Physics 2102 Physics 2102 Lecture: 08 THU 11 FEB Lecture: 08 THU 11 FEB

Capacitance I Capacitance I

Physics 2102

Jonathan Dowling

25.1–4

QuickTime™ and a decompressor

are needed to see this picture.

Capacitors and Capacitors and CapacitanceCapacitance

Capacitor: any two conductors, one with charge +Q, other with charge –Q

+Q

–Q

Uses: storing and releasing electric charge/energy. Most electronic capacitors:micro-Farads (F),pico-Farads (pF) — 10–12 FNew technology: compact 1 F capacitors

Uses: storing and releasing electric charge/energy. Most electronic capacitors:micro-Farads (F),pico-Farads (pF) — 10–12 FNew technology: compact 1 F capacitors

Potential DIFFERENCE between conductors = V

Q = CCV where C = capacitance

Units of capacitance: Farad (F) = Coulomb/Volt

CapacitanceCapacitance

• Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors

• e.g. Area of conductors, separation, whether the space in between is filled with air, plastic, etc.

+Q–Q

(We first focus on capacitorswhere gap is filled by AIR!)

Electrolytic (1940-70)Electrolytic (new)

Paper (1940-70)

Tantalum (1980 on) Ceramic (1930 on)Mica (1930-50)

Variableair, mica

CapacitorsCapacitors

Parallel Plate CapacitorParallel Plate Capacitor

-Q

E field between the plates: (Gauss’ Law)

Relate E to potential difference V:

What is the capacitance C ?

Area of each plate = ASeparation = dcharge/area = = Q/A

∫ ⋅=d

xdEV0

rr

A

Qddx

A

Qd

00 0 εε∫ ==

d

A

V

QC 0ε==

d

A

V

QC 0ε==

A

QE

00 εε

==A

QE

00 εε

==

We want capacitance: C = Q/V 

€

Units :C2

Nm2

m2

m

⎡

⎣ ⎢

⎤

⎦ ⎥=

C2

Nm

⎡

⎣ ⎢

⎤

⎦ ⎥=

CC

J

⎡ ⎣ ⎢

⎤ ⎦ ⎥=

C

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥ √

€

Units :C2

Nm2

m2

m

⎡

⎣ ⎢

⎤

⎦ ⎥=

C2

Nm

⎡

⎣ ⎢

⎤

⎦ ⎥=

CC

J

⎡ ⎣ ⎢

⎤ ⎦ ⎥=

C

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥ √

+Q

Capacitance and Your iPhone!Capacitance and Your iPhone!

d

A

V

QC 0ε==

Parallel Plate Capacitor — Parallel Plate Capacitor — ExampleExample

• 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?

Lesson: difficult to get large valuesof capacitance without specialtricks!

Lesson: difficult to get large valuesof capacitance without specialtricks!

C = ε0A/d

= (8.85 x 10–12 F/m)(0.25 m2)/(0.001

m)

= 2.21 x 10–9 F

(Very Small!!)

€

Units :C2

Nm2

m2

m

⎡

⎣ ⎢

⎤

⎦ ⎥=

C2

Nm

⎡

⎣ ⎢

⎤

⎦ ⎥=

CC

J

⎡ ⎣ ⎢

⎤ ⎦ ⎥=

C

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥≡ F[ ] = Farad

€

Units :C2

Nm2

m2

m

⎡

⎣ ⎢

⎤

⎦ ⎥=

C2

Nm

⎡

⎣ ⎢

⎤

⎦ ⎥=

CC

J

⎡ ⎣ ⎢

⎤ ⎦ ⎥=

C

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥≡ F[ ] = Farad

IsolatedIsolated Parallel Plate Capacitor Parallel Plate Capacitor

• A parallel plate capacitor of capacitance C is charged using a battery.

• Charge = Q, potential difference = V.• Battery is then disconnected. • If the plate separation is INCREASED,

does Potential Difference V:

(a) Increase?(b) Remain the same?(c) Decrease?

+Q –Q

• Q is fixed!• C decreases (=ε0A/d)• V=Q/C; V increases.

Parallel Plate Capacitor & Parallel Plate Capacitor & BatteryBattery

• A parallel plate capacitor of capacitance C is charged using a battery.

• Charge = Q, potential difference = V.

• Plate separation is INCREASED while battery remains connected.

+Q –Q

• V is fixed by battery!• C decreases (=ε0A/d)• Q=CV; Q decreases• E = Q/ε0A decreases

Does the Electric Field Inside:(a) Increase?(b) Remain the Same?(c) Decrease?

Spherical CapacitorSpherical Capacitor

What is the electric field insidethe capacitor? (Gauss’ Law)

Radius of outer plate = bRadius of inner plate = a

Concentric spherical shells:Charge +Q on inner shell,–Q on outer shellRelate E to potential difference

between the plates:

204 r

QE

πε=

∫ ⋅=b

a

rdEVrr b

a

b

ar

kQdr

r

kQ⎥⎦⎤

⎢⎣⎡−==∫ 2 ⎥⎦

⎤⎢⎣⎡ −=

bakQ

11

Spherical CapacitorSpherical Capacitor

What is the capacitance?C = Q/V =

Radius of outer plate = bRadius of inner plate = a

Concentric spherical shells:Charge +Q on inner shell,–Q on outer shell

Isolated sphere: let b >> a,

⎥⎦⎤

⎢⎣⎡ −

=

baQ

Q11

4 0πε

)(

4 0

ab

ab

−=

πε

aC 04πε=

Cylindrical CapacitorCylindrical Capacitor

What is the electric field in between the plates? Gauss’ Law!

Relate E to potential differencebetween the plates:

Radius of outer plate = bRadius of inner plate = a

cylindrical Gaussiansurface of radius r

Length of capacitor = L+Q on inner rod, –Q on outer shell

rL

QE

02πε=

∫ ⋅=b

a

rdEVrr

b

a

b

a L

rQdr

rL

Q⎥⎦

⎤⎢⎣

⎡==∫

00 2ln

2 πεπε

€

=Q

2πε0Llnb

a

⎛

⎝ ⎜

⎞

⎠ ⎟

QuickTime™ and a decompressorare needed to see this picture.

€

C =Q /V =2πε0L

lnb

a

⎛

⎝ ⎜

⎞

⎠ ⎟

€

C =Q /V =2πε0L

lnb

a

⎛

⎝ ⎜

⎞

⎠ ⎟

SummarySummary• Any two charged conductors form a capacitor.

•Capacitance : C= Q/V

•Simple Capacitors:

Parallel plates: C = ε0 A/d

Spherical: C = 4π e0 ab/(b-a)

Cylindrical: C = 2π 0 L/ln(b/a)]

Capacitors in Parallel: Capacitors in Parallel: V=ConstantV=Constant

• An ISOLATED wire is an equipotential surface: V=Constant

• Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge!

• VAB = VCD = V

• Qtotal = Q1 + Q2

• CeqV = C1V + C2V

• Ceq = C1 + C2

• Equivalent parallel capacitance = sum of capacitances

A B

C D

C1

C2

Q1

Q2

CeqQtotal

V = VAB = VA –VB

V = VCD = VC –VD

VA VB

VC VD

V=V

€

Cparallel =C1 +C2

PAR-V (Parallel V the Same)

Capacitors in Series: Capacitors in Series: Q=ConstantQ=Constant

• Q1 = Q2 = Q = Constant

• VAC = VAB + VBC

A B C

C1 C2

Q1 Q2

Ceq

Q = Q1 = Q2

21 C

Q

C

Q

C

Q

eq

+=

€

1

Cseries

=1

C1

+1

C2

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

Isolated Wire:Q=Q1=Q2=Constant

SERI-Q (Series Q the Same)

Capacitors in parallel and in Capacitors in parallel and in seriesseries

• In series : 1/Cser = 1/C1 + 1/C2

Vser = V1  + V2

Qser= Q1 = Q2

C1 C2

Q1 Q2

C1

C2

Q1

Q2

• In parallel : Cpar = C1 + C2

Vpar = V1 = V2

Qpar = Q1 + Q2 Ceq

Qeq

Example: Parallel or Example: Parallel or Series?Series?

What is the charge on each capacitor?

C1=10 F

C3=30 F

C2=20 F

120V

• Qi = CiV • V = 120V = Constant• Q1 = (10 F)(120V) = 1200 C • Q2 = (20 F)(120V) = 2400 C• Q3 = (30 F)(120V) = 3600 C

Note that:• Total charge (7200 C) is

shared between the 3 capacitors in the ratio C1:C2:C3

— i.e. 1:2:3

Parallel: Circuit Splits Cleanly in Two (Constant V)

€

Cpar =C1 +C2 +C3 = 10 + 20 + 30( )μF = 60μF

Example: Parallel or Example: Parallel or SeriesSeries

What is the potential difference across each capacitor?

C1=10F C3=30FC2=20F

120V

• Q = CserV• Q is same for all capacitors• Combined Cser is given by:

€

1

Cser

=1

(10μF)+

1

(20μF)+

1

(30μF)

• Ceq = 5.46 F (solve above

equation)

• Q = CeqV = (5.46 F)(120V) = 655

C

• V1= Q/C1 = (655 C)/(10 F) = 65.5

V

• V2= Q/C2 = (655 C)/(20 F) = 32.75

V

• V3= Q/C3 = (655 C)/(30 F) = 21.8

V

Note: 120V is shared in the ratio of INVERSE capacitances i.e. (1):(1/2):(1/3)

(largest C gets smallest V)

Series: Isolated Islands (Constant Q)

Example: Series or Example: Series or Parallel?Parallel?

In the circuit shown, what is the charge on the 10F capacitor?

In the circuit shown, what is the charge on the 10F capacitor?

10 F

10F 10V

10F

5F5F 10V

• The two 5F capacitors are in parallel

• Replace by 10F • Then, we have two 10F

capacitors in series• So, there is 5V across the 10 F capacitor of interest

• Hence, Q = (10F )(5V) = 50C

Neither: Circuit Compilation Needed!