Physics 208, Lecture 6
Today’s Topics
„
Capacitance (Ch. 26.1-3)
„ Capacitors and Capacitance
„ Calculating Capacitance for
parallel-plate, cylindrical, spherical capacitors.
„ Combinations of capacitors
„ Hope you
o have
ha e pre
previewed!
ie ed!
About Exam 1
‰ When and where
ƒ Monday Sept. 27th 5:30-7:00 pm
ƒ 2301,
2301 2241 Chamberlin (room allocation to be anno
announced)
nced)
‰ Format
ƒ Close book
ƒ One 8x11 formula sheet allowed, must be self prepared, no photo
copying/download-printing of solutions, lecture slides, etc.
ƒ 20-25 multiple choice questions
ƒ Bring a calculator (but no computer). Only basic calculation
functionality can be used.
ƒ Bring
g a B2 pe
pencil
c for
o Sca
Scantron.
to
‰ Special requests:
ƒ Must have been approved by now.
ƒ All specially arranged tests (e.g.
(e g those at alternative time) are
held in our 202 labs. (for approved requests only)
Chapters Covered
‰ Chapter 23: Electric Fields
‰ Chapter
Ch t 24:
24 G
Gauss’s
’ Law
L
‰ Chapter 25: Electric Potential
I will not post past/sample exams as none that I can find are
representative. Often those can be misleading.
I will use Thursday’s
Thursday s lecture to review for the test.
(and will show a few sample test questions to help you get familiar
with the test style)
Exercise: Parallel Plates
‰ Find the potential difference between the two large
conductor plates of area A
E=0 +Q E=0
and
d separation
ti
d
++++++++++++++
See board
d
E=σ/ε0
----------------------¾ Answer
E=0 -Q E=0
ΔV= Qd/(ε
( 0A))
Note: ΔV is proportional to Q
Realistic case
Capacitors
‰ A generic
i capacitor:
it
two conductors
oppositely charged
ΔV ∝ Q
‰ Capacitor are very useful devices:
ƒ Timing control, noise filters, energy buffer, frequency
generator/selector/filter sensors
generator/selector/filter,
sensors, memories
memories...
flo
ow of e-
Demo: Charging A Pair of Parallel Conductors
ΔV=V+-V-
Uncharged
Charging
Capacitance
‰ ΔV ∝ Q Æ Q=CΔV Æ C is called capacitance
‰ C= Q/ΔV: amount of charge per unit of potential diff.
¾ Unit: Farad (F) = 1 Coulomb/Volt
¾ Parallel-plate: C=ε0A/d
¾ Cylindrical and Spherical: see examples in text
ƒ Cylindrical:
ƒ Spherical:
C
C
=
=
2 k
k
e
e
l
ln( b / a )
ab
(b − a )
-Q
+Q
Charging A Capacitor
+Q ΔV=Q/C
+q ΔV=q/C
++++++++
++++++++++++
-dq
-----------
Uncharged
---------------------
-q
q
-Q
Q
Charging
Charged
du=ΔVdq
du
ΔVdq
‰ Electric p
potential energy
gy gained:
g
U
=
∫
du
=
∫
( − Δ V )( − dq ) =
∫
Q
0
q
dq
C
=
1
Q
2
2
/ C
ÎAfter charging the capacitor stores potential energy:
U= ½ Q2/C
Discharging A Capacitor
+Q ΔV=Q/C
+++++++++++
+q ΔV=q/C
++++++++
- dq
--------------------
-----------
-q
q
-Q
Q
Charged
Uncharged
dis-Charging
du=ΔVdq
du
ΔVdq
‰ Potential energy
gy released:
U
=
∫
dU
=
∫
Δ V ( − dq ) =
∫
0
Q
−
q
dq
C
=
1
Q
2
2
/ C
Îthe originally charged capacitor has potential energy:
U= ½ Q2/C
Combinations of Capacitors In Series
Charge conservation: Q1=Q2(=Q)
-Q
Q1
+Q
Q2
-Q2
Effective Capacitance
C=Q/ΔV Î 1/C = 1/C1+1/C2
+Q1
C1ΔV1 =Q
C2ΔV2 =Q
Q
C1
C2
ΔV1 ΔV2
effectively
C
ΔV=ΔV1+ΔV2
ΔV
1/Cseries= 1/C1+1/C2+1/C3+…
Note: Cseries always < Ci
Combinations of Capacitors In Parallel
C1ΔV1 =Q1
C2ΔV2 =Q2
ΔV1 Q
C1 1
ΔV
C2
Q2
ΔV2
C
Q=Q1+Q2
effectively
ΔV
ΔV
ΔV1 =ΔV2 =ΔV
(why?)
Effective Capacitance
p
C=Q/ΔV Î C = C1+ C2
Cparallel= C1+C
C2+C
C3+…
Note: Cparallel always > Ci
Quick Quiz/exercise:
Combination of Capacitors
p
‰ What is the effective capacitance for this combination?
(C1=1μF
1μF, C2=2μF
2μF, C3=3μF)
3μF)
1. C=6μF
C1
C3
2. C=3μF
μ
C2
Î 3. C=1.5μF
4. None of above