Transcript V 0

Capacitor Circuits
Series
1
1

C
i Ci
Parallel
C   Ci
i
Thunk some more …
C1
V
C2
(12+5.3)pf
C3
C1=12.0 uf
C2= 5.3 uf
C3= 4.5 ud
dW  dq  Ed
Gauss

q 1
E

0 A 0
So….
q 1
dW 
d  dq
A 0
Q
W U  
0
d
q2 d
q2 1
qdq 

A 0
2 A 0
2 ( A 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
A Closer Look at this stuff..
q
Consider this capacitor.
No dielectric experience.
Applied Voltage via a battery.
++++++++++++
V0
C0
-q
------------------
C0 
0 A
d
q  C0V0 
0 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 (charged),
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


0
E0 
q   
  
0 A  0 
A little sheet from the past..
-q’
+q’
- -q
-
+
q+
+
Esheet

q'


2 0 2 0 A
q'
0
2xEsheet
0
q'
Esheet / dialectric  2 

2 0 A  0 A
Some more sheet…
Edielectricch arg e
q
E 0
0 A
so
q  q'
Enet 
0 A
 q'

0 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
V0
k
and
V  Ed
V0  E0 d
thus
V 1 E
 
V0 k E0
E
E0
k
Add Dielectric to Capacitor
Vo
• Original Structure
+
-
+
V0
• Disconnect Battery
+
• Slip in Dielectric
Note: Charge on plate does not change!
SUMMARY OF RESULTS
V
V0
E
E0
k
C  kC0
k
APPLICATION OF GAUSS’ LAW
E0 
q
0 A
q  q ' E0
E

0 A
k
E
q
k 0 A
and
q  q' 
q
k
New Gauss for Dielectrics
k
E

d
A


sometimes
  k 0
q free
0