homework set #1ee 6318 / phys 6383 spr. 2001goeckner/plasma_sci... · homework set #1ee 6318 / phys...

Homework Set #1 EE 6318 / Phys 6383 Spr. 2001Due Monday Jan 29, 01

Read Lieberman Chapter 1

Unless otherwise noted, each homework problem will be graded out of 10 points.

Problem 1Show that if in one dimension

f v( )dvÚ = nthen

const = n m2pkT

Ê Ë

ˆ ¯

1/ 2

.

Also show that if in three dimensionsf v ( )dv3Ú = n

then

const = n m2pkT

Ê Ë

ˆ ¯

3/ 2

.

Here n is the particle density.

Set 1 Problem 1 Solution:Consider

n = f v( )dvÚ = A exp -mv2

2kTÊ Ë Á

ˆ ¯ ˜ dvÚ

= A 2kTm

exp -x 2( )dxÚNow let us look at I = exp -x2( )dxÚ .

I2 = exp -x2( )dx-•

•

Ú exp -y 2( )dy-•

•

Ú= exp -x2 - y2( )ÚÚ dxdy; transform coordinates

= exp -r 2( )rdrdqÚÚ= 2p exp -r 2( )rdr

0

•

Ú= p exp -r 2( )d r2( )Ú= -pexp -r2( )

0

•

= pTherefore

n = A 2kTm

I

= A 2pkTm

fl

A = n m2pkT

Going to the 3dimensional case is trivial

n = f v( )d3vÚÚÚ = A exp-m vx

2 + vy2 + vz

2( )2kT

Ê

Ë Á

ˆ

¯ ˜ dvxdvydvzÚÚÚ

= A exp-m vx

2( )2kT

Ê

Ë Á

ˆ

¯ ˜ dvxÚ exp

-m vy2( )

2kT

Ê

Ë Á

ˆ

¯ ˜ dvyÚ exp

-m vz2( )

2kTÊ

Ë Á

ˆ

¯ ˜ dvzÚ

= A 2kTm

Ê Ë

ˆ ¯

3/ 2

I3

fl

A = n m2pkT

Ê Ë

ˆ ¯

3/ 2

Problem 2Show that in three dimensions that

E =32

kT

Set 1 Problem 2 Solution:We showed in class that in 1-Dimension

Ewhere

E =12

m v2 f v( )dv-•

•

Ú

=12

m v2n m2pkT

exp mv2

2kTÊ Ë Á

ˆ ¯ ˜ dv

-•

•

Ú

=12

kT

In 3-dimensions we have

E =12

m v2 f v( )d3vÚÚÚ

=12

m vx2 + vy

2 + vz2( )n m

2pkTÊ Ë

ˆ ¯

3/ 2

expm vx

2 + vy2 + vz

2( )2kT

Ê

Ë Á

ˆ


=12

m vx2( )n m

2pkTÊ Ë

ˆ ¯

3/ 2

expm vx

2 + vy2 + vz

2( )2kT

Ê

Ë Á

ˆ


+12

m vy2( )n m

2pkTÊ Ë

ˆ ¯

3/ 2

expm vx

2 + vy2 + vz

2( )2kT

Ê

Ë Á

ˆ


+12

m vz2( )n m

2pkTÊ Ë

ˆ ¯

3/ 2

expm vx

2 + vy2 + vz

2( )2kT

Ê

Ë Á

ˆ


=12

m vx2( )n m

2pkTÊ Ë

ˆ ¯

1/ 2

expm vx

2( )2kT

Ê

Ë Á

ˆ

¯ ˜ dvxÚÚÚ

+12

m vy2( )n m

2pkTÊ Ë

ˆ ¯

1/ 2

expm vy

2( )2kT

Ê

Ë Á

ˆ

¯ ˜ dvyÚÚÚ

+12

m vz2( )n m

2pkTÊ Ë

ˆ ¯

1/ 2

expm vz

2( )2kT

Ê

Ë Á

ˆ

¯ ˜ dvzÚÚÚ

=32

m v2( )n m2pkT

Ê Ë

ˆ ¯

1/ 2

expm v2( )2kT

Ê

Ë Á

ˆ

¯ ˜ dvÚÚÚ

=32

kT

Homework Set #2 EE 6318 / Phys 6383 Spr. 2001Due Wednesday Jan 31, 01

1) Determine the tube length necessary to maintain 75% effective pump speed in laminar fluid andmolecular flow.

Solution Problem 1 set 2

First, we know that1

Seff

=1C

+1

Spump

.

Thus we require that C = 600 l / s . Now we need to two equations.

Fluid: C =p

128hd4

LP1 + P2

2; h = gas vis cosity

Molecular: C =pd2

4

RT2pMmolar

È

Î Í

˘

˚ ˙

1/ 2

1 +34

Ld

Ê Ë

ˆ ¯ z

; where 1 £ z £1.12 is a fudge factor

(Yes, I dropped the square-root in my notes before…) Further, we need to make some simplifyingapproximations

First for fluid flow, we will assume that the main chamber is at atmosphere, 760 Torr or 1013.0mBar. Further, we will assume that the back of the pump is at 0.0 mBar. NowDP = Pchamber - Ppump end = 0( ) =

ISeff

=IC

+I

Spump

fl

I = Pchamber * 0.75 * Spump

and

Ppump top -Ppump end = 0( ) =I

Spump

fl

Ppump top = 0.75 * Pchamber = 760 mBarThus

C = 600 l / s =p

128104

L1773

2; (letting h = 1)

L =p

128104 1773

2 * 600= 362 cm

Now, for Molecular flow, we need to know the Knudsen number

Kn =lmfp

mean- free pathfor all collisions}

dcharacteristic dis tan ceacross system

{=

<1 Molecular~ 1 Transition

>> 1 Fluid

Ï

Ì Ô

Ó Ô

Here, we will assume thatd ª10 cm

fl

lmfp =1

ns≥10 cm

assu min g s ª1E - 16 cm2 (which we can get from a hard sphere approx.)fi n £ 1E15 cm-3

Using Loschmidt’s number, we findPchamber ª 30 mTorr or 0.0377 mBarand hencePpump top ª 22.5 mTorr or 0.0283 mBarWe can now pump this into our equation for molecular conductance

C = 600 L / s =p102

4

RT2pMmolar

È

Î Í

˘

˚ ˙

1/ 2

1 +34

L10

Ê Ë

ˆ ¯

; letting z ª1

fl

L =p103

3

RT2pMmolar

È

Î Í

˘

˚ ˙

1/2

C-

403

Using Mmolar = 28g / mole (N2 ) R = 83.14 mBar L/Mole/K and T = 275 K (room temperature).fi L ª 38 cm

2) Determine the throughput (G • A ) for the system given a pressure of 100 mTorr in the mainchamber.

Solution Problem 2 Set 2

DP = Pchamber - Ppump end = 0( ) =I

Seff

=IC

+I

Spump

fl

I = GA = Pchamber * 0.75 * Spump

Pchamber =100 mTorr fi 3E15 particles / l

fl

I ª 4.5E17 particles / s

Homework Set #3 EE 6318 / Phys 6383 Spr. 2001Due Monday Feb. 5, 01


Problems 2.1, 2.2, 2.3Solution problem 2.1

Maxwell’s equations in the point form are:— Ÿ E = -∂ tB— Ÿ H = J free + ∂tD— • D = r free

— • B = 0

Taking the divergence of the second equation gives:

— • — Ÿ H( ) = 0 = — • J free + — • ∂tD

fl

— • J free = -— •∂ tD= -∂ t— • D= -∂ tr free

Solution problem 2.2Poisson’s equation is-—2f =

re

The physical system looks like

f(x)

X

s0 LL-s0

f0

0

neni

where we have circled important points. We know that the electric field is continuous and thus thefirst derivative of the potential must also be continuous. Further, we know that the potential is zeroat x=0 and x=L. Finally, we know that in the center of the discharge that-—2f = 0 =

re

because ne = ni .

