1. Fun with Capacitors - Part II

2. 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.

3. Capacitor Circuits

4. 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

5. 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

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

7. 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

8. So….
Sorta like (1/2)mv2

9. What's Happening?
DIELECTRIC

10. Polar Materials (Water)

11. Apply an Electric Field
Some LOCAL ordering Larger Scale Ordering

12. Adding things up..
Net effect REDUCES the field

13. Non-Polar Material

14. Non-Polar Material
Effective Charge is
REDUCED

15. 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)

16. 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).

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

19. 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.

20. 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.

21. 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

22. 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.

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

24. 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.

25. Slip in a Dielectric Almost, but not quite, filling the space
Gaussian Surface V0 q -q
-q’
+q’ E
E’ from induced
charges E0

26. A little sheet from the past..
-q’ q’ -
+ + + -q q
xEsheet

27. Some more sheet…

28. A Few slides back No Batteryq0
q=C0Vo
When the dielectric is inserted, no charge
is added so the charge must be the same. + - V0 V qk

29. From this last equation

30. Another look + - Vo

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

32. 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

33. SUMMARY OF RESULTS

34. APPLICATION OF GAUSS’ LAW

35. New Gauss for Dielectrics