Transcript Lecture Set 5 - Capacitance Part II

Fun with Capacitors - Part II
What’s up doc?
 Today – More on Capacitors
 Friday
 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
Series
1
1

C
i Ci
Parallel
C   Ci
i
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
V
C2
(12+5.3)pf
C3
C1=12.0 uf
C2= 5.3 uf
C3= 4.5 ud
More on the Big C
E=e0A/d
+dq
+q
-q
 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
dW  dq  Ed
Gauss

q 1
E

e0 A e0
So….
q 1
dW 
d  dq
A e0
Q
W U  
0
d
q2 d
q2 1
qdq 

Ae 0
2 Ae 0
2 ( Ae 0 )
d
or
Q 2 C 2V 2 1
U

 CV 2
2C
2C
2
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 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
C0
V
V0
Charge to V0 and then disconnect from
The battery.
Connect the two together
C0 will lose some charge to the capacitor with the dielectric.
We can measure V with a voltmeter (later).
Checking the idea..
q0  C0V0
V
q1  C0V
q2  CV
q0  q1  q2
C0V0  C0V  CV
 V0

C  C0   1  kC0
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.
+
V
-
-
For this case, we insert the
dielectric but hold the voltage
constant,
+
+
q=CV
V
Remember – We hold V
constant with the battery.
-
since C  kC0
qk kC0V
THE EXTRA CHARGE
COMES FROM 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
V0
-
When the dielectric is inserted, no charge
is added so the charge must be the same.
+
qk  kC0V
qk
V
-
q0  C0V0  qk  kC0V
or
V
V0
k
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..
q
Consider this virgin capacitor.
No dielectric experience.
Applied Voltage via a battery.
++++++++++++
V0
C0
-q
------------------
C0 
e0 A
d
q  C0V0 
e0 A
d
V0
Remove the Battery
q
++++++++++++
V0
The Voltage across the
capacitor remains V0
q remains the same as
well.
-q
-----------------The capacitor is fat (charged),
dumb and happy.
Slip in a Dielectric
Almost, but not quite, filling the space
Gaussian Surface
q
V0
-q
++++++++++++
- - - - - - - -
-q’
+ + + + + +
+q’
------------------
E0
E
E’ from induced
charges
in..small..gap
q
E

d
A


e0
E0 
q   
  
e0 A  e0 
A little sheet from the past..
-q’
+q’
- -q
-
+
q+
+
Esheet

q'


2e 0 2e 0 A
q'
0
2xEsheet
0
q'
Esheet / dialectric  2 

2e 0 A e 0 A
Some more sheet…
Edielectricch arg e
q
E 0
e0 A
so
q  q'
E
e0 A
 q'

e0 A
A Few slides back
No Battery
+
V0
-
+
q=C0Vo
q0
When the dielectric is inserted, no charge
is added so the charge must be the same.
qk  kC0V
qk
V
-
q0  C0V0  qk  kC0V
or
V
V0
k
From this last equation
V
V
0
k
and
V  Ed
V0  E0 d
thus
V 1 E
 
V0 k E0
E
E0
k
Another look
Vo
+
-
Parallel  Plate
e0 A
C0 
d
e 0 AV0
Q0  C0V0 
d
Electric  Field
V0
E0 
d
Q0 e 0V0
0 

A
d
Add Dielectric to Capacitor
Vo
• Original Structure
+
-
+
V0
• Disconnect Battery
+
• Slip in Dielectric
Note: Charge on plate does not change!
What happens?
o +
o -
i 
i 
E0
V0 1
E

k
d k
and
V0
V  Ed 
k
Potential Difference is REDUCED
by insertion of dielectric.
Q
Q
C 
 kC0
V V0 / k 
Charge on plate is Unchanged!
Capacitance increases by a factor of k
as we showed previously
SUMMARY OF RESULTS
V
V0
E
E0
k
C  kC0
k
APPLICATION OF GAUSS’ LAW
E0 
q
e0 A
q  q ' E0
E

e0 A
k
E
q
ke 0 A
and
q  q' 
q
k
New Gauss for Dielectrics
k
E

d
A


sometimes
e  ke 0
q free
e0