Let us first assume that -—f = E ≠ 0 in the region s ≤ X ≤ L. First E must be a constant andcontinuous. Second, by symmetry E L 2 - b( ) = -E L 2 + b( ) . (This means that if there is anelectric field which points from one side toward the center, then on the other side the electric fieldmust also point toward the center – which is in the opposite direction.) Combining these two ideasrequires that f = const = f0 s £ X £ L - s . We can also see this if we just solve the differentialequation. Using all pieces gives-—2f =

re

fl

∂ x2f =

-ee

ni X £ s0 s £ X £ L - s

-ee

ni L - s £ X

Ï

Ì Ô Ô

Ó Ô Ô

Integrating once and then twice gives

∂ xf =

-ee

ni X + C1 X £ sC2 s £ X £ L - s

-ee

niX + C3 L - s £ X

Ï

Ì

Ô Ô

Ó

Ô Ô

f =

-e

2eniX2 + C1X + D1 X £ sC2 X + D2 s £ X £ L - s

-e

2eniX 2 + C3 X + D3 L - s £ X

Ï

Ì

Ô Ô

Ó

Ô Ô

We know that at X = 0, X = L f = 0 . Further at X = L/2, f = f0 . Thus solving for our D’s gives

f =

-e

2eniX2 + C1X X £ s

C2 X - C2 L + f0 s £ X £ L - s-

e2e

niX 2 + C3 X +e

2eniL2 - C3 L L - s £ X

Ï

Ì

Ô Ô

Ó

Ô Ô

Now matching the inside boundaries and using our symmetry argument above to set C2 = 0. (Wecould do this in an algebraic fashion but this is easier.)C1 =

f0

s+

e2e

nis

C3 = -f0

s+

e2e

ni 2L - s( )

or

f =

-e2e

niX 2 +f0

sX +

e2e

nisX X £ sf0 s £ X £ L - s

-e

2eniX 2 -

f0

sX +

e2e

ni 2L - s( )X +f0

sL -

e2e

ni L2 - sL( )Ê Ë

ˆ ¯

L - s £ X

Ï

Ì

Ô Ô

Ó

Ô Ô

Now, let’s look at the first derivative, and set them equal at the boundaries.f0 =

e2e

nis2 so

f =

-e

2eniX

2 +ee

nisX X £ se

2enis

2 s £ X £ L - s

-e

2eniX 2 +

ee

ni L - s( )X -e2e

ni L - s( )2 +e2e

nis2 L - s £ X

Ï

Ì

Ô Ô Ô

Ó

Ô Ô Ô

or

f =

-f0

s2 X 2 +2f0

sX X £ s

f0 s £ X £ L - s-

f0

s2 X 2 + 2 f0

s2 L - s( )X -f0

s 2 L - s( )2 + f0 L - s £ X

Ï

Ì

Ô Ô

Ó

Ô Ô

Likewise

E = -—f =

-ee

ni s - X( ) X £ s0 s £ X £ L - s

ee

ni X - L - s( )( ) L - s £ X

Ï

Ì Ô Ô

Ó Ô Ô

- or - E = -—f =

-2f0

s2 s - X( ) X £ s0 s £ X £ L - s

2f0

s 2 X - L - s( )( ) L - s £ X

Ï

Ì Ô Ô

Ó Ô Ô

a) and b) Letting s=L/8 gives

-1

-0.5

0

0.5

1

0 0.5 1X/L

phi/

phi0

-20

-10

0

10

20

E/ph

i0

phi (x)/phi0E/phi0

c) Fromf0 =

e2e

nis 2 ª 4kTe

fl

s = 8 ekTe

eni

Ê

Ë Á

ˆ

¯ ˜

1/2

ª The order of a few (~ 3) Debye lengths

Solution problem 2.3This is of course trivial.From class or the book, we know

mn ∂ v∂t

+ v • —r vÊ Ë Á

ˆ ¯ ˜ = DM c

momentumlost viacollisions

1 2 3 - m v f c

momentum changevia particle gain/ loss

1 2 4 3 4 - —r • P + qn E + v Ÿ B( )

Here the external fields are zero (or the charge is zero) and the collisions are ignored or assume to beunimportant. Thus

mn ∂ v∂t


ˆ ¯ ˜ = -—r • P .

In addition, we assume time independence. Thusmn v • —r v( ) = -—r • P

fl

12

mn—r • v v = -—r • P

fl

—r • 12

mn v v + PÊ Ë

ˆ ¯

= 0 - in 1- D

fl

12

mn v 2 + p x( ) = const

To measure the ∂flow in a pipe I might use a flow meter but then I am tired. (Use the pressure!)

Homework Set #4 EE 6318 / Phys 6383 Spr. 2001Due Wednesday Feb 7, 01


Problems 2.5, 2.6 (Note problem 2.5 is SPEED not velocity!)

Solution to problem 2.5

In 1-D the average speed is

v =v f v( )dv

-•

•

Úf v( )dv

-•

•

Ú=

1n

v f v( )dv-•

•

Ú

=m

2pkTÊ Ë

ˆ ¯

1/ 2

v exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙ dv

-•

•

Ú

= 2 m2pkT

Ê Ë

ˆ ¯

1/2

v exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙ dv

0

•

Ú - even function

= 2 m2pkT

Ê Ë

ˆ ¯

1/ 2 12

exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙ dv2

0

•

Ú

=2kTmp

Ê Ë

ˆ ¯

1/ 2

exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙ d

m v( )2

2kTÊ

Ë Á

ˆ

¯ ˜ 0

•

Ú

=2kTmp

Ê Ë

ˆ ¯

1/ 2

exp -u[ ]d u( )0

•

Ú

=2kTmp

Ê Ë

ˆ ¯

1/ 2

where

f v( ) = n m2pkT

Ê Ë

ˆ ¯

1/ 2

exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙

Now doing this in 3-D

v =v f v( )4pv2dv

0

•

Úf v( )4pv 2dv

0

•

Ú

=1n

v f v( )4pv2 dv0

•

Ú

=m

2pkTÊ Ë

ˆ ¯

3/ 2

exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙ 4pv2dv

0

•

Ú

= 4pm

2pkTÊ Ë

ˆ ¯

3/ 2

v 3 exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙ dv

0

•

Ú

= 4pm

2pkTÊ Ë

ˆ ¯

3 /2 12

v2 exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙ dv2

0

•

Ú

= 2 2kTmp

Ê Ë

ˆ ¯

1/ 2 m v( )2

2kTÊ

Ë Á

ˆ

¯ ˜ exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙ d

m v( )2

2kTÊ

Ë Á

ˆ

¯ ˜ 0

•

Ú

= 2 2kTmp

Ê Ë

ˆ ¯

1/ 2

u exp -u[ ]d u( )0

•

Ú

= 2 2kTmp

Ê Ë

ˆ ¯

1/ 2

-u exp -u[ ]0•

+ exp -u[ ]d u( )0

•

Ú[ ]=

8kTmp

Ê Ë

ˆ ¯

1/ 2

where

f v( ) = n m2pkT

Ê Ë

ˆ ¯

3/ 2

exp -m v( )2

2kTÈ

Î Í

˘

˚ ˙

Solution to problem 2.6a) First, we have Boltzmann’s equation

ne = n0 exp eF

kTe

Ê

Ë Á

ˆ

¯ ˜ and ni = n0 exp eF

kTi

Ê

Ë Á

ˆ

¯ ˜

and Poisson’s equation—2F = -

re

=ee

ne - ni( )In spherical coordinates,—2F = r -1∂r

2 rF( )( ) + (r2 sin(q ))-1∂q (sin(q)∂qF) + (r sin(q ))-2∂f2F( )

