diferentes leyes
TRANSCRIPT
How Electronics Started! And JLab Hits the Wall!
In electronics, a vacuum diode or tube is
a device used to amplify, switch, other-
wise modify, or create an electrical sig-
nal by controlling the movement of elec-
trons in a low-pressure space. Almost all
depend on the thermal emission of elec-
trons. The figure shows the components
of the device. A demonstration of how
they work is here.
The physics here also determines the
limits on the amount of beam that can be
produced in an accelerator.
Vacuum Diode – p.1/21
The Child-Langmuir Law
In a vacuum diode, electrons are boiled off a hot cathode at zero potentialand accelerated across a gap to the anode at a positive potential V0. Thecloud of moving electrons within the gap (called the space charge) buildsup and reduces the field at the cathode surface to zero. From then on asteady current flows between the plates.
Suppose two plates are large relative to the separation (A >> d2 in thefigure) so that edge effects can be ignored and V , ρ, and the electronspeed v are functions of only x.
d
Electron
Cathode (V=0) Anode(V=V )0
A
x
Vacuum Diode – p.2/21
The Child-Langmuir Law
1. What is Poisson’s equation for the region between the plates?
2. Assuming the electrons start from rest at the cathode, what is theirspeed at point x?
3. In the steady state I, the current, is independent of x. How are ρ andv related?
4. Now generate a differential equation for V by eliminating ρ and v andsolve this equation for V as a function of x, V0, and d. Make a plot tocompare V (x) and the potential without the space charge.
5. What are ρ and v as functions of x?
6. Show that
I = KV3/2
0
and find K. This the Child-Langmuir law.
d
Electron
Cathode (V=0) Anode(V=V )0
A
x
Vacuum Diode – p.3/21
Why Should You Care? The Physics 132 Picture
vd
E
vdpositive negative+
Area = A
i
d
t∆
I = JA = nqvdA = AρdV
I = 1
RV
V = IR
2007-10-16 11:35:02
Current (A)0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
Vo
ltag
e (V
)
0
1
2
3
4
5
6
7
8
9
10
Ohm’s Law
Ω 8 ±slope = 46
Ω = 46.5 measR
Vacuum Diode – p.4/21
Why Should You Care, Part Deaux?
January 11, 2006
Vacuum Diode – p.5/21
Coulomb’s Law and Superposition
Measuring the Electrostatic Forceof Charges.
Use a torsion pendulum and charged spheres.
0.04
0.00 0.05 0.10 0.15 0.20 0.250
50
100
150
200
250
300
350
H L
Results
V
0.3
0 1 2 3 4 5 6 7
0
20
40
60
80
100
H L
Results
Vacuum Diode – p.6/21
Coulomb’s Law and Superposition
Measuring the Electrostatic Forceof Charges.
Use a torsion pendulum and charged spheres.
Evidence for Coulomb’s Law.
Fix charge and vary distance.
n=1.99 ± 0.04
0.00 0.05 0.10 0.15 0.20 0.250
50
100
150
200
250
300
350
rHmL
ΘHd
egL
Intermediate Lab Results
V
0.3
0 1 2 3 4 5 6 7
0
20
40
60
80
100
H L
Results
Vacuum Diode – p.6/21
Coulomb’s Law and Superposition
Measuring the Electrostatic Forceof Charges.
Use a torsion pendulum and charged spheres.
Evidence for Coulomb’s Law.
Fix charge and vary distance.
n=1.99 ± 0.04
0.00 0.05 0.10 0.15 0.20 0.250
50
100
150
200
250
300
350
rHmL
ΘHd
egL
Intermediate Lab Results
Evidence for Superposition.
Fix distance and vary charge on one sphere.
Q µ V
Quadratic term: 0.5 ± 0.3
0 1 2 3 4 5 6 7
0
20
40
60
80
100
V HkVL
ΘHd
egL
Intermediate Lab Results
Vacuum Diode – p.6/21
Electric Field of a Ring
A ring of radius r as shown in the figure has a positive charge distributionper unit length with total charge q. Calculate the electric field ~E along theaxis of the ring at a point lying a distance x from the center of the ring.Get your answer in terms of r, x, q. What happens as x → ∞?
r
xq
Q
Vacuum Diode – p.7/21
Electric Field of a Ring
A ring of radius r as shown in the figure has a positive charge distributionper unit length with total charge q. Calculate the electric field ~E along theaxis of the ring at a point lying a distance x from the center of the ring.Get your answer in terms of r, x, q. What happens as x → ∞?
r
xq
Q
Vacuum Diode – p.7/21
Electric Field Lines
Point Charges Electric Dipole Positive Charges
Vacuum Diode – p.8/21
Properties of Electric Field Lines
Field lines start from positive charges (sources) andend at negative ones (sinks).
