Last time…
Equipotential
lines
Capacitance and
capacitors
Thur. Oct. 8, 2009
Physics 208 Lecture 11
1
V  Q
C
1
Parallel plate capacitor
V  Q /C
C
o A
d
-Q
Geometrical factor
determined from electric
fields
Energy stored in parallel-plate capacitor
1
1 o A
1
2
2
U  CV  
Ed  Ado E 2
2
2 d
2
+Q
Area
A
Area
A
Energy density
Thur. Oct. 8, 2009
1
U /Ad  o E 2
2
Physics 208 Lecture 11
d
2
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
-
d
+
+
pull
+
A) Charge Q has decreased
Cap. isolated  Q constant
B) Capacitance C has increased
C = 0A/d  C decreases
C) Electric field E has increased
E = (Q/A)/0  E constant
D) Voltage difference V between
plates has increased
V= Ed  V increases
E) None of these
Thur. Oct. 8, 2009
Physics 208 Lecture 11
3
Quick Quiz
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
+
+
-
1) Increases
pull
-
2) Decreases
d
+
+q
pull
+
3) Stays the same
Thur. Oct. 8, 2009
Physics 208 Lecture 11
4
Different geometries of capacitors
+Q
-Q
+Q
-Q
A
L
d
Parallel plate
capacitor
Q o A
C

V
d
Thur. Oct. 8, 2009
Spherical
capacitor
1 1 1
Q
C
 4 o   
a b 
V
Physics 208 Lecture 11
Cylindrical
capacitor
C
Q
2o L

V ln b /a
5
Combining Capacitors — Parallel

Connect capacitors together with metal wire
Ceq
C2
C1
“Equivalent” capacitor
Potential difference V
Both have same V
Need different charge
Q1  C1 /V Q2  C2 /V
Total charge Qeq  Q1  Q2
Ceq 

Thur. Oct. 8, 2009
Qeq
V

Physics 208 Lecture 11
Q1  Q2
 C1  C2  Ceq
V

6
Combining Capacitors — Series
VA
C1
Q
Vm
Q
VA
-Q
C2
-Q
-Q
VB
VB
V  VA  VB
 V1  V2
Q Q
Q
V  

C1 C1 Ceq
V1  VA Vm  Q/C1
V2  Vm VB  Q/C2
Q on each is same
Thur. Oct. 8, 2009
Ceq
Q

Physics 208 Lecture 11
1
1
1
 
Ceq C1 C2
7
Thur. Oct. 8, 2009
Physics 208 Lecture 11
8
Current in a wire:
not electrostatic equilibrium

Battery produces
E-field in wire

Thur. Oct. 8, 2009
Charge moves in
response to E-field
Physics 208 Lecture 11
9
Electric Current

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 particles:
current in direction of particle motion is positive
- charge particles:
current in direction of particle motion is negative
Thur. Oct. 8, 2009
Physics 208 Lecture 11
10
Quick Quiz

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
Thur. Oct. 8, 2009
Constant force qE
Produces constant accel. qE/m
Velocity increases v(t)=qEt/m
Charge / time crossing plane
increases with time
Physics 208 Lecture 11
11
But experiment says…


Current constant in time
Proportional to voltage
1
I V
R


Also written J 


R = resistance (unit Ohm = )
1

V
J = current density = I / (cross-section area)
 = resistivity = R x (cross-section area) / (length)


Thur. Oct. 8, 2009
Resistivity is independent of shape
Physics 208 Lecture 11
12
Charge motion with collisions




Wire not empty space, has various fixed objects.
Charge carriers accelerate, then collide.
After collision, charged particle reaccelerates.
Result: average “drift” velocity vd
Thur. Oct. 8, 2009
Physics 208 Lecture 11
13
Current and drift velocity

This average velocity
called drift velocity

This drift leads to a current


e  
v d   E
m 
 e 2
I  en e Av d  n e
m

Current density J

Thur. Oct. 8, 2009

AE

Conductivity
n e e 2
J  I/A  
E  E
m
Physics 208 Lecture 11
Electric field
14
What about Ohm’s law?

Current density proportional to electric field
J  E


Current proportional to current density through
geometrical factor
Electric field proportional to electric potential through
geometrical factor
I  JA  AE 
Thur. Oct. 8, 2009
A
L
EL  V /R
Physics 208 Lecture 11
L
L
R

A
A
15
Resistivity

Resistivity


A
R
L
Independent of
sample geometry
SI units Ω-m
Thur. Oct. 8, 2009
Physics 208 Lecture 11
16
Resistors
Circuits
Physical layout
Schematic layout
Thur. Oct. 8, 2009
Physics 208 Lecture 11
17
Quick Quiz
Which bulb is brighter?
A. A
B. B
C. Both the same
Current through each must be same
Conservation of current (Kirchoff’s current law)
Charge that goes in must come out
Thur. Oct. 8, 2009
Physics 208 Lecture 11
18
I2
Current conservation
Iin
I1
I3
I1=I2+I3
I1
I3
Iout
Iout = Iin
Thur. Oct. 8, 2009
I2
I1+I2=I
Physics 208 Lecture 11
3
19
Quick Quiz
How does brightness of bulb B compare to that of A?
A. B brighter than A
B. B dimmer than A
C. Both the same
Battery maintain constant potential difference
Extra bulb makes extra resistance -> less current
Thur. Oct. 8, 2009
Physics 208 Lecture 11
20
Resistors in Series


I1 = I 2 = I
Potentials add


V = V1 + V2 = IR1 + IR2 =
= I (R1+R2)
The equivalent resistance
Req = R1+R2
2 resistors in series:
RL
Like summing lengths
R
R
=
Thur. Oct. 8, 2009
2R
L
R
A
Physics 208 Lecture 11
21
Quick Quiz
What happens to the brightness of the
bulb B when the switch is closed?
A. Gets dimmer
B. Gets brighter
C. Stays same
D. Something else
Battery is constant voltage,
not constant current
Thur. Oct. 8, 2009
Physics 208 Lecture 11
22
Resistors in Parallel



V = V1 = V2
I = I 1 + I 2 (lower resistance path
has higher current)
R
R
Equivalent Resistance
R/2
Add areas
L
R
A
Thur. Oct. 8, 2009
Physics 208 Lecture 11
23
Quick Quiz
What happens to the brightness of the
bulb A when the switch is closed?
A. Gets dimmer
B. Gets brighter
C. Stays same
D. Something else
Thur. Oct. 8, 2009
Physics 208 Lecture 11
24