Capacitor circuits
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Capacitor Circuits. Thunk some more …. C1=12.0 u f C2= 5.3 u f C3= 4.5 u d. C 1 C 2. (12+5.3)pf. V. C 3. So…. Sorta like (1/2)mv 2. What's Happening?. DIELECTRIC. Polar Materials (Water). Apply an Electric Field.

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Capacitor Circuits

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Capacitor circuits

Capacitor Circuits


Thunk some more

Thunk some more …

C1=12.0 uf

C2= 5.3 uf

C3= 4.5 ud

C1 C2

(12+5.3)pf

V

C3


Capacitor circuits

So….

Sorta like (1/2)mv2


Capacitor circuits

What's Happening?

DIELECTRIC


Polar materials water

Polar Materials (Water)


Apply an electric field

Apply an Electric Field

Some LOCAL ordering Larger Scale Ordering


Adding things up

Adding things up..

- +

Net effect REDUCES the field


Non polar material

Non-Polar Material


Non polar material1

Non-Polar Material

Effective Charge is

REDUCED


We can measure the c of a capacitor later

We can measure the C of a capacitor (later)

C0 = Vacuum or air Value

C = With dielectric in place

C=kC0

(we show this later)


How to check this

How to Check This

Charge to V0 and then disconnect from

The battery.

C0

V0

Connect the two together

V

C0 will lose some charge to the capacitor with the dielectric.

We can measure V with a voltmeter (later).


Checking the idea

V

Checking the idea..

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


Messing with capacitors

Messing with Capacitors

The battery means that the

potential difference across

the capacitor remains constant.

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

q=CV

since C  kC0

qk kC0V

THE EXTRA CHARGE COMES FROM THE BATTERY!

+

V

-

+

-

+

-

+

V

-

Remember – We hold V constant with the battery.


Another case

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

No Battery

q0

+

-

+

-

q0 =C0Vo

When the dielectric is inserted, no charge

is added so the charge must be the same.

V0

V

qk


A closer look at this stuff

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

q

V0

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

-q

A Closer Look at this stuff..

Consider this capacitor.

No dielectric experience.

Applied Voltage via a battery.

C0


Remove the battery

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

q

V0

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

-q

Remove the Battery

The Voltage across the

capacitor remains V0

q remains the same as

well.

The capacitor is (charged),


Slip in a dielectric almost but not quite filling the space

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

q

- - - - - - - -

-q’

+q’

V0

+ + + + + +

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

-q

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

Gaussian Surface

E

E’ from induced

charges

E0


A little sheet from the past

A little sheet from the past..

-q’ +q’

-

-

-

+++

-q

q

0 2xEsheet 0


Some more sheet

Some more sheet…


A few slides back no battery

A Few slides backNo Battery

q=C0Vo

When the dielectric is inserted, no charge

is added so the charge must be the same.

q0

+

-

+

-

V0

V

qk


From this last equation

From this last equation


Add dielectric to capacitor

Original Structure

Disconnect Battery

Slip in Dielectric

Vo

+

-

+

-

+

-

Add Dielectric to Capacitor

V0

Note: Charge on plate does not change!


Summary of results

SUMMARY OF RESULTS


Application of gauss law

APPLICATION OF GAUSS’ LAW


New gauss for dielectrics

New Gauss for Dielectrics


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