Assuming spherical symmetry gives—2F = r -1∂r

2 rF( )( )

=en0

e1+

eF

kTe

+ O eF

kTe

Ê

Ë Á

ˆ

¯ ˜

2Ê

Ë Á

ˆ

¯ ˜ -1 -

eF

kTi

- O eF

kTi

Ê

Ë Á

ˆ

¯ ˜

2Ê

Ë Á

ˆ

¯ ˜

Ê

Ë Á Á

ˆ

¯ ˜ ˜

ªe2n0F

e1

kTe

-1kTi

Ê

Ë Á ˆ

¯ ˜

So that

∂r2 rF( ) ª

e2n0rFe

1kTe

-1

kTi

Ê

Ë Á ˆ

¯ ˜

fl

∂r2 u( ) ª

e2n0ue

1kTe

-1

kTi

Ê

Ë Á ˆ

¯ ˜ where l =e2n0

e1

kTe

-1

kTi

Ê

Ë Á ˆ

¯ ˜ Ê

Ë Á ˆ

¯ ˜

-1/ 2

is the debye length

= u / l2

fl

u = u0e± r / l

fl

F = F0

re- r / l (drop the ' +' for physical reasons)

b)

l =ke

e2n0

TeTi

Te - Ti

Ê

Ë Á

ˆ

¯ ˜

Ê

Ë Á

ˆ

¯ ˜

1/ 2

if Te > Ti then

ªke

e2 n0

TeTi

Te

Ê

Ë Á

ˆ

¯ ˜

Ê

Ë Á

ˆ

¯ ˜

1/2

=ke

e2n0Ti

Ê

Ë Á ˆ

¯ ˜

1/ 2

Æ the low temperature component is the important part

Homework Set #5 EE 6318 / Phys 6383 Spr. 2001Due Monday Feb. 12, 01


Problems 3.1, 3.2Solution to problem 3.1

Lab Frame Center of Mass Frame

In the CM frame the total momentum is zero.Thus

˜ v 2˜ v 1

=¢ ˜ v 2¢ ˜ v 1

=m1

m2; ˜ represents the CM frame

Likewise, because of conservation of energy˜ v 2 = ¢ ˜ v 2˜ v 1 = ¢ ˜ v 1

We can now look at the transition from lab frame to CM frame. This requires simply subtracting thevector, Vcm

In the lab frame we have

tanq1 =¢ v 1y

¢ v 1x Lab Frame

=v1 - Vcm sinQ

Vcm + v1 - Vcm cosQCM Frame

=sinQ

Vcm

v1 - Vcm

+ cosQ

But from above,˜ v 2˜ v 1

=¢ ˜ v 2¢ ˜ v 1

=m1

m2, so

tanq1 =sinQ

m1

m2

+ cosQ

We can do the same thing for q2 .

tanq2 =¢ v 2y

¢ v 2 x Lab Frame

=Vcm sin Q

Vcm - Vcm cosQCM Frame

=sin Q

1- cosQ

If energy is not conserved than we get the equations in the book.

tanq1 =¢ v 1y

¢ v 1x Lab Frame

=¢ ˜ v 1 sin Q

Vcm + ¢ ˜ v 1 cosQCM Frame

=sin Q

Vcm

¢ ˜ v 1+ cosQ

tanq2 =¢ v 2y

¢ v 2 x Lab Frame

=¢ ˜ v 2 sinQ

Vcm - ¢ ˜ v 2 cosQCM Frame

=sin Q

Vcm

¢ ˜ v 2- cosQ

Now

˜ v rel = v rel = v1 = Vcmm1 + m2

m1

Ê

Ë Á

ˆ

¯ ˜

and

˜ ¢ v rel = ¢ v rel = ˜ ¢ v 1 - ˜ ¢ v 2

= ˜ ¢ v 1 +Think about it}

˜ ¢ v 2

=m2

m1

Ê

Ë Á

ˆ

¯ ˜ ˜ ¢ v 2 + ˜ ¢ v 2

=m1 + m2

m1

Ê

Ë Á

ˆ

¯ ˜ ˜ ¢ v 2

or

=m1 + m2

m1

Ê

Ë Á ˆ

¯ ˜ m1

m2

Ê

Ë Á ˆ

¯ ˜ ˜ ¢ v 1

.

So…tanq1 =

sinQv rel

¢ ˜ v rel

m1

m2

Ê

Ë Á

ˆ

¯ ˜ + cosQ

tanq2 =sinQ

v rel

¢ ˜ v rel

- cosQ

Solution for Problem 3.2

From the book, the CM differential cross section isI v,Q( ) =

14

pa12 (Yes, I know that the book is not clear on this. See my notes.)

so that

I v1,q1( ) = I vrel ,Q( ) sinQ

sinq1

dQ

dq1

This means that we must determine sin Q

sinq1 and dQ

dq1

First dQ

dq1.

We know from above thattanq1 =

sinQm1

m2

+ cosQ

Further, we knowbdb = I v1,q1( )sinq1dq1 = I vrel ,Q( )sin QdQso thatI v1,q1( ) = I v rel ,Q( ) sinQ

sinq1

dQ

dq1

=14

pa12sinQ

sinq1

dQ

dq1

Now,d

dq1

tanq1 =d

dq1

sin Q

m1 m2 + cosQ

fl

sec2 q1 =cosQ

m1 m2 + cosQ

dQ

dq1_ sin2 Q

m1 m2 + cosQ( )2dQ

dq1

Looking at the left-hand sidesec2 q1 = 1+ tan2 q1

= 1+sinQ

m1 m2 + cosQ

Ê

Ë Á ˆ

¯ ˜

2

=m1 m2 + cosQ( )2

+ sin2 Q

m1 m2 + cosQ( )2

=m1 m2( )2

+ 2cosQ m1 m2 +1m1 m2 + cosQ( )2

The right-hand side becomescosQ

m1 m2 + cosQ

dQ

dq1_ sin2 Q

m1 m2 + cosQ( )2dQ

dq1=

cosQ

m1 m2 + cosQ_ sin2 Q

m1 m2 + cosQ( )2

Ê

Ë Á

ˆ

¯ ˜

dQ

dq1

=m1 m2 cosQ + cos2 Q - sin2 Q

m1 m2 + cosQ( )2

Ê

Ë Á

ˆ

¯ ˜

dQ

dq1

=m1 m2 cosQ + cos2Q

m1 m2 + cosQ( )2

Ê

Ë Á

ˆ

¯ ˜

dQdq1

Combining the two sidesdQ

dq1=

m1 m2( )2+ m1 m2 2cosQ + 1

m1 m2 cosQ + cos2Q

Unfortunately, this would give I v1,q1( ) entirely in terms of Q , which is notparticularly useful. We need to determine q1 in terms of Q and translate. (On apositive note, we can quickly determine the form to go the other way:dq1

dQ=

m1 m2 cosQ + cos2Q

m1 m2( )2+ m1 m2 2cosQ + 1

.)

Thus we go back to the beginningtanq1 =

sinQm1

m2+ cosQ

tanq1m1

m2+ cosQ

Ê

Ë Á

ˆ

¯ ˜ = sinQ

m1

m2sinq1 = sinQ cosq1 - cosQsinq1 = sin Q -q1( )

m1

m2=

sin Q -q1( )sinq1

ddQ

m1

m2= 0 =

ddQ

sin Q -q1( )sinq1

=cos Q -q1( )

sinq11 -

dq1

dQÊ Ë

ˆ ¯

-sin Q - q1( )

sin2 q1cosq1

dq1

dQ

cos Q -q1( )sinq1

=cos Q -q1( )sinq1

sin2 q1+

sin Q -q1( )cosq1

sin2 q1

È

Î Í

˘

˚ ˙

dq1

dQ

cos Q -q1( ) =sinQsinq1

dq1

dQdQ

dq1=

sin Q

sinq1

1cos Q -q1( )

Plugging this into I v1,q1( ) = I vrel ,Q( ) sinQ

sinq1

dQ

dq1 gives

I v1,q1( ) = I vrel ,Q( ) sin2 Q

sin2 q1

1cos Q - q1( )

Now we need sin Q

sinq1.