Are symmetrical around point charges.
Density of field lines is related to strength of field.
Direction of field is tangent to the field line.
Field lines never cross!
Vacuum Diode – p.9/21
Electric Flux
ΦE = EA ΦE = EA cos θ = ~E · ~A ΦE =∮
~E · ~dA
Vacuum Diode – p.10/21
An Example of Electric Flux
A nonuniform magnetic field is described by
~E = (3.0 N/C − m)xx + (4.0 N/C)y
pierces the Gaussian cube shown in the figure. What is theflux through the cube? Note the orientation of thecoordinate system.
Vacuum Diode – p.11/21
Gauss’s Law
What is the flux from a point charge q at the origin through asphere centered at the origin of radius r?
Vacuum Diode – p.12/21
Differential Surface and Volume Elements
Cartesian Coordinates Cylindrical Coordinates Spherical Coordinates
y
z
x
x
y
z
(x,y,z)
θ
φ
θ
φ
dA = dxdy dA = ρdφdz dA = r2d cos θdφ
dτ = dxdydz dτ = ρdφdzdρ dτ = r2d cos θdφdr
Vacuum Diode – p.13/21
Applying Gauss’s Law: Nuclear ~E
The nucleus of a gold atom has a radius R = 6.2 × 10−15 mand a positive charge q = Ze where Z = 79 is the atomicnumber. Assume the gold nucleus is spherical and thecharge is uniformly distributed in the volume. What is theelectric field for any r?
0 5 10 15 20 250.0
0.5
1.0
1.5
2.0
2.5
3.0
H L
E
Vacuum Diode – p.14/21
Applying Gauss’s Law: Nuclear ~E
The nucleus of a gold atom has a radius R = 6.2 × 10−15 mand a positive charge q = Ze where Z = 79 is the atomicnumber. Assume the gold nucleus is spherical and thecharge is uniformly distributed in the volume. What is theelectric field for any r?
0 5 10 15 20 250.0
0.5
1.0
1.5
2.0
2.5
3.0
rHfmL
EH1
021NCL
Electric field of a gold nucleus
Vacuum Diode – p.14/21
An Example of Applying ∇× ~E
A nonuniform magnetic field is described by
~E = (3.0 N/C − m)xx + (4.0 N/C)y
pierces the Gaussian cube shown in the figure. Is this aconservative field?
Vacuum Diode – p.15/21
Applying V
1. What is the potential energy of a point charge?
2. What is the potential inside and outside a uniformlycharged sphere of total charge q and radius R?
3. What is the electric field for the previous question?
Vacuum Diode – p.16/21
The Child-Langmuir Law
In a vacuum diode, electrons are boiled off a hot cathode at potential zeroand accelerated across a gap to the anode at a positive potential V0. Thecloud of moving electrons within the gap (called the space charge) buildsup and reduces the field at the cathode surface to zero. From then on asteady current flows between the plates.
Suppose two plates are large relative to the separation (A >> d2 in thefigure) so that edge effects can be ignored and V , ρ, and the electronspeed v are functions of only x.
d
Electron
Cathode (V=0) Anode(V=V )0
A
x
Vacuum Diode – p.17/21
The Child-Langmuir Law
1. What is Poisson’s equation for the region between the plates?
2. Assuming the electrons start from rest at the cathode, what is theirspeed at point x?
3. In the steady state I, the current, is independent of x. How are ρ andv related?
4. Now generate a differential equation for V by eliminating ρ and v andsolve this equation for V as a function of x, V0, and d. Make a plot tocompare V (x) and the potential without the space charge.
5. What are ρ and v as functions of x?
6. Show that
I = KV3/2
0
and find K. This the Child-Langmuir law.
d
Electron
Cathode (V=0) Anode(V=V )0
A
x
Vacuum Diode – p.18/21
Using Mathematica To Solve the DE
Vacuum Diode – p.19/21
The Child-Langmuir Law: Results
Comparing the Child-Langmuir Law with the no-space-charge solution.
Blue - ’132’ picture
Red - Child-Langmuir
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
V Hunits of VmaxL
Cur
rentHu
nits
ofI m
axL
Vacuum Diode – p.20/21
More Child-Langmuir Law Results
Comparing the Child-Langmuir Law in vacuum with a copper wire.
Red - copper wire
Blue - vacuum
0.0 0.2 0.4 0.6 0.8 1.0100
105
108
1011
1014
V HmegavoltsL
Cur
rent
Den
sityHAm
2L
Wires vs. Vacuum
JCU =1
ρ
V
dρ = 1.7×10−8 Ω−m JCL =
4ǫ09d2
√
2e
meV 3/2 d = 0.1 m
Vacuum Diode – p.21/21