Capacitor Circuits

ii

i i

CC

ParallelCC

Series11

Thunk some more …

C1 C2

V

C3

C1=12.0 fC2= 5.3 fC3= 4.5 d

(12+5.3)pf

So….

2222

0

2

0

2

0 0

0

00

21

22

)(

122

1

1

CVCVC

CQU

ordA

qAdqqdq

AdUW

dqdAqdW

AqE

GaussEddqdW

Q

Sorta like (1/2)mv2

DIELECTRIC

Polar Materials (Water)

Apply an Electric Field

Some LOCAL ordering Larger Scale Ordering

Adding things up..

- +Net effect REDUCES the field

Non-Polar Material

Non-Polar Material

Effective Charge isREDUCED

We can measure the C of a capacitor (later)

C0 = Vacuum or air Value

C = With dielectric in place

C=C0

(we show this later)

How to Check This

Charge to V0 and then disconnect fromThe battery.C0 V0

Connect the two togetherV

C0 will lose some charge to the capacitor with the dielectric.We can measure V with a voltmeter (later).

Checking the idea..

V

00

0

000

210

2

01

000

1 CVVCC

CVVCVCqqq

CVqVCqVCq

Note: When two Capacitors are the same (No dielectric), then V=V0/2.

Messing with Capacitors

+

V-

+

V-

+

-

+

-

The battery means that thepotential difference acrossthe capacitor remains constant.

For this case, we insert the dielectric but hold the voltage constant,

q=CV

since C C0

qC0V

THE EXTRA CHARGE COMES FROM THE BATTERY!

Remember – We hold V constant with the battery.

Another Case We charge the capacitor to a voltage

V0. We disconnect the battery. We slip a dielectric in between the

two plates. We look at the voltage across the

capacitor to see what happens.

No Battery

+

-

+

-

q0

q

q0 =C0Vo

When the dielectric is inserted, no chargeis added so the charge must be the same.

0

0000

0

VV

orVCqVCq

VCq

V0

V

A Closer Look at this stuff..Consider this capacitor.No dielectric experience.Applied Voltage via a battery.

C0

00

00

00

VdAVCq

dAC

++++++++++++

------------------

V0

q

-q

Remove the Battery

++++++++++++

------------------

V0

q

-q

The Voltage across thecapacitor remains V0

q remains the same aswell.

The capacitor is (charged),

Slip in a DielectricAlmost, but not quite, filling the space

++++++++++++

------------------

V0

q

-q

- - - - - - - -

+ + + + + +

-q’

+q’

E0

E

E’ from inducedcharges

Gaussian Surface

000

0

....

AqE

qd

gapsmallin

AE

A little sheet from the past..

+++

---q-q

-q’ +q’

Aq

AqE

AqE

dialectricsheet

sheet

00/

00

'2

'2

2'

2

0 2xEsheet 0

Some more sheet…

AqqEnet

soAqE

AqE echdielectric

0

00

0arg

'

'

A Few slides backNo Battery

+

-

+

-

q0

q

q=C0Vo

When the dielectric is inserted, no chargeis added so the charge must be the same.

0

0000

0

VV

orVCqVCq

VCq

V0

V

From this last equation

0

00

00

0

1

EE

EE

VVthus

dEVEdV

and

VV

Add Dielectric to Capacitor

• Original Structure

• Disconnect Battery

• Slip in Dielectric

+

-

Vo

+

-

+

-

V0

Note: Charge on plate does not change!

SUMMARY OF RESULTS

0

0

0

EE

CC

VV

APPLICATION OF GAUSS’ LAW

qqq

andAqE

EAqqE

AqE

'

'

0

0

0

00

New Gauss for Dielectrics

0

0

sometimes

qd freeAE