From m1

m2=

sin Q -q1( )sinq1

sin Q -q1( )sinq1

cosq1 =m1

m2cosq1

sin Q -q1( )sinq1

cosq1 + cos Q -q1( ) =m1

m2cosq1 + cos Q -q1( )

Thussin Q

sinq1=

m1

m2cosq1 + cos Q -q1( )

I v1,q1( ) = I vrel ,Q( ) m1

m2cosq1 + cos Q -q1( )È

Î Í

˘

˚ ˙

21

cos Q -q1( )

Now we need to get cos Q -q1( )Using

sin Q - q1( ) =m1

m2

sinq1

1 - cos2 Q - q1( ) =

fl

cos Q -q1( ) = 1-m1

m2sinq1

Ê

Ë Á

ˆ

¯ ˜

2È

Î Í Í

˘

˚ ˙ ˙

1/ 2

I v1,q1( ) = I vrel ,Q( ) m1

m2cosq1 + 1 -

m1

m2sinq1

Ê

Ë Á

ˆ

¯ ˜

2È

Î Í Í

˘

˚ ˙ ˙

1/ 2È

Î

Í Í

˘

˚

˙ ˙

2

1 -m1

m2sinq1

Ê

Ë Á

ˆ

¯ ˜

2È

Î Í Í

˘

˚ ˙ ˙

-1/ 2

Which is a mess…



Problems 3.5, 3.6, 3.7 (Extra credit 3.9)Solution to Problem 3.5Equation 3.3.3

I vrel ,Q( ) =b0

4sin2 Q 2( )È

Î Í

˘

˚ ˙

2

Equation 3.3.4s90 =

14

pb02

Now the total scattering cross section is determined bys = 2p I v,q( )

0

p

Ú sinqdq

The total scattering cross section for scattering a large angle iss = 2p I v,q( )

p / 2

p

Ú sinqdq

so that

s lrg = 2pb0

4sin2 Q 2( )È

Î Í

˘

˚ ˙

2

sin QdQp / 2

p

Ú

= 8pb0

4sin2 Q 2( )È

Î Í

˘

˚ ˙

2

sin Q 2( )cos Q 2( )d Q 2( )p / 2

p

Ú

= 8pb0

2

16sin3 u( )È

Î Í

˘

˚ ˙ cos u( )d u( )

p / 4

p / 2

Ú

= 8pb0

2

16sin3 u( )È

Î Í

˘

˚ ˙ d sin u( )( )

p / 4

p / 2

Ú

=b0

2

2p

1sin2 u( )

È

Î Í

˘

˚ ˙

p / 4

p / 2

=b0

2

2p 1-

12

È Î Í

˘ ˚ ˙

=14

pb02

Solution to Problem 3.7a) Straight out of the class notes

Example: The differential scattering cross section off of a hard sphere collision in the CM frame.

We know from above that in the center of mass frame that- ˜ v 2˜ v 1

=˜ v 2˜ v 1

=- ¢ ˜ v 2

¢ ˜ v 1=

¢ ˜ v 2¢ ˜ v 1

=m1

m2

,

which can be found from zero total momentum.Likewise, we know that for a hard sphere, the energy lost to the internal structure is assumed to bezero so that˜ v 2 = ¢ ˜ v 2˜ v 1 = ¢ ˜ v 1Now in the CM frame, this implies that we have a collision that looks like

Let us look closely at the moment of impact.

Because the force is only along a12 , and ˜ v 1 = ¢ ˜ v 1 in the CM frame, this forces L1 = L2 . Thus,

a1 = a2 = a andQ = p - 2a

= p - 2arcsin ba12

Ê

Ë Á

ˆ

¯ ˜

Thus we again find

b = a12 sin a( ) = a12 sin p - Q

2Ê Ë

ˆ ¯ = a12 cos Q

2Ê Ë

ˆ ¯ ; and

db = a12 cos a( )daFrom above we havebdb = I v1,q1( )sinq1dq1 = I vrel ,Q( )sin QdQso

bdb = a12 sin a( ) • a12 cos a( )da

=a12

2

2sin 2a( )da

=a12

2

2sin Q - p( )d Q

2Ê Ë

ˆ ¯

= -a12

2

4sin Q( )dQ

= I v rel ,Q( )sin QdQ

rearranging and normalizing

I vrel ,Q( ) =a12

2

4

b)ssc = 2p I v,q( )sinqdq

0

p

Ú Equation 3.1.14

sm = 2p 1 - cosq( )I v,q( )sinqdq0

p

Ú Equation 3.1.15Using I v,q( ) from above gives

ssc = 2pa12

2

4sinQdQ

0

p

Ú

= -12

pa122 cosQ( )0

p

= pa122

sm = 2p 1 - cosQ( )a12

2

4sinQdQ

0

p

Ú

= pa12

2

2sinQ - sin QcosQ( )dQ

0

p

Ú

= pa122 - p

a122

2sin Q( )d sin Q( )

0

p

Ú

= pa122 - p

a122

2sin2 Q( )0

p

= pa122

c)If a = a p

1/ 3 where a = 11.08a03 so that a ª 2.24a0 .

Thus ssc ª 4.4Å2 = 4.4E -16cm2

For a 5 eV electron the velocity is12

mv2 = kT

fl

vc

=2kTm

ª10

.5E6= 0.0045

fl

v = 0.134E9 cm / s

For 20 mTorr X Loschmit’s numbern ª 20 X 3E13 cm-3

= 6E14 cm-3

Sol =

1ns

= 3.78 cmandn = nsv

= 35.5 MHz

d)The fractional energy that a moving particle loses when it collides with a second particle is given byEquation 3.2.18z L =

2m1m2

m1 + m2( )2 1- cos Q( )( ) .

The fractional energy lost when averaged over all angles is given by Equation 3.2.19

z L Q=

2m1m2

m1 + m2( )2

1 - cos Q( )( )I v rel ,Q( )2p sinQdQ0

p

ÚI v rel ,Q( )2p sinQdQ

0

p

Ú

=2m1m2

m1 + m2( )2s m

s sc

For hard sphere collisions

I vrel ,Q( ) =a12

2

4So thatsm = 1- cos Q( )( )I vrel ,Q( )2p sinQdQ

0

p

Ú

=a12

2

41- cos Q( )( )2p sinQdQ

0

p

Ú

=a12

2

42p sinQdQ

0

p

Ú - cos Q( )( )2p sin QdQ0

p

Ú[ ]=

a122

42p sinQdQ

0

p

Ú

= I v rel ,Q( )2p sin QdQ0

p

Ú= s sc

Thus,z L Q

=2m1m2

m1 + m2( )2

The total energy lost per collision is then simplyE zL Q

= E 2m1m2

m1 + m2( )2

Now, the particles are going to collide at a frequency ofn = vs scng

where ng is the target gas density.Thus the power lost for a single electron is simplyE zL Q

n = vsscng E 2m1m2

m1 + m2( )2

= sscng12

m1v3 2m1m2

m1 + m2( )2

= a122 ng

p

2v3 2m1

2m2

m1 + m2( )2

adding all the electrons, which is simply the average of above times the electron density, (and letting1 -> e and 2 -> i) gives

pei = ne E z L Qn

= ngs sc

22me

2mi

me + mi( )2 ne v3

= ngs sc

22me

2mi

me + mi( )2 ne4pme

2pkTe

Ê

Ë Á

ˆ

¯ ˜

3/ 2

v3 exp -me v( )2

2kTe

È

Î Í

˘

˚ ˙ v2 dv

0

•

Ú

ª ngnesscme

2

mi

4pme

2pkTe

Ê

Ë Á ˆ

¯ ˜

3/ 2

v5 exp -me v( )2

2kTe

È

Î Í

˘

˚ ˙ dv

0

•

Ú

ª ngnessckTe( )2

mi

354 2

- by integration by parts

which is not the same as equation 3.5.8 which has a mass ratio term.

