Last time…
Parallel plate capacitor
Equipotential
lines
-Q
ΔV = Q /C
εA
C= o
d
€
€
Geometrical factor
determined from electric
fields
+Q
Energy stored in parallel-plate capacitor
1
1ε A
1
2
2
U = C (ΔV ) = o ( Ed) = ( Ad )εo E 2
2
2 d
2
Capacitance and
capacitors
Thur. Oct. 8, 2009
ΔV =
1
Q
C
Physics 208 Lecture 11
Energy density
€
1
d
Physics 208 Lecture 11
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 ) = εo E 2
2
Thur. Oct. 8, 2009
Area
A
Area
A
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 = ε0A/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
3) Stays the same
+
d
+q
+
-
+
-
+
pull
E) None of these
Thur. Oct. 8, 2009
Physics 208 Lecture 11
3
Thur. Oct. 8, 2009

+Q
-Q
A
+Q
Parallel plate
capacitor
Q εo A
C=
=
ΔV
d
Thur. Oct. 8, 2009
€
€
 1 1
Q
C=
= 4 πεo  − 
 a b
ΔV
Physics 208 Lecture 11
€
Ceq
C2
Both have same ΔV
Cylindrical
capacitor
−1
Connect capacitors together with metal wire
C1
L
d
Spherical
capacitor
4
Combining Capacitors — Parallel
Different geometries of capacitors
-Q
Physics 208 Lecture 11
Need different charge
Q2 = C2 /ΔV
Q1 = C1 /ΔV
Q
2πεo L
C=
=
ΔV ln(b /a)
5
€
Thur. Oct. 8, 2009
“Equivalent” capacitor
Potential difference ΔV
Total charge Qeq = Q1 + Q2
Q
Q + Q2
Ceq = eq = 1
= C1 + C2 = Ceq
ΔV
ΔV
€
Physics 208 Lecture 11
€
6
€
1
Combining Capacitors — Series
VA
Vm
Q
VA
C1
Q
-Q
-Q
-Q
VB
VB
ΔV = VA − VB
= ΔV1 + ΔV2
Q Q
Q
ΔV = +
=
C1 C1 Ceq
ΔV1 = VA − Vm = Q /C1
ΔV2 = Vm − VB = Q /C2
€
€
Ceq
Q
C2
Q on each is same
€
Thur. Oct. 8, 2009
Physics 208 Lecture 11
1
1
1
= +
Ceq C1 C2
7
Thur. Oct. 8, 2009
Physics 208 Lecture 11
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€
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. 8, 2009
Physics 208 Lecture 11

9
+ charge particles:
current in direction of particle motion is positive
- charge particles:
current in direction of particle motion is negative
Thur. Oct. 8, 2009

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

Produces constant accel. qE/m
C. is constant in time
Velocity increases v(t)=qEt/m
Current constant in time
Proportional to voltage
I=


Constant force qE
B. decreases with time
D. Depends on field
€
Charge / time crossing plane
increases with time
Physics 208 Lecture 11


R = resistance (unit Ohm = Ω)
1
V
ρ
J = current density = I / (cross-section area)
ρ = resistivity = R x (cross-section area) / (length)

11
1
V
R
Also written J =
€
Thur. Oct. 8, 2009
10
But experiment says…
Quick Quiz

Physics 208 Lecture 11
Thur. Oct. 8, 2009
Resistivity is independent of shape
Physics 208 Lecture 11
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 11
13

e 2τ
I = (−en e A)v d =  −n e
m

Current density J
€
Thur. Oct. 8, 2009
eτ
vd =   E
m

A E

Conductivity
2
J = I/A = −
Thur. Oct. 8, 2009
nee τ
E = σE
m
Physics 208 Lecture 11
Electric field
14
€
What about Ohm’s law?

Resistivity
Current density proportional to electric field
J = σE

€

I = JA = σAE =
€

Current proportional to current density through
geometrical factor
Electric field proportional to electric potential through
geometrical factor
σA
( EL) = V /R
L
Thur. Oct. 8, 2009
R=
A
L
Independent of
sample geometry

ρ=R

SI units Ω-m
€
L
L
=ρ
σA
A
Physics 208 Lecture 11
Resistivity
15
Thur. Oct. 8, 2009
Physics 208 Lecture 11
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
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Physics 208 Lecture 11
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Physics 208 Lecture 11
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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
Battery maintain constant potential difference
I2
Iout = Iin
Extra bulb makes extra resistance -> less current
I1+I2=I3
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Physics 208 Lecture 11
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

ΔV = ΔV1 + ΔV2 = IR1 + IR2 =
= I (R1+R2)
A.  Gets dimmer
B.  Gets brighter
The equivalent resistance
Req = R1+R2
R
C.  Stays same
D.  Something else
2 resistors in series:
R∝L
Like summing lengths
R
=
2R
R=ρ
Thur. Oct. 8, 2009
20
What happens to the brightness of the
bulb B when the switch is closed?
I1 = I2 = I
Potentials add

Physics 208 Lecture 11
Quick Quiz
Resistors in Series

Thur. Oct. 8, 2009
L
A
Battery is constant voltage,
not constant current
Physics 208 Lecture 11
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Thur. Oct. 8, 2009
Physics 208 Lecture 11
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€
Resistors in Parallel

ΔV = ΔV1 = ΔV2
I = I 1 + I 2 (lower resistance path
has higher current)

Equivalent Resistance

Quick Quiz
What happens to the brightness of the
bulb A when the switch is closed?
R
R
A.  Gets dimmer
B.  Gets brighter
R/2
C.  Stays same
D.  Something else
Add areas
L
R=ρ
A
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Physics 208 Lecture 11
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Thur. Oct. 8, 2009
Physics 208 Lecture 11
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€
4