Fun with Capacitors - Part II

What’s up doc? Today – More on Capacitors Friday Monday 7:30 AM problem session 9:30 AM Quiz More on Capacitors Monday Who knows … let’s see how far we can get.

Capacitor Circuits

A Thunker If a drop of liquid has capacitance 1.00 pF, what is its radius? STEPS Assume a charge on the drop. Calculate the potential See what happens

Anudder Thunker Find the equivalent capacitance between points a and b in the combination of capacitors shown in the figure. V(ab) same across each

Thunk some more … C1 C2 (12+5.3)pf V C3 C1=12.0 uf C2= 5.3 uf C3= 4.5 ud C1 C2 (12+5.3)pf V C3

More on the Big C +q -q E=e0A/d +dq 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

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. 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 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 + - q0 =C0Vo When the dielectric is inserted, no charge is added so the charge must be the same. V0 V qk

Another Way to Think About This There is an original charge q on the capacitor. If you slide the dielectric into the capacitor, you are adding no additional STORED charge. Just moving some charge around in the dielectric material. If you short the capacitors with your fingers, only the original charge on the capacitor can burn your fingers to a crisp! The charge in q=CV must therefore be the free charge on the metal plates of the capacitor.

A Closer Look at this stuff.. Consider this virgin capacitor. No dielectric experience. Applied Voltage via a battery. ++++++++++++ ------------------ V0 q -q C0

Remove the Battery ------------------ The Voltage across the ++++++++++++ ------------------ V0 q -q The Voltage across the capacitor remains V0 q remains the same as well. The capacitor is fat (charged), dumb and happy.

Slip in a Dielectric Almost, but not quite, filling the space Gaussian Surface ++++++++++++ ------------------ V0 q -q - - - - - - - - + + + + + + -q’ +q’ 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 back No Battery q0 q=C0Vo When the dielectric is inserted, no charge is added so the charge must be the same. + - V0 V qk

From this last equation

Another look + - Vo

Add Dielectric to Capacitor + - Vo Original Structure Disconnect Battery Slip in Dielectric + - V0 + - Note: Charge on plate does not change!

Potential Difference is REDUCED by insertion of dielectric. What happens? + - si + si - so Potential Difference is REDUCED by insertion of dielectric. Charge on plate is Unchanged! Capacitance increases by a factor of k as we showed previously

SUMMARY OF RESULTS

APPLICATION OF GAUSS’ LAW

New Gauss for Dielectrics