Homework Set #7 EE 6318 / Phys 6383 Spr. 2001Due Monday Feb 26, 01

Homework CHEN: 2.1, 2.3, 2.4, 2.5 and 2.7Lieberman: None

Solution to Chen Problem 2.1From the back of Chen

Solution to Chen Problem 2.3vEŸ B =

E Ÿ BB2 = 1000 m / s and B = 1T

fl

E = vEŸ B B =1000 V / m

Solution to Chen Problem 2.7From Maxwell’s equations

D • dSEnclosingsurface

ÚÚ = r freedtVolumeÚÚÚ

eE r z2pr = -enezpr2 r £ aa2 r ≥ a

Ï Ì Ó

; E z = Eq = 0

fl

E r = -ene

2err 2 r £ aa2 r ≥ a

Ï Ì Ó

at r = aE r = -

enea2e

= 9035 V / m

ThusvE ŸB =

EB

= 4.5 km / s


Homework CHEN: NoneLieberman: 4.1, 4.2 and 4.6

Solution to Problem 4.1Almost copying out of the class notes:

From Maxwell’s equations we have,e— • E = r = e ni - ne( )(We will ignore the induced magnetic field.) Then our equation of motion (momentumconservation) and continuity (particle conservation) becomeContinuity Equation

—r • n v( ) +∂n∂t

= 0

Energy Equation

qn E( ) = mn ∂ v∂t


ˆ ¯ ˜

Here we have five items that are changing with time, E, ne ni <ve> and <vi>. We will expand each ofthese items to produce a time average term, denoted with a ‘0’ and an oscillating term denoted with a‘1’. ThusE = E0 + E1 - but E0 ª 0!ne = ne0 + ne1 - but ne0 ª ni0 ª n0

v e = ve 0 + v e1 - but ve 0 ª 0!ni = ni 0 + ni1

v i = vi 0 + v i1 - but vi 0 ª 0!Then our conservation of momentum (energy) equation for either species becomesiwv1 = ±

em

E1

Doing the same thing for the two continuity equationsin0bv1 = iwn1

Finally, we solve Poisson’s Equatione— • E = e ni - ne( )

e— • E1 = e ni0 + ni1 - ne 0 + ne1( )( )-ibeE1 = e ni1 - ne1( )This gives us five equations and five unknownsin0bve1 = iwne1 , in0bvi1 = iwni1 , -ibeE1 = e ni1 - ne1( ) , iwve1 = -

eme

E1 , iwvi1 =e

mi

E1.

Plugging the last two into the first two givesne1 = ib en0

mew2 E1 ,ni1 = -ib n0e

miw2 E1 .

Placing these into Poisson’s Eqn. to eliminate ni1 and ne1 gives

w2 =n0e2

mee+

n0e2

mie

Ê

Ë Á

ˆ

¯ ˜ ,

Solution to Problem 4.2From the equation of motion

mq

d2rdt2 = E

— • mq

d2rdt2 = — • E =

re

Assuming r is a function of ei wt - bx( ) then

r = -w 2 meq

— • r but w 2 =n0e2

me and q = -e so

r = n0e— • r(I guess that I did this backwards!)

Solution to Problem 4.6Copied verbatim from the class notes:

Now, let us assume that we are dealing with (positive) ions. Here, however, the electric field isdetermined by the electrons, not the ions. Thus, we need to replace the electric field with thegradient of the potential and use Boltzmann’s relation on the electron density. Thus, our equationsbecome—r • nv( ) +

∂n∂t

= 0

fl

n0 bv1 = wn1

(as before)

min∂ v∂t


ˆ ¯ ˜ = qn E( ) -gkT—rn

= -en —f( ) - gkTi—rn

mi n0 + n1( )∂ v0

= 0}+ v1

Ê

Ë Á ˆ

¯ ˜

∂t+ v0

= 0}+ v1

Ê

Ë Á ˆ

¯ ˜ • —r v0

=0}+ v1

Ê

Ë Á ˆ

¯ ˜

Ê

Ë

Á Á Á Á

ˆ

¯

˜ ˜ ˜ ˜

= -e n0 + n1( )— f0

= 0}+ f1

Ê

Ë Á ˆ

¯ ˜ -gkTi— n0 + n1( )

min0∂ v1( )

∂t= -en0—f1 - gkTi—n1

fl

min0wv1 = en0bf1 + gkTibn1

Now, we can’t use Poisson’s equation but rather we assume that the change in local density can bemodeled with Boltzmann’s equation. E.g. that the local ion density is the same as the local electrondensity and that the electron density is given by

ni ª ne = n0eef / kTe (This approximation causes some small error)

= n0ee f0

=0}+f1

Ê

Ë Á Á

ˆ

¯ ˜ ˜

/ kTe

= n0eef1 / kTe

ª n0 1+ef1

kTe

Ê

Ë Á

ˆ

¯ ˜

fl

n1 = n0ef1

kTe

This gives us our three equations and three unknownsn1 = n0

ef1

kTe, n0bv1 = wn1 , min0wv1 = en0 bf1 +gkTibn1

Putting together the first two givesv1 =

wb

ef1

kTe.

Plugging this and the first into the third givesw2

b 2 =kTe + gkTi

mi

This is the ion-acoustic or ion-sound waves.

The group velocity isdwdb

=kTe + gkTi

mi

ªkTe

mi

Homework Set #9 EE 6318 / Phys 6383 Spr. 2001Due Monday March 12, 01


Solution to Problem 5.1This is a somewhat poorly worded problem. After looking at the problem for a while, I think thatwhat Lieberman is looking for is for us to assume that the ion diffusion is zero, e.g. Ti = 0. In thiscase the solution is comes from the basic equationGs = ns vs = -Ds—rns + nsm sE - for species ' s'whereD =

gkTmnm

m =q

mnm

- note that I have used my definition for m

ThenGi = nivi = nim iE - for ions

fl

E = Gi

nimi

Ge = nev e = -De—rne + nem eE - for electrons

fl

Ge = -De—rne + nem eGi

nimi

Let ne = ni = n and Ge = Gi = G . Then

G = -De—rn +me

m i

G

fl

G = Demi

m e - m i

Ê

Ë Á

ˆ

¯ ˜ —rn

THE OTHER POSSIBLE SOLUTION IS TO ASSUME

Ge = 0 = -De—rne + nem eE - for electrons

fl

E =De

nem e

—rne

for the electrons. For the ions, we will assume

Gi = nimiE - for ions

fl

Gi = nimiDe

nem e

—rne

fl

G =m i

m e

De—rn

EITHER ONE IS ACCEPTABLE.

