physics 202, lecture 7 - icecube.wisc.edukarle/courses/phys202/202lecture07.pdf · 1 physics 202,...
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Physics 202, Lecture 7Today’s Topics
About Exam 1
Capacitance (Ch. 26-II) Review Energy storage in capacitors Dielectric materials, electric dipoles Dielectrics and Capacitance
Next lecture: More information about Exam 1. Current and Resistance (Ch. 27-I)
About the First Midterm Exam The exam will be on Monday Oct 1 at 5:30-7:00pm.Rooms: 2650 and 3650 Humanities
McBurney students: contact me for special arrangements. If conflicts with exam time due to an evening course:
• Stay tuned for special arrangements.• Be sure to have informed me via email by Friday at the latest.
An 8½ x 11” double-sided formula sheet is allowed. Has to be self prepared: hand written or printed, but no Xerox.
Bring a calculator: but do not use programming functionality. Absolutely no communication functionality will be allowed.
Format: word problems. More about this Thursday.
2
First Midterm Exam Chapters Covered:
Chapter 23: Electric Fields Chapter 24: Gauss’s Law Chapter 25: Electric Potential Chapter 26.1-26.3: Capacitance
Exceptions: you will not be responsible for the material coveredin sections 24.5, 25.7, 25.8.
Office hours: I will be available in my office (5215 CH)on Friday afternoon, from 1-2:30, 3:30-5:30. TA’s are alsoavailable in the helproom as always.
Review sessions: Sunday, Sept 30, 1-3 PM, Rennebohm (this room) Saturday, Sept 29, 7 PM, place TBA
Capacitors: Summary
• Definition:
• Capacitance depends on geometry:
d
A
- - - - -+ + + +
Parallel Plates
a
b L
r
+Q
-Q
Cylindrical
a
b
+Q-Q
Spherical
C has units of “Farads” or F (1F = 1C/V) ε o has units of F/m
C !Q
"V
C =!oA
d
C = 4!"o
ab
b # a
C =2!"
oL
lnb
a
#$%
&'(
3
Capacitors in Series and Parallel
Parallel:
Series:
ΔV
C2
C1
ΔV
Q=Q1+Q2
Cp ΔV=Q
ΔV1 = ΔV2 = ΔV
ΔV1 ΔV2ΔV=ΔV1+ΔV2
Q
ΔV
C1 C2CSΔV=Q
Q1 = Q2 = Q
Cp = C1 + C2
Cs=
1
C1
+1
C2
!
"#$
%&
'1
ΔV
Example: Ch. 26 #27, 28 (board)
Energy of a Capacitor• How much energy is stored in a charged capacitor?
– Calculate the work provided (usually by a battery) to charge acapacitor to +/- Q:
Incremental work dW needed to add charge dq to capacitor at voltage V:- +
• In terms of the voltage V:
• The total work W to charge to Q is then given by:
Two ways to write W
dW = V (q)idq =q
C
!"#
$%&idq
W =1
Cqdq =
0
Q
!1
2
Q2
C
W =1
2CV
2
4
Capacitor Variables
• In terms of the voltage V:
• The total work to charge capacitor to Q equals the energy Ustored in the capacitor:
You can do one of two things to a capacitor :
• hook it up to a battery specify V and Q follows
• put some charge on it specify Q and V follows
U =1
2CV
2
U =1
Cqdq =
0
Q
!1
2
Q2
C
Q = CV
V =Q
C
Example (I)
d
A
- - - - -+ + + +• Suppose the capacitor shown here is charged
to Q. The battery is then disconnected.
• Now suppose the plates are pulled further apart to a finalseparation d1.
• How do the quantities Q, C, E, V, U change?
• How much do these quantities change?.. See board.
• Q:• C:• E:• V:• U:
remains the same.. no way for charge to leave.
increases.. add energy to system by separating
decreases.. capacitance depends on geometry
increases.. since C ↓, but Q remains same (or d ↑ but E the same)remains the same... depends only on charge density
Answers: C1=d
d1
C V1=d1
dV
U1=
d1
dU
5
• Suppose the battery (V) is keptattached to the capacitor.
• Again pull the plates apart from d to d1.
• Now what changes?
• C:• V:• Q:• E:• U:
decreases (capacitance depends only on geometry)must stay the same - the battery forces it to be Vmust decrease, Q=CV charge flows off the plate
d
A
- - - - -+ + + +V
• How much do these quantities change?.. See board.
Answers:
U1=
d
d1
U
C1=
d
d1
C
E1=
d
d1
E
must decrease ( , )
E =!
E0
E =V
D
must decrease ( ) U =
1
2CV
2
Example (II)
Where is the Energy stored?• Claim: energy is stored in the electric field itself.
• The electric field is given by:
⇒• The energy density u in the field is given by:
Units:
• Consider the example of a constant field generated by a parallelplate capacitor:
++++++++ +++++++
- - - - - - - - - - - - - -- Q
+Q
u =U
volume=
U
Ad=
1
2!
0E
2
E =!
"0
=Q
"0A
U =1
2!
0E
2Ad
U =1
2
Q2
C=
1
2
Q2
( A!0
/ d)
J
m3
6
Question A parallel plate capacitor is holding a charge q. The
capacitor is not connected to a battery.If plates are pulled apart, what happens to the stored energy?
IncreasesDecreasesStays the same
- - - - - - -
+ + + + + + + +q
-qd
pull
pull
Arguments: volume increases, E same work done to it when
plates are being pulled apart
Dielectrics• Empirical observation:
Inserting a non-conducting material (dielectric) between theplates of a capacitor changes the VALUE of the capacitance.
• Definition:The dielectric constant of a material is the ratio of thecapacitance when filled with the dielectric to that without it:
κ values are always > 1 (e.g., glass = 5.6; water = 80)INCREASE the capacitance of a capacitorThey permit more energy to be stored on a given capacitor:
! =C
C0
! =C
C0
!U =CV
2
2="C
0V
2
2="U
7
Dielectric Materials Dielectrics are electric insulators:
Charges are not freely movable, but can still have smalldisplacements in an external electric field
Atomic view: composed of permanent (or inducible)electric dipoles
- -
++
Permanent dipole
O--
H+ H+
P
+ - + -
+ - + -
Inducible dipole E>0
E=0
Electric Dipole in External E Field Electric dipole moment p.
Electric dipole moment in constant E field
+
-
d-q
+q
!F! = 0
!" =!p #!E
U = $!pi
!E
Net Force
Net Torque
Potential energy
!p = q
!d
8
Quick Quiz 1 Which of the following configurations has the
highest potential energy?
Points at 45o
Points towards the E field
Points against the E field
Points normal to the E field
E
E
E
E
Quick Quiz 2 An electric dipole moment initially points at 45o with
respect to the x axis. When an external E field in thepositive x direction is applied, what will happen?
No change
Points towards the E field
Points against the E field
Points normal to the E field
E
E
E
E
9
Dielectrics In External Field Alignment of permanent dipoles in external field
Induced field by non-permanent dipoles
Note: induced field always opposite to the external field E0
Applyingexternal E field
Zeroexternal field Equilibrium
Insert Dielectrics In Between Conductor Plates
10
+++++++++++++++
Parallel Plate Example
• Deposit a charge Q on parallel platesfilled with vacuum (air)—capacitance C0
• Disconnect from battery• The potential difference is V0 = Q / C0.
Now insert material with dielectric constant κ .Charge Q remains constantCapacitance increases C = κ C0
Voltage decreases from V0 to:
Q+++++++++++++++
- - - - - - - - - - - - - - -
E0V0
Electric field decreases also:
- - - - - - - - - - - - - - -
Q
V E
+- +
-
+- +
-
+-
+-+-
Note: The field onlydecreases when thecharge is held constant!
V =Q
C=
Q
!C0
=V
!
E =V
d=V0
d!=E0
!
Example: Ch. 26 #47 (board)
Dielectric Constant For Various Materials
11
Use of Dielectric Material Non-conducting dielectric material can be inserted
in between conductor ends to increase capacitance.
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