thur. oct. 8, 2009physics 208 lecture 111 last time… equipotential lines capacitance and...
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Thur. Oct. 8, 2009 Physics 208 Lecture 11 1
Last time…
Equipotential lines
Capacitance and capacitors
€
ΔV =1
CQ
Thur. Oct. 8, 2009 Physics 208 Lecture 11 2
Parallel plate capacitor
+Q
-Q
d
€
ΔV = Q /C
€
C =εoA
d Geometrical factor determined from electric fields
€
U =1
2C ΔV( )
2=
1
2
εoA
dEd( )
2=
1
2Ad( )εoE 2
Energy stored in parallel-plate capacitor
€
U / Ad( ) =1
2εoE 2
Energy density
AreaA
AreaA
Thur. Oct. 8, 2009 Physics 208 Lecture 11 3
An isolated parallel plate capacitor has charge Q and potential V. The plates are pulled apart. Which describes the situation afterwards?
A) Charge Q has decreased
B) Capacitance C has increased
C) Electric field E has increased
D) Voltage difference V between plates has increased
E) None of these
+Q-Q+
+
+
+
-
-
-
-
dpullpull
E = (Q/A)/0 E constant
V= Ed V increases
C = 0A/d C decreases
Quick Quiz
Cap. isolated Q constant
Thur. Oct. 8, 2009 Physics 208 Lecture 11 4
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?
1) Increases
2) Decreases
3) Stays the same
+q
-q+
+
+
+
-
-
-
-
dpullpull
Quick Quiz
Thur. Oct. 8, 2009 Physics 208 Lecture 11 5
Different geometries of capacitors
Parallel plate capacitor
Spherical capacitor
Cylindrical capacitor
+Q
-QL
€
C =Q
ΔV=
2πεoL
ln b /a( )
€
C =Q
ΔV=
2πεoL
ln b /a( )
€
C =Q
ΔV=
εoA
d
€
C =Q
ΔV=
εoA
d
+Q
-Q
d
A
€
C =Q
ΔV= 4πεo
1
a−
1
b
⎛
⎝ ⎜
⎞
⎠ ⎟−1
€
C =Q
ΔV= 4πεo
1
a−
1
b
⎛
⎝ ⎜
⎞
⎠ ⎟−1
Thur. Oct. 8, 2009 Physics 208 Lecture 11 6
Combining Capacitors — Parallel Connect capacitors together with metal wire
C1 C2Ceq
Both have same ΔV
Need different charge
€
Q1 = C1 /ΔV
€
Q2 = C2 /ΔV
“Equivalent” capacitorPotential difference ΔVTotal charge
€
Qeq = Q1 + Q2
€
Ceq =Qeq
ΔV=
Q1 + Q2
ΔV= C1 + C2 = Ceq
Thur. Oct. 8, 2009 Physics 208 Lecture 11 7
Combining Capacitors — Series
21
111CCCeq
+=21
111CCCeq
+=
C1
C2
Ceq
VA
VB
Vm
€
ΔV1 = VA −Vm = Q /C1
VA
VB
€
ΔV2 = Vm −VB = Q /C2
€
ΔV = VA −VB
= ΔV1 + ΔV2
Q
Q
-Q
-Q
€
ΔV =Q
C1
+Q
C1
=Q
CeqQ on each is same
Thur. Oct. 8, 2009 Physics 208 Lecture 11 8
Thur. Oct. 8, 2009 Physics 208 Lecture 11 9
Current in a wire: not electrostatic equilibrium
Battery produces E-field in wire
Charge moves in response to E-field
Thur. Oct. 8, 2009 Physics 208 Lecture 11 10
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 11
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
Constant force qE
Produces constant accel. qE/m
Velocity increases v(t)=qEt/m
Charge / time crossing plane increases with time
Thur. Oct. 8, 2009 Physics 208 Lecture 11 12
But experiment says…
Current constant in time Proportional to voltage
R = resistance (unit Ohm = )
Also written
J = current density = I / (cross-section area) = resistivity = R x (cross-section area) / (length)
Resistivity is independent of shape
€
I =1
RV
€
J =1
ρV
Thur. Oct. 8, 2009 Physics 208 Lecture 11 13
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 14
Current and drift velocity
This average velocity called drift velocity
€
I = −ene A( )vd = −ne
e2τ
mA
⎛
⎝ ⎜
⎞
⎠ ⎟E
€
vd =eτ
m
⎛
⎝ ⎜
⎞
⎠ ⎟E
Current density J
€
J = I / A = −nee
2τ
mE = σ E
This drift leads to a current
Conductivity
Electric field
Thur. Oct. 8, 2009 Physics 208 Lecture 11 15
What about Ohm’s law?
Current density proportional to electric field
€
J = σ E
€
I = JA = σAE =σA
LEL( ) = V /R
Current proportional to current density through geometrical factor
Electric field proportional to electric potential through geometrical factor
€
R =L
σA= ρ
L
A
Thur. Oct. 8, 2009 Physics 208 Lecture 11 16
Resistivity
Resistivity
SI units Ω-m
€
=RA
LIndependent of sample geometry
Thur. Oct. 8, 2009 Physics 208 Lecture 11 17
Resistors
Schematic layout
Circuits
Physical layout
Thur. Oct. 8, 2009 Physics 208 Lecture 11 18
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 19
Current conservation
Iin
Iout
Iout = Iin
I1
I2
I3I1=I2+I3
I2
I3
I1
I1+I2=I3
Thur. Oct. 8, 2009 Physics 208 Lecture 11 20
Quick QuizHow 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 21
Resistors in Series I1 = I2 = I Potentials add
ΔV = ΔV1 + ΔV2 = IR1 + IR2 =
= I (R1+R2) The equivalent resistance
Req = R1+R2
R
R=
2R
2 resistors in series:R LLike summing lengths
€
R = ρL
A
Thur. Oct. 8, 2009 Physics 208 Lecture 11 22
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 23
Resistors in Parallel ΔV = ΔV1 = ΔV2
I = I 1 + I 2 (lower resistance path has higher current)
Equivalent Resistance
R/2
R R
Add areas
€
R = ρL
A
Thur. Oct. 8, 2009 Physics 208 Lecture 11 24
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