f07 practice midterm solutions

8/8/2019 F07 Practice Midterm Solutions

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Physics 505 Fall 2007

Practice Midterm Solutions

This midterm will be a two hour open book, open notes exam. Do all three problems.

1. A rectangular box has sides of lengths a, b and c

c

x

y

z

a

b

a) For the Dirichlet problem in the interior of the box, the Greens function may be

expanded as

G(x,y,z ; x , y , z ) =

m =1

n =1

gmn (z, z )sinmx

asin

mxa

sinny

bsin

nyb

Write down the appropriate differential equation that gmn (z, z ) must satisfy.

Note that sin kx satises the completeness relation

m =1

sinmx

asin

mxa

=a2

(x x )

Hence the Greens function equation

2x G( x, x ) = 4

3 ( x x )has an expansion

m,n

2x gmn (z, z )sin

mxa

sinmx

asin

nyb

sinny

b

= 4(z z )4ab

m,n

sinmx

asin

mxa

sinny

bsin

nyb

Working out the x and y derivatives on the left-hand side yields

m,n

d2

dz 2 ma

2

nb

2

gmn (z, z )sinmx

asin