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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C1=12.0 uf
C2= 5.3 uf
C3= 4.5 ud
C1 C2
(12+5.3)pf
V
C3
Sorta like (1/2)mv2
What's Happening?
DIELECTRIC
Some LOCAL ordering Larger Scale Ordering
 +
Net effect REDUCES the field
Effective Charge is
REDUCED
C0 = Vacuum or air Value
C = With dielectric in place
C=kC0
(we show this later)
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
Note: When two Capacitors are the same (No dielectric), then V=V0/2.
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.
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
Consider this capacitor.
No dielectric experience.
Applied Voltage via a battery.
C0
++++++++++++
q
V0

q
The Voltage across the
capacitor remains V0
q remains the same as
well.
The capacitor is (charged),
++++++++++++
q
       
q’
+q’
V0
+ + + + + +

q
Gaussian Surface
E
E’ from induced
charges
E0
q’ +q’



+++
q
q
0 2xEsheet 0
q=C0Vo
When the dielectric is inserted, no charge
is added so the charge must be the same.
q0
+

+

V0
V
qk
Original Structure
Disconnect Battery
Slip in Dielectric
Vo
+

+

+

V0
Note: Charge on plate does not change!