Lecture 9 jan31 - Lecture#9 Energy storage in Capacitors...

Exam Locations:
Vallery (7, 8, 9): AUD B Angell
Kurdak (10,11,12): 182 Dennison
Krisch (13,14,15): 170 Dennison
Lecture #9
Energy storage in Capacitors
Dielectrics
Good luck on the exam tonight!
Bring pencils, note card, calculator, student ID
“Is that Your Final Answer?”-----the Scantron always is.
Results should be on SAMS this weekend

Energy “stored” in a capacitor
(Demo: The “Killer Capacitor”)
Separating + from - charge requires work by an external force:
Moving charge Q through ΔV =V requires ΔU = W
ext
= QV (NOT HERE!)
(This is indeed the work done by the battery!)
But V of capacitor builds up as q builds up: V = q/C
Thus, dU = Vdq = dW
ext
= (q/C)dq
2
2
0
2
0
2
1
2
1
2
1
2
1
CV QV
C
Q
C
q
C
qdq
U
Q
Q
q
= = = = = ∆
∫
=
So why the factor
of ½ here???
Question #1>>>>>

Question #1
The work done by the battery to transfer Q from one
conductor of the capacitor to the other is QV. The
energy stored in the capacitor is only (½)QV. So what
happened to the other (1/2)QV of energy?