How Electronics Started! and Jlab Hits the Wall!

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 electrons in a low-pressure space. Almost all depend on the thermal emission of electrons. 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 potential and accelerated across a gap to the anode at a positive potential V0. The cloud of moving electrons within the gap (called the space charge) builds up and reduces the field at the cathode surface to zero. From then on a steady current flows between the plates. Suppose two plates are large relative to the separation (A>>d2 in the figure) so that edge effects can be ignored and V , ρ, and the electron speed v are functions of only x.

Electron A

Cathode (V=0) Anode(V=V0 )

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 their speed at point x? 3. In the steady state I, the current, is independent of x. How are ρ and v related? 4. Now generate a differential equation for V by eliminating ρ and v and solve this equation for V as a function of x, V0, and d. Make a plot to compare V (x) and the potential without the space charge. 5. What are ρ and v as functions of x? d 6. Show that 3/2 Electron A I = KV0

and ﬁnd K. This the Child-Langmuir law. x

Cathode (V=0) Anode(V=V0 )

Vacuum Diode – p.3/21 Why Should You Care? The Physics 132 Picture

∆ positive vd t negative +

A I = JA = nqvdA = ρd V

vd Area = A E Ohm's Law

i 10

d 9 ± Ω 8 slope = 46 8

Voltage (V) Ω Rmeas = 46.5 7

1 6 I = V R 5

V = IR 3

0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 Current (A) 2007-10-16 11:35:02

Vacuum Diode – p.4/21 Why Should You Care, Part Deaux?

January 11, 2006

Vacuum Diode – p.5/21 Results 350 300 0.04 250 200 150 100 50 0 0.00 0.05 0.10 0.15 0.20 0.25 HL

Results 100 0.3 80 V 60

0 1 2 3 4 5 6 7 HL

Coulomb's Law and Superposition

Measuring the Electrostatic Force of Charges.

Use a torsion pendulum and charged spheres.

Vacuum Diode – p.6/21 Results 100 0.3 80 V 60

0 1 2 3 4 5 6 7 HL

Coulomb's Law and Superposition

Measuring the Electrostatic Force of Charges.

Use a torsion pendulum and charged spheres.

Intermediate Lab Results Evidence for Coulomb’s Law. 350 300 n=1.99 ± 0.04 250

Fix charge and vary distance. L 200 deg H 150 Θ 100 50 0 0.00 0.05 0.10 0.15 0.20 0.25 rHmL

Vacuum Diode – p.6/21 Coulomb's Law and Superposition

Measuring the Electrostatic Force of Charges.

Use a torsion pendulum and charged spheres.

Intermediate Lab Results Evidence for Coulomb’s Law. 350 300 n=1.99 ± 0.04 250

Fix charge and vary distance. L 200 deg H 150 Θ 100 50 0 0.00 0.05 0.10 0.15 0.20 0.25 rHmL

Intermediate Lab Results 100 Evidence for Superposition. Quadratic term: 0.5 ± 0.3 80 Q µ V 60 Fix distance and vary charge on one sphere. L

deg 40 H Θ 20

0 1 2 3 4 5 6 7 V HkVL

Vacuum Diode – p.6/21 Electric Field of a Ring

A ring of radius r as shown in the ﬁgure has a positive charge distribution per unit length with total charge q. Calculate the electric ﬁeld E~ along the axis 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 Q x q

Vacuum Diode – p.7/21 Electric Field of a Ring

r Q x 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) and end 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~ H

Vacuum Diode – p.10/21 An Example of Electric Flux

A nonuniform magnetic ﬁeld is described by

E~ = (3.0 N/C − m)xxˆ + (4.0 N/C)ˆy pierces the Gaussian cube shown in the ﬁgure. What is the ﬂux through the cube? Note the orientation of the coordinate system.

Vacuum Diode – p.11/21 Gauss's Law

What is the ﬂux from a point charge q at the origin through a sphere centered at the origin of radius r?

Vacuum Diode – p.12/21 Differential Surface and Volume Elements

Cartesian Coordinates Cylindrical Coordinates Spherical Coordinates

z (x,y,z) φ y θ x z φ θ x y

dA = dxdy dA = ρdφdz dA = r2d cos θdφ dτ = dxdydz dτ = ρdφdzdρ dτ = r2d cos θdφdr

Vacuum Diode – p.13/21 3.0 2.5 2.0 1.5

E 1.0 0.5 0.0 0 5 10 15 20 25 HL

Applying Gauss's Law: Nuclear E~

The nucleus of a gold atom has a radius R = 6.2 × 10−15 m and a positive charge q = Ze where Z = 79 is the atomic number. Assume the gold nucleus is spherical and the charge is uniformly distributed in the volume. What is the electric ﬁeld for any r?

Vacuum Diode – p.14/21 Applying Gauss's Law: Nuclear E~

3.0

L 2.5 C

N 2.0

21 1.5 10 H

E 1.0 0.5 0.0 0 5 10 15 20 25 rHfmL

Vacuum Diode – p.14/21 An Example of Applying ∇× E~

A nonuniform magnetic ﬁeld is described by

E~ = (3.0 N/C − m)xxˆ + (4.0 N/C)ˆy pierces the Gaussian cube shown in the ﬁgure. Is this a conservative 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 uniformly charged 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 zero and accelerated across a gap to the anode at a positive potential V0. The cloud of moving electrons within the gap (called the space charge) builds up and reduces the field at the cathode surface to zero. From then on a steady current flows between the plates. Suppose two plates are large relative to the separation (A>>d2 in the figure) so that edge effects can be ignored and V , ρ, and the electron speed v are functions of only x.

Electron A

Cathode (V=0) Anode(V=V0 )

Vacuum Diode – p.17/21 The Child-Langmuir Law

and ﬁnd K. This the Child-Langmuir law. x

Cathode (V=0) Anode(V=V0 )

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.

1.0

Blue - '132' picture 0.8

L Red - Child-Langmuir max I 0.6 units of H 0.4 Current 0.2

0.0 0.0 0.2 0.4 0.6 0.8 1.0

V Hunits of VmaxL

Vacuum Diode – p.20/21 More Child-Langmuir Law Results

Comparing the Child-Langmuir Law in vacuum with a copper wire.

Wires vs. Vacuum

1014 L 2 Red - copper wire m 11 A

H 10 Blue - vacuum

108

Current Density 105

100 0.0 0.2 0.4 0.6 0.8 1.0 V HmegavoltsL

V ǫ0 e 1 × −8 − 4 2 3/2 JCU = ρ = 1.7 10 Ω m JCL = 2 V d = 0.1 m ρ d 9d rme

Vacuum Diode – p.21/21