Solution to Problem 5.2From∂nS

∂t= -G0 = -— • GS = -—r • -DS—rnS + nSmSE( )

= DS—r2nS - nSm S—r • E

= DA—r2nS

—r2nS = ∂ x

2nS = -G0

DA

Note that we have to have a loss term, - G0, as we are sourcing the volume at the same rate.Integrating twice gives

∂ x2nSdxÚ = -

G0

DA

dxÚ

fl

∂xnS = -G0

DA

x + C1 - integrate again

fl

nS = -G0

2DA

x2 + C1x + C2

applying the boundary conditionsnS x = -l / 2

= -G0

8DA

l2 - C1l2

+ C2 = 0

fl

C2 = C1l2

+G0

8DA

l2

nS x = l / 2= -

G0

8DA

l2 + C1l2

+ C1l2

+G0

8DA

l2 = 0

fl

C1 = 0So

nS = -G0

2DA

x2 +G0

8DA

l2

nmax =G0

8DA

l2 = n0

Solution to Problem 5.4From Equation 5.2.15 we known = n0 cos bx( ), b = n iz / D( )1/ 2

which gives rise tou =

G

n=

-Dn

dndx

=Db

n0 cos bx( )n0 sin bx( )

= Db tan bx( )Thus we find Equation 5.2.22

uB = Db tan b ¢ l 2

Ê Ë

ˆ ¯

Now because flux is conserved we can determine the density at the sheath edge

n x =± ¢ l 2

= nS =G

u x =± ¢ l 2

= n0 cos bx( )x =

± ¢ l 2

= n0 cos b¢ l

2Ê Ë

ˆ ¯

This means that we need to determine b ¢ l 2 . From 5.2.22 this is simplyb ¢ l 2

= arctan uB

DbÊ Ë Á

ˆ ¯ ˜ .

So

nS = n0 cos arctan uB

DbÊ Ë Á

ˆ ¯ ˜

Ê

Ë Á ˆ

¯ ˜

From a handy dandy Trig formula book we findcos arctan x( )( ) =

11+ x2

sonS

n0=

1

1+uB

2

D2b 2

Homework Set #10 EE / Phys 6383 Spr. 2003Due Monday March 19, 03


Solution to Problem 6.1We know from the book that (equation 6.3.11)

-F( )3/ 4 =32

eGi

eÊ Ë

ˆ ¯

1/2 Mi

2eÊ Ë

ˆ ¯

1/ 4

r

=32

eMiGi2

2e2Ê Ë Á

ˆ ¯ ˜

1/ 4

r

Following the derivation in the book, except we will let12

Mi vi2 - vB

2( ) = -e F( )

Thus,

Gi = nivi = ni vB2 -

2eMi

F

Assume that the potential structure is the same as Child’s law, given above. Thus

Gi = nivi = ni vB2 +

2eMi

32

Ê Ë

ˆ ¯

4 / 3 eMiGi2

2e2Ê

Ë Á

ˆ

¯ ˜

1/ 3

r4/ 3

fl

ni = Gi vB2 +

81e4Gi2

4e 2Mi2

Ê

Ë Á

ˆ

¯ ˜

1/3

r4/ 3Ê

Ë Á

ˆ

¯ ˜

-1/ 2

If vB = 0 , then the equation collapses to the form in the book. This equation however does not havethe singularity at r=0.Don’t worry about the plots.

Solution to Problem 6.3The basic equations are conservation of flux combined with conservation of energy to produceequation 6.1.2

ni = niS 1 -2e

miviS2 F

È

Î Í

˘

˚ ˙

-1/ 2

The Boltzmann relation for the electron densityne = neSe

e F / kTe

and for the negative ions in the sheathni - = ni - Se

eF / kTi

and Poisson’s equation for the electric potential—2f = -

re

=ee

ne + ni - - ni +( )

where ni - is the negative ion density and ni + is the positive ion density. Further, we have quasi-neutrality at the sheath edge such thatneS + ni - S = ni+ S .This implies thatni - = ni +S - neS( )eeF / kTi

Plugging all of this into Poisson’s equation gives

—2f = -re

=ee

neSe-e F / kTe + ni + S - neS( )e- eF/ kTi- - ni +S 1 -

2emivi + S

2 FÈ

Î Í

˘

˚ ˙

-1/ 2Ê

Ë Á

ˆ

¯ ˜

Multiplying by —F and integrating gives

—F —2F( ) • dl0

l

Ú = —Fee

neSeeF/ kTe + ni + S - neS( )eeF / kTi- - ni + S 1-2e

mivi +S2 F

È

Î Í

˘

˚ ˙

-1/ 2Ê

Ë Á

ˆ

¯ ˜ • dl

0

l

Ú

12 — —F( )2 • dl

0

l

Ú =ee

— neSkTe

eeeF/ kTe +

kTi

eni+ S - neS( )eeF/ kTi- + 2ni + S

E i +S

e1 -

eFE i +S

È

Î Í

˘

˚ ˙

1/ 2Ê

Ë Á

ˆ

¯ ˜ • dl

0

l

Ú

fl

—F( )2

0

l=

2ee

neSkTe

eeeF/ kTe +

kTi

eni +S - neS( )eeF/ kTi- + 2ni + S

E i +S

e1 -

eF

E i + S

È

Î Í

˘

˚ ˙

1/ 2Ê

Ë Á

ˆ

¯ ˜

0

l

Assuming that —F r = 0( ) = 0 and F r = 0( ) = 0 ,

—F l( )( )2=

2ee

neSkTe

eeeF / kTe - neS

kTe

e+

kTi

eni + S - neS( )eeF / kTi- -

kTi

eni+ S - neS( ) + 2ni+ S

E i + S

e1-

eF

E i + S

È

Î Í

˘

˚ ˙

1/ 2

-1Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô

Ê

Ë Á Á

ˆ

¯ ˜ ˜

Now we want the electric field to be a real positive thus the term in the bracket must be positive.Taking the second order expansion gives

0 £ neSkTe

eeeF / kTe - neS

kTe

e+

kTi

eni +S - neS( )eeF/ kTi- -

kTi

eni + S - neS( ) + 2ni + S

E i +S

e1 -

eF

E i + S

È

Î Í

˘

˚ ˙

1/ 2

-1Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô

Ê

Ë Á Á

ˆ

¯ ˜ ˜

£

neSkTe

e1 +

eF l( )kTe

+e2F2 l( )2k2Te

2

Ê

Ë Á

ˆ

¯ ˜ -1

È

Î Í Í

˘

˚ ˙ ˙

+kTi

eni+ S - neS( ) 1+

eF l( )kTi -

+e2F2 l( )2k 2Ti -

2Ê

Ë Á

ˆ

¯ ˜ -1

È

Î Í Í

˘

˚ ˙ ˙

+2ni +SE i+ S

e1-

eF l( )2E i +S

-e2F2 l( )8E i +S

2Ê

Ë Á

ˆ

¯ ˜ -1

Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô

Ê

Ë

Á Á Á Á Á Á Á Á Á

ˆ

¯

˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜

≥

neS 1+eF l( )2kTe

Ê

Ë Á

ˆ

¯ ˜

È

Î Í Í

˘

˚ ˙ ˙

+ ni + S - neS( ) 1 +eF l( )2kTi-

Ê

Ë Á

ˆ

¯ ˜

È

Î Í Í

˘

˚ ˙ ˙

-ni + S 1 +eF l( )4E i+ S

Ê

Ë Á

ˆ

¯ ˜

Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô

Ê

Ë


ˆ

¯

˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜

- remember F l( ) < 0

≥

neSeF l( )2kTe

Ê

Ë Á

ˆ

¯ ˜

È

Î Í Í

˘

˚ ˙ ˙

+ ni + S - neS( ) eF l( )2kTi -

Ê

Ë Á

ˆ

¯ ˜

È

Î Í Í

˘

˚ ˙ ˙

-ni + SeF l( )4E i +S

Ê

Ë Á

ˆ

¯ ˜

Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô

Ê

Ë


ˆ

¯

˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜

£neS

kTe

+ni +S - neS( )

kTi -

-ni +S

2E i + S

Thus,

vi + S ≥ni + S

Mi

Ê

Ë Á

ˆ

¯ ˜

1/ 2neS

kTe

+ni+ S - neS( )

kTi -

Ê

Ë Á

ˆ

¯ ˜

-1/ 2

Solution to Problem 6.5A) Letvi = m iE; m i =

eMinin

Assume particle conservation, which means that the collision term in the continuity equation is zero.Further, we are examining a DC process so the time derivatives are zero. This means that flux isconserved. Thus

vi =viSniS

ni

= miE

fl

ni =viSniS

miE

Now we can employ Poisson’s equation, under the assumption that the electron density is zero—2F = -— • E = -

ee

ni = -ee

viSniS

miEfl

E— • E = E dEdr

=ee

viSniS

mi

fl

EdE =ee

viSniS

mi

dr

fl

12

dE2 =ee

viSniS

mi

dr = const dl - integrating

12

E2 =ee

viSniS

mi

r

E =2ee

viSniS

mi

r

Now we can get the potential

—F = -E = -2ee

viSniS

m i

r

fl

F = -23

2ee

viSniS

mi

r3/ 2

= Vwallrs

Ê Ë

ˆ ¯

3/ 2

; Vwall = -23

2ee

viSniS

mi

s3/ 2

From this you can get the current etc.Ji = m i

9e8s3 Vwall

2

s =8e9e

viSniS

mi

Ê

Ë Á

ˆ

¯ ˜

-1/3

-V( )wall2/ 3

=8e9e

kTe

Mi

niS

mi

Ê

Ë Á ˆ

¯ ˜

-1/ 3

-V( )wall2/ 3

B) Let

li = 330 p( )-1 cm; p in Torr; p =10 Torr

=1

3300cm so

n in =

vi

speed}

li

=

8kTi

pmi

Ê

Ë Á

ˆ

¯ ˜

1/ 2

li

=1.4895E9( )1/ 2

1 / 3300= 127.3 MHz

Then plugging this into our sheath width, we find s = 0.747 cm (This last number is wrong but Idon’t have the time to recalculate it.)

Doing the same with the Childs Law sheath has a width of s = 1.02 cm

Homework Set #11 EE / Phys 6383 Spr. 2003Due Monday April 9, 03


Solution to Problem 6.6Part one: A probe whose collection area is 2mm X 2mm X 2 sides collects a current of 100 µA. Thismeans that we are collecting a current density of 0.125E-2 A/cm2. According to our simple theoryfor Langmuir probes,

Ji = 0.61en0kTe

Mi

Mi ª 40GeV / c2 If kTe = 2 eV, then

Ji = 2E -14 n0 (A cm)n0 = 6E10 cm-3

Part two:

s =2

3lD

2eVwall

kTe

Ê

Ë Á ˆ

¯ ˜

3/ 4

lD =ekTe

en0= 743 kTe

n0= 4.29E - 3cm

sos = 0.19 mmnow s2 << A so our assumption is OK

Solution to Problem 6.7Part a)Eqn 6.6.2112

m vr2 + vf

2( ) + q F p - Vp( ) =12

m ¢ v r2 + ¢ v f

2( )Eqn 6.6.22svf = a ¢ v fCombining these equations we find

¢ v r2 = vr

2 + vf2( ) +

2 qm

Fp - Vp( ) -s2

a2 vf2

¢ v f =sa

vf

For the particle to reach the probe the final radial velocity at r=a must be zero or negative, e.g. stillheading toward the probe. Thus ¢ v r £ 0 and ¢ v r

2 ≥ 0 . The minimum radial energy must be zero.Thus

vf2 1-

s2

a2Ê Ë Á

ˆ ¯ ˜ + vr

2 +2 qm

Fp -Vp( ) ≥ 0

vf2 £

vr2 +

2 qm

Fp - Vp( )s2

a2 -1Ê Ë Á

ˆ ¯ ˜

or

vf £ vf 0 =vr

2 +2 qm

F p - Vp( )s2

a2 -1Ê

Ë Á

ˆ

¯ ˜

È

Î

Í Í Í Í

˘

˚

˙ ˙ ˙ ˙

1/ 2

This gives us the limits on the possible angular velocities. (The limits on the radial velocity must bethat the particle head toward the probe.)Now we need to use the approximation that Lieberman used. First, that the probe tip is muchsmaller than the sheath width, a << s .

fi vf0 ªas

vr2 +

2 qm

Fp - Vp( )È

Î Í

˘

˚ ˙

1/ 2

Now we will assume that the potential energy is much larger than the initial kinetic energy. Thus,

fi vf0 ªas

2 qm

F p - Vp( )È

Î Í

˘

˚ ˙

1/ 2

.

The final thing that we will assume is that the angular velocity is less than the thermal velocity.(This is a weak assumption.)

Now, we can get and deal with Eqn 6.6.26I = qGA = qA dvz-•

•

Ú vr-•

0

Ú dvr dvf f vr ,vf ,vz( )- vf 0

vf 0

Ú

= qA vr-•

0

Ú dvr dvf f vr ,vf( )-vf 0

vf 0

Ú

= q2psd vr-•

0

Ú dvr dvf f vr ,vf( )-vf 0

vf 0

Úand Eqn 6.6.27

f vr ,vf( ) = nsm

2pkTe

Ê

Ë Á

ˆ

¯ ˜ exp

m vr2 + vf

2( )2kTe

Ê

Ë Á

ˆ

¯ ˜

so that

I = qsdnsmkTe

Ê

Ë Á

ˆ

¯ ˜ vrdvr-•

0

Ú dvf exp-m vr

2 + vf2( )

2kTe

Ê

Ë Á

ˆ

¯ ˜

-vf 0

v f0

Ú

= qsdnsmkTe

Ê

Ë Á

ˆ

¯ ˜ exp -mvr

2

2kTe

Ê

Ë Á

ˆ

¯ ˜ vrdvr

-•

0

Ú dvf exp-mvf

2

2kTe

ª 06 7 8 Ê

Ë

Á Á Á

ˆ

¯

˜ ˜ ˜

-vf 0

vf 0

Ú

= qsdnsmkTe

Ê

Ë Á

ˆ

¯ ˜ exp -mvr

2

2kTe

Ê

Ë Á

ˆ

¯ ˜ vrdvr-•

0

Ú vf -v f0

vf 0

= 2vf 0qsdnsm

kTe

Ê

Ë Á

ˆ

¯ ˜ exp -mvr

2

2kTe

Ê

Ë Á

ˆ

¯ ˜ vrdvr-•

0

Ú

= 2vf 0qsdns exp -mvr2

2kTe

Ê

Ë Á

ˆ

¯ ˜ d

mvr2

2kTe

Ê

Ë Á

ˆ

¯ ˜

-•

0

Ú= 2vf 0qsdns

= 2qsdnsas

2 qm

F p - Vp( )È

Î Í

˘

˚ ˙

1/ 2

Part b)This part is simpleFirst the total current to the double probe is such that there is no net current. ThenI1i - I1e = - I2i - I2e( )Now the current through the probe isI1i - I1e - I2i + I2e = - I(Perhaps California has put all of their charge carriers in their plasmas and not in their power grid.But I think that Lieberman has an extra 2.)so thatI1i - I1e = -

12

I

I2i - I2e =12

I

Now we the Poisson relationI1e = A1 Jsate

eV1 / kTe

I2e = A2 JsateeV2 / kTe

dividingI1e

I2e

=I1i + 1

2 II2i - 1

2 I

=A1

A2ee V1- V2( ) / kTe =

A1

A2eeV/ kTe

Letting A1 = A2 and I1i=I2i=Ii, which gives

Ii + 12 I( ) = Ii - 1

2 I( )eeV / kTe

Ii + 12 I( )e-eV / 2kTe = Ii - 1

2 I( )eeV / 2kTe

Ii eeV / 2 kTe - e-eV / 2kTe( ) = 12 I e- eV/ 2kTe + eeV/ 2 kTe( )

I = 2IieeV / 2kTe - e-eV / 2kTe

e-eV / 2kTe + eeV / 2kTe

Ê

Ë Á

ˆ

¯ ˜

= 2Ii tanh eV / 2kTe( )

Part c)dIdV V = 0

=d2Ii tanh eV / 2kTe( )

dVV =0

=2eIi

2kTe

sec h2 eV / 2kTe( )V= 0

=eIi

kTe

Solution to Problem 6.8The emission current observed for a heated probe is given by

Ie =Ie 0g Vp - fp( )exp

-e Vp -f p( )kTwall

È

Î Í Í

˘

˚ ˙ ˙

Vp < fp

Ie 0 Vp > fp

Ï

Ì Ô

Ó Ô

where g Vp - fp( ) is a function due to orbital motion.Likewise the electron current due to collection of electrons from the plasma is given by

IC =IC0 exp

-e Vp - fp( )kTe

È

Î Í Í

˘

˚ ˙ ˙

Vp < fp

IC 0 ¢ g Vp - fp( ) Vp > fp

Ï

Ì Ô Ô

Ó Ô Ô

Assuming strong emission, the current is space charge limited by a double sheath, e.g. a potential dipbetween the probe and the plasma. (Think of this as an unattached sheath.) This dip is only a fewtimes the probe temperature, Twall. Further Twall << Te. The g functions can be approximated as

g = 1 +e f f - fp( )

kTwall

È

Î Í Í

˘

˚ ˙ ˙

1/ 2

¢ g @ 1 +e f f -fp( )

kTe

È

Î Í Í

˘

˚ ˙ ˙

1/2

ª 1; kTe >> e f f - fp( )

Now looking for when the current match, e.g. determine the floating potential.

IC0 = IC 0 ¢ g Vp - fp( )

= Ie 0g Vp - fp( )exp-e Vp -f p( )

kTwall

È

Î Í Í

˘

˚ ˙ ˙

= Ie0 1 +e f f - fp( )

kTwall

È

Î Í Í

˘

˚ ˙ ˙

1/ 2

exp-e Vp - fp( )

kTwall

È

Î Í Í

˘

˚ ˙ ˙

fl

ln Ie0

IC 0

Ê

Ë Á ˆ

¯ ˜ =e Vp - fp( )

kTwall

-12

ln 1 +e f f - fp( )

kTwall

È

Î Í Í

˘

˚ ˙ ˙

(It helps if you learn to look up references!)

Homework Set #12 EE / Phys 6383 Spr. 2003Due Monday April 16, 03


Solution to Problem 11.1The admittance is simply one over the impedance. (This problem appears to be putting the cartbefore the horse. – at no point do we show why each of the terms correspond to the given values!)At any rate,Yp =

1Zp

= iwep A / dwhere d is the plasma slab thickness, A is the cross sectional area and e p is the dielectric constant.From class,

iwe p = iwe0 +e0w ps

2

iw +n m( )È

Î Í Í

˘

˚ ˙ ˙

= iwe0 + s p[ ]ThenYp =

1Zp

= iw Ad

e0 1 -w ps

2

w 2 - inmw( )È

Î Í Í

˘

˚ ˙ ˙

- let C0 =Ad

e0

= iwC0 - iwC0w ps

2

w 2 - inmw( )È

Î Í Í

˘

˚ ˙ ˙

- let Lp-1 = w ps

2 C0

= iwC0 +1

Lp iw +n m( )È

Î Í Í

˘

˚ ˙ ˙

- let Rp = Lpnm

= iwC0 +1

iwLp + Rp( )

Assumption (b) states

wpe2 >> w 2 1+

nm2

w 2Ê Ë Á

ˆ ¯ ˜

1/2

Thus,Yp = iwC0 +

1iwLp + Rp( )

= iwC0 - iwC0w ps

2

w 2 - inmw( )È

Î Í Í

˘

˚ ˙ ˙

= iwC0 - iwC0w ps

2

w w - inm( )

= iwC0 - iwC0wps

2

w w2 +nm2( )1/ 2

= iwC0 1 -w ps

2

w 2 1+nm

2

w 2Ê Ë Á

ˆ ¯ ˜

1/ 2

È

Î

Í Í Í Í Í

˘

˚

˙ ˙ ˙ ˙ ˙

ª -iwC0wps

2

w 2 1 +nm

2

w 2Ê Ë Á

ˆ ¯ ˜

1/ 2 - via (b)

=1

iwLp + Rp( )

Solution to Problem 11.2 THIS IS ALL SCREWED UP AND I DON’T HAVE TIME TO FIX IT!(SANKET – DO NOT GRADE)

us = uB 1 +plD

2li

Ê

Ë Á

ˆ

¯ ˜

-1/ 2

11.2.53

Ji = ensus ª1.68e02eM

Ê Ë

ˆ ¯

1/ 2 V 3/ 2li1/ 2

sm5/ 2 11.2.54

J1 ª1.52 e0wsm

V1 11.2.55

V ª 0.78V1 11.2.56

Sstoc ª 0.59 2li s0

p2l D2

Ê

Ë Á ˆ

¯ ˜

1/ 2

1+plD

2li

Ê

Ë Á

ˆ

¯ ˜

1/ 2

mns v eu02 11.2.58

Plugging 11.2.53 into 11.2.58


p2l D2

Ê

Ë Á

ˆ

¯ ˜

1/ 2uB

us

mns v eu02

= 0.59 2li s0

p2l D2

Ê

Ë Á ˆ

¯ ˜

1/ 21us

kTe

MÊ Ë

ˆ ¯

1/ 2

mnsv eu02

Further

u02 =

J12

e2ns2

v e =8kTpm

Ê Ë

ˆ ¯

1/ 2

lD =ekTe2 n0

Ê

Ë Á

ˆ

¯ ˜

12

s0 =J1

ewns


p2l D2

Ê

Ë Á ˆ

¯ ˜

1/ 2 1us

kTe

MÊ Ë

ˆ ¯

1/ 2

mnsv eu02

= 0.592li

J1

ewns

p2 ekTe

e2n0

Ê

Ë Á

ˆ

¯ ˜

Ê

Ë

Á Á Á Á

ˆ

¯

˜ ˜ ˜ ˜

1/ 2

1us

kTe

MÊ Ë

ˆ ¯

1/ 2

mns8kTe

pmÊ Ë

ˆ ¯

1/ 2 J12

e2ns2

= 0.59 2lin0

wnse

Ê

Ë Á

ˆ

¯ ˜

1/21us

8mkTe

p3Ê Ë

ˆ ¯

1/ 2 J15/ 2

M1/ 2e3/ 2ns3/ 2

= I can' t find the right steps!

ª 0.6 mkTe

Ê Ë

ˆ ¯

1/ 2

e0V1w2

Solution to Problem 11.3The total impedance (series) isZp =

1iwCs

+ iwLp

Resonance occurs when the imaginary part is zero.

Im Zp = Im 1iwCs

+ iwLpÊ

Ë Á

ˆ

¯ ˜

= Im -iwCs

+iw 2LpCs

wCs

Ê

Ë Á

ˆ

¯ ˜

= 0

fl

1 = w2 LpCs

fl

wres =1 LpCs

From the book,Lp = w pe

-2C0-1

Cs =A

2sm

e0

C0 =Ad

e0

wres = 1 LpCs

= 1 w pe-2d / 2sm

= w pe 2sm dPlugging in the numbersnres = wres / 2p ª 628 MHz

homework set #1ee 6318 / phys 6383 spr. 2001goeckner/plasma_sci... · homework set #1ee 6318 / phys...

Documents