Last time…
Parallel plate capacitor
Equipotential
lines
-Q
"V = Q /C
"A
C= o
d
!
!
Geometrical factor
determined from electric
fields
+Q
Energy stored in parallel-plate capacitor
1
1# A
1
2
2
U = C ("V ) = o ( Ed) = ( Ad )#o E 2
2
2 d
2
Capacitance and
capacitors
Thur. Oct. 9, 2007
"V =
1
Q
C
Physics 208 Lecture 12
Energy density
!
1
Thur. Oct. 9, 2007
d
Physics 208 Lecture 12
2
!
!
Quick Quiz
Quick Quiz
An isolated parallel plate capacitor has charge Q and potential V.
The plates are pulled apart.
+
Which describes the situation afterwards?
-Q
+Q
-
pull
1
U /( Ad ) = "o E 2
2
-
d
An isolated parallel plate capacitor has
a charge q. The plates are then pulled
further apart. What happens to the
energy stored in the capacitor?
-q
+
+
pull
+
-
A) Charge Q has decreased
Cap. isolated ⇒ Q constant
B) Capacitance C has increased
C = ε 0A/d ⇒ C decreases
C) Electric field E has increased
E = (Q/A)/ε0 ⇒ E constant
D) Voltage difference V between
plates has increased
V= Ed ⇒ V increases
+
+
-
pull
1) Increases
-
2) Decreases
d
+q
+
pull
+
3) Stays the same
E) None of these
Thur. Oct. 9, 2007
Physics 208 Lecture 12
3
Thur. Oct. 9, 2007

+Q
-Q
A
+Q
Thur. Oct. 9, 2007
!
!
Spherical
capacitor
& 1 1)
Q
C=
= 4 #$o ( % +
' a b*
"V
Physics 208 Lecture 12
!
Ceq
C2
Both have same ΔV
Cylindrical
capacitor
%1
Connect capacitors together with metal wire
C1
L
d
Parallel plate
capacitor
Q #o A
C=
=
"V
d
4
Combining Capacitors — Parallel
Different geometries of capacitors
-Q
Physics 208 Lecture 12
Need different charge
Q2 = C2 /"V
Q1 = C1 /"V
Q
2#$o L
C=
=
"V ln(b /a)
5
Q on each is same
!
Thur. Oct. 9, 2007
“Equivalent” capacitor
Potential difference V
Total charge Qeq = Q1 + Q2
Q
Q + Q2
Ceq = eq = 1
= C1 + C2 = Ceq
"V
"V
!
Physics 208 Lecture 12
!
6
!
1
Combining Capacitors — Series
VA
Vm
Q
VA
C1
Q
-Q
Ceq
Q
C2
-Q
-Q
VB
VB
"V = VA # VB
"V1 = VA # Vm = Q /C1
= "V1 + "V2
Q Q
Q
"V = +
=
C1 C1 Ceq
"V2 = Vm # VB = Q /C2
!
!
!
Thur. Oct. 9, 2007
Physics 208 Lecture 12
1
1
1
= +
Ceq C1 C2
7
Thur. Oct. 9, 2007
Physics 208 Lecture 12
8
!
Current in a wire:
not electrostatic equilibrium
Electric Current


Battery produces
E-field in wire

Electric current = I = amount of charge per unit time flowing
through a plane perpendicular to charge motion

SI unit: ampere 1 A = 1 C / s

Depends on sign of charge:

Charge moves in
response to E-field
Thur. Oct. 9, 2007
Physics 208 Lecture 12

9
+ charge particles:
current in direction of particle motion is positive
- charge particles:
current in direction of particle motion is negative
Thur. Oct. 9, 2007

An infinite number of positively charged particles are
uniformly distributed throughout an otherwise empty
infinite space.
A spatially uniform positive electric field is applied.
The current due to the charge motion
A. increases with time
B. decreases with time
C. is constant in time
D. Depends on field

Current constant in time
Proportional to voltage
I=


Constant force qE
Produces constant accel. qE/m
Velocity increases v(t)=qEt/m
Charge / time crossing plane
increases with time
!
Physics 208 Lecture 12


R = resistance (unit Ohm = Ω)
1
V
"
J = current density = I / (cross-section area)
ρ = resistivity = R x (cross-section area) / (length)

11
1
V
R
Also written J =
!
Thur. Oct. 9, 2007
10
But experiment says…
Quick Quiz

Physics 208 Lecture 12
Thur. Oct. 9, 2007
Resistivity is independent of shape
Physics 208 Lecture 12
12
2
Charge motion with collisions
Current and drift velocity
Wire not empty space, has various fixed objects.
Charge carriers accelerate, then collide.
After collision, charged particle reaccelerates.
Result: average “drift” velocity vd






This average velocity
called drift velocity
This drift leads to a current
!

Physics 208 Lecture 12
13
$
e 2#
I = ("en e A)v d = & "n e
m
%
Current density J
!
Thur. Oct. 9, 2007
#e"&
vd = % ( E
$m'
'
A) E
(
Conductivity
2
J = I/A = "
Thur. Oct. 9, 2007
nee #
E = $E
m
Physics 208 Lecture 12
Electric field
14
!
What about Ohm’s law?

Resistivity
Current density proportional to electric field
J = "E

!

I = JA = "AE =
!

Current proportional to current density through
geometrical factor
Electric field proportional to electric potential through
geometrical factor
"A
( EL) = V /R
L
Thur. Oct. 9, 2007
R=
A
L
Independent of
sample geometry

"=R

SI units Ω-m
!
L
L
=#
"A
A
Physics 208 Lecture 12
Resistivity
15
Thur. Oct. 9, 2007
Physics 208 Lecture 12
16
!
Resistors
Circuits
Quick Quiz
Which bulb is brighter?
A. A
B. B
Physical layout
C. Both the same
Current through each must be same
Schematic layout
Conservation of current (Kirchoff’s current law)
Charge that goes in must come out
Thur. Oct. 9, 2007
Physics 208 Lecture 12
17
Thur. Oct. 9, 2007
Physics 208 Lecture 12
18
3
I2
Current conservation
Iin
Quick Quiz
How does brightness of bulb B compare to that of A?
I1
I3
I1 =I2 +I3
A. B brighter than A
B. B dimmer than A
I1
C. Both the same
I3
Iout
Iout = I in
Thur. Oct. 9, 2007
Battery maintain constant potential difference
I2
Extra bulb makes extra resistance -> less current
I1 +I2 =I3
Physics 208 Lecture 12
19
Thur. Oct. 9, 2007
Physics 208 Lecture 12
20
4