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


Thunk some more …

C1=12.0 uf

C2= 5.3 uf

C3= 4.5 ud

C1 C2

(12+5.3)pf

V

C3


So….

Sorta like (1/2)mv2


What's Happening?

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 is

REDUCED


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

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).


V

Checking the idea..

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


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

  • 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

+

-

+

-

q0 =C0Vo

When the dielectric is inserted, no charge

is added so the charge must be the same.

V0

V

qk


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

q

V0

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

-q

A Closer Look at this stuff..

Consider this capacitor.

No dielectric experience.

Applied Voltage via a battery.

C0


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

q

V0

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

-q

Remove the Battery

The Voltage across the

capacitor remains V0

q remains the same as

well.

The capacitor is (charged),


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

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..

-q’ +q’

-

-

-

+++

-q

q

0 2xEsheet 0


Some more sheet…


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


Original Structure

Disconnect Battery

Slip in Dielectric

Vo

+

-

+

-

+

-

Add Dielectric to Capacitor

V0

Note: Charge on plate does not change!


SUMMARY OF RESULTS


APPLICATION OF GAUSS’ LAW


New Gauss for Dielectrics


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