Capacitance PHY 2049 Chapter 25 Chapter 25 Capacitance In this chapter we will cover the following topics: -Capacitance C of a system of two isolated.

Capacitance

PHY 2049 Chapter 25

Chapter 25 Capacitance

In this chapter we will cover the following topics:

-Capacitance C of a system of two isolated conductors. -Calculation of the capacitance for some simple geometries.

-Methods of connecting capacitors (in series , in parallel). -Equivalent capacitance. -Energy stored in a capacitor. -Behavior of an insulator (a.k.a. dielectric) when placed in the electric field created in the space between the plates of a capacitor.

-Gauss’ law in the presence of dielectrics.

(25 - 1)

Capacitors

Capacitor

Composed of two metal plates. Each plate is charged

one positive one negative

Stores energy

SYMBOL

A simple Capacitor

TWO PLATES

Battery

WIRES

INSIDE THE DEVICE

What is STORED in the capacitor?

An Electric Field Energy Charge All three None of these

Two Charged Plates(Neglect Fringing Fields)

d

Air or Vacuum

Area A

- Q +QE

V=Potential Difference

Symbol

ADDED CHARGE

Where is the charge?

d

Air or Vacuum

Area A

- Q +QE

V=Potential Difference

------

++++++

AREA=A

=Q/A

One Way to Charge: Start with two isolated uncharged plates. Take electrons and move them from the +

to the – plate through the region between. As the charge builds up, an electric field

forms between the plates. You therefore have to do work against the

field as you continue to move charge from one plate to another.

Capacitor

More on Capacitorsd

Air or Vacuum

Area A

- Q +QE

V=Potential Difference

GaussianSurface

000

0

0

0

)/(

0

AQ

A

QE

EAQ

QEAAEA

qd

Gauss

AE

Same result from other plate!

DEFINITION - Capacity The Potential Difference is

APPLIED by a battery or a circuit.

The charge q on the capacitor is found to be proportional to the applied voltage.

The proportionality constant is C and is referred to as the CAPACITANCE of the device.

CVq

orV

qC

UNITSUNITS A capacitor which

acquires a charge of 1 coulomb on each plate with the application of one volt is defined to have a capacitance of 1 FARAD

One Farad is one Coulomb/Volt

CVq

orV

qC

Continuing…

d

AC

sod

AVEAAq

V

qC

0

00

The capacitance of a parallel plate capacitor depends only on the Area and separation between the plates.

C is dependent only on the geometry of the device!

After the switch is closed, how much charge passed through the capacitor?

C/V V/C CV C+V

V

S

P

N

n̂

The plates have area and are separated

by a distance . The upper plate has a

charge and the lower plate a charge -

A

d

q q

Capacitance of a parallel plate capacitor

We apply Gauss' law using the Gaussian surface S shown in the figure.

The electric flux cos 0 .

From Gauss' law we have:

The potential difference between the positive ano o o

EA EA

q q qEA E

A

V

d the negative plate is

given by: cos0

The capacitance /

o

o

o

qdV Eds E ds Ed

A

Aq qC

V qd A d

oA

Cd

(25 - 6)

Units of 0

mpFmF

andm

Farad

Voltm

CoulombVoltCoulombm

Coulomb

Joulem

Coulomb

Nm

Coulomb

/85.8/1085.8 120

2

2

2

2

0

pico

Simple Capacitor Circuits Batteries

Apply potential differences

Capacitors Wires

Wires are METALS. Continuous strands of wire are all at the same

potential. Separate strands of wire connected to circuit

elements may be at DIFFERENT potentials.

NOTE

Work to move a charge from one side of a capacitor to the other is = qEd.

Work to move a charge from one side of a capacitor to the other is qV

Thus qV = qEd E=V/d As before

TWO Types of Connections

SERIES

PARALLEL

Parallel Connection

VCEquivalent=CE

321

321

321

33

22

1111

)(

CCCC

therefore

CCCVQ

qqqQ

VCq

VCq

VCVCq

E

E

E

Series Connection

V C1 C2

q -q q -q

The charge on eachcapacitor is the same !

Series Connection Continued

21

21

21

111

CCC

or

C

q

C

q

C

q

VVV

V C1 C2

q -q q -q

More General

ii

i i

CC

Parallel

CC

Series

11

Example

C1 C2

V

C3

C1=12.0 fC2= 5.3 fC3= 4.5 d

(12+5.3)pf

series

(12+5.3)pf

More on the Big C We move a charge

dq from the (-) plate to the (+) one.

The (-) plate becomes more (-)

The (+) plate becomes more (+).

dW=Fd=dq x E x d+q -q

E=0A/d

+dq

So….

2222

0

2

0

2

0 0

0

00

2

1

22

)(

1

22

1

1

CVC

VC

C

QU

ord

Aq

A

dqqdq

A

dUW

dqdA

qdW

A

qE

Gauss

EddqdW

Q

Not All Capacitors are Created Equal

Parallel Plate

Cylindrical Spherical

Spherical Capacitor

02

0

2

0

4)(

4

r

qrE

qEr

qd

Gauss

AE

Calculate Potential Difference V

drr

qV

EdsV

a

b

platepositive

platenegative

20

.

.

1

4

(-) sign because E and ds are in OPPOSITE directions.

Continuing…

ab

ab

V

qC

ab

abq

ba

qV

r

q

r

drqV

b

a

0

00

02

0

4

4

11

4

)1

(44

Lost (-) sign due to switch of limits.