25-2. Capacitance25-2. Capacitance

charge storage, energy storage

CVq

Capacitor: Any two isolated conductors

Given

As a Result

25-2. Capacitance25-2. Capacitance

00

1

A

qEEE

EdVV if

if VVV Potential (or Voltage)d

A

qEdV

0

Capacitanced

AC 0

FV

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

25-2. Capacitance25-2. Capacitance

+q -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

dA

qEdV

0

00

1

A

qE

d

A

V

QC 0

FV

C

25-2. Capacitance25-2. Capacitance

d

A

V

QC 0

Capacitance Capacitance and Your iPodand Your iPod

25-2. Capacitance25-2. Capacitance

• 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 = 0 A/d

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

= 2.21 x 10–9 F

(Very Small!!)

FV

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

25-2. Capacitance25-2. Capacitance

Does the capacitance of a capacitor increase, decrease, or remain the same?

(a)When the charge q on it is doubled

(b)When the potential difference across it tripled

CVq

Determine what is NOT changed.

25-2. Capacitance25-2. Capacitance

• 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 (=0A/d)• Q=CV; Q decreases• E = Q/ 0A decreases

Does the Electric Field Inside:

(a) Increase?

(b) Remain the Same?

(c) Decrease?

25-3. Calculating the Capacitance25-3. Calculating the Capacitance

Cylindrical Capacitor

abh

Cln

2 0

a

b

h

qdrrh

qdrEV

b

a

b

aln

2

1

2 00

0

)2(

qhrE

25-3. Calculating the Capacitance25-3. Calculating the Capacitance

Spherical Capacitor

0

2 )4(

qrE

abh

qdr

r

qdrEV

b

a

b

a

11

2

1

4 02

0

baa

C1

4 0

25-3. Calculating the Capacitance25-3. Calculating the Capacitance

Spherical Capacitor

baa

C1

4 0

aC 04

ab b

Capacitance of a single conducting sphere

25-4. C in Parallel and in Series 25-4. C in Parallel and in Series

V: the same

Parallel Series Q: the same

25-4. C in Parallel and in Series 25-4. C in Parallel and in Series

V: the sameParallel

• A wire is an equipotential surface.

• VAB = VCD = V

• Qtotal = Q1 + Q2

• CeqV = C1V + C2V

• Equivalent parallel capacitance = sum of capacitances

A B

C D

C1

C2

Q1

Q2

CeqQtotal

21 CCCeq

25-4. C in Parallel and in Series 25-4. C in Parallel and in Series

Q: the sameSeries

• Q1 = Q2 = Q (WHY??)

• VAC = VAB + VBC

AB C

C1 C2

Q1 Q2

Ceq

Q

21 C

Q

C

Q

C

Q

eq

21

111

CCCeq

25-4. Example 125-4. Example 1

10 F

30 F

20 F

120V

• Q = CV; V = 120 V

• 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

What is the charge on each capacitor?

25-4. Example 225-4. Example 2

10 F 30 F20 F

120V

• Q = CV; Q is same for all capacitors• Combined C is given by:

)30(

1

)20(

1

)10(

11

FFFCeq

• Ceq = 5.46 F• Q = CV = (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

What is the potential difference across each capacitor?

25-4. Example 325-4. Example 3

What is the charge on the 10F capacitor?

10 F

5 F5 F 10V

10 F

10 F10V

• Then, we have two 10F capacitors in series

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

• Hence, Q = (10F )(5V) = 50C

25-4. Example 425-4. Example 4

(N-1) Capacitors with A and d: Parallel Connection

+

–

25-4. Example 525-4. Example 5

What is the charge on each capacitor?

10F (C1)

20F (C2)

• 10F capacitor is initially charged to 120V.• 20F capacitor is initially uncharged.• Switch is closed, equilibrium is reached.

Initial charge on 10F = (10F)(120V)= 1200C

After switch is closed, let charges = Q1 and Q2.

Charge is conserved: Q1 + Q2 = 1200C

And, Vfinal is same:

2

2

1

1

C

Q

C

Q

22

1

QQ

• Q1 = 400C• Q2 = 800C• Vfinal= Q1/C1 = 40 V

25-5. Energy Stored25-5. Energy Stored

2

2

1CVU

• Start out with uncharged capacitor• Transfer small amount of charge dq from

one plate to the other until charge on each plate has magnitude Q

• How much work was needed?dq

Q

VdqU0

Q

C

Qdq

C

q

0

2

2 2

2CV

25-5. Energy Stored in an E-Field25-5. Energy Stored in an E-Field

• Energy stored in capacitor:U = Q2/(2C) = CV2/2 • View the energy as stored in ELECTRIC FIELD• For example, parallel plate capacitor: • Energy DENSITY = energy/volume = u

General expression for any region with

vacuum (or air)

Add

AQ

AdC

Q

Volume

Uu

1

2

1

2 0

22

20

0

2

0 2

1

2E

A

Qu

25-6. Capacitor with a Dielectric25-6. Capacitor with a Dielectric

Dielectric: Noncoducting material

d

AC

d

AC 00

HW5-8

Breakdown Potential: Dielectric Strength (kV/mm) X

Length (mm)

00

25-6. Capacitor with a Dielectric25-6. Capacitor with a Dielectric

00

d

AC

d

AC 00

2

2

1CVU C

QU

2

2

1or ??

25-6. Capacitor with a Dielectric25-6. Capacitor with a Dielectric

00 d

AC

d

AC 00

25-6. Capacitor with a Dielectric25-6. Capacitor with a Dielectric

00 d

AC

d

AC 00

25-6. Example25-6. Example

• Capacitor has charge Q, voltage V• Battery remains connected while

dielectric slab is inserted.

• Do the following increase, decrease or stay the same:– Potential difference?– Capacitance?– Charge?– Electric field?

dielectric slab

25-6. Example25-6. Example

• Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E

• Battery remains connected• V is FIXED; Vnew = V (same)

• Cnew = C (increases)• Qnew = (C)V = Q (increases).• Since Vnew = V, Enew = E (same)

dielectric slab