVdVVdqdW ==

CQCVCVdVW22

1 22 === ∫

Work

Q V

dV dq

C

Electrostatic Energy : 221

21

2

2CVQV

CQU ===

Parallel Plate with Area AdAC 0ε

=

( )2020

21

21 Ed

dAV

dAU ⋅

ε=⋅

ε=

AdEU 202

1ε=

202

1 EudA

UVolU

ε==⋅

=Energy Density :(Energy per unit Volume)

202

1 Eu ε=* Energy density of any electrostatic field :

Example. Rewiring Two Charged Capacitors

Connect both switches

+ −

− +

Q1 = C1V1

Q2 = C2V2

21 QQQ −=

21 CCCeq +=

21

21

21 CCQQ

CCQV

+−

=+

=∆21

2211

CCVCVC

+−

=

021

21 VCCCCV ∆

+−

=∆⇒021 VVV ∆==If

2021

202

201 ))((

21)(

21)(

21 VCCVCVCUi ∆+=∆+∆=

2

21

21⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=⎟⎟⎠

⎞⎜⎜⎝

⎛⇒

CCCC

UU

i

f

20

21

2212

021 )()(21))((

21 V

CCCCVCCU f ∆

+−

=∆+=

25-6. Capacitor with a Dielectric• Dielectric Materials

0=Er 0≠E

r

1

00

>

=−=

κκQqQQeff

κ0EE =

AQE0

0

0 ε=

εσ

=r

dA

QV0

0 ε=∆

−−−−−−−

− Q0

+++++++

dAC 00 ε=

0

202

000 2)(

21

CQVCU =∆=

κκεσ

εκε0

00

0

0

1 EA

QA

QE eff

d ====+ Q0

Dielectric Materials

AdQVVd

0

00 1εκκ

==∆ +++++++

−−−−−−−

− + − + − + − +− + − + − + − +− + − + − + − +− + − + − + − +− + − + − + − +

+ Q0 − Q0

000 C

dA

VQC

dd κκε

==∆

=

Increase of Capacitance

κκ

κ

020

20

02

1)(21

21)(

21

UVC

VkCVCU dd

=∆=

⎟⎠⎞

⎜⎝⎛ ∆

=∆=

1

00

>

=−=

κκQqQQeff

Materials Dielectric Constant κ Dielectric Strength (V/m)

Vacuum Air (dry) Bakelite Fused quartz Pyrex glass Polystyrene Teflon Neoprene rubber Nylon Paper Strontium titanate Water Silicone Oil

1.00000 1.00059 4.9 3.78 5.6 2.56 2.1 6.7 3.4 3.7

233 80 2.5

---- 3 × 106 24 × 106 8 × 106 14 × 106 24 × 106 60 × 106 12 × 106 14 × 106 16 × 106

8 × 106 ----

15 × 106

• Dielectric Constant and Dielectric Strength

Example

++++++

−−−−−−

− +− +− +− +− +

+ Q0 − Q0

κ

ld

I II III

∆V0

∆Vd

x⎟⎠⎞

⎜⎝⎛ −−=

+−=∆

κεε

εκε11

1)(

00

00

AQld

AQ

lA

QldA

QVd

κ

εlld

AVQC

d +−=

∆= 0

AQE00 ε

=εσ

=

I ; II ;

AQ

AQ

E eff

00

1εκε

==

III ;

AQE00 ε

=εσ

=

25-8. Dielectrics and Gauss’s Law• Gauss Law in vacuum

qAEAdE ==⋅∫ 000 εεrr

000 ε

σε

==A

qE

κεε qqqEAAdE =′−==⋅∫ 00

rr

κεσ

ε 00

=′−

=AqqE

• Gauss Law in dielectric material

qAdE =⋅∴ ∫rr

κε0

qAdD =⋅∫rr

EEDrrr

εκε == 0