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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

http://micro.magnet.fsu.edu/electromag/java/diode1/index.html

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



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


A

x


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


Why Should You Care, Part Deaux?

January 11, 2006


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





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





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


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



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


Electric Field Lines

Point Charges Electric Dipole Positive Charges


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!


Electric Flux

ΦE = EA ΦE = EA cos θ = ~E · ~A ΦE =∮

~E · ~dA


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.


Gauss’s Law

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


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


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


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


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?


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?



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


A

x



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


A

x


Using Mathematica To Solve the DE


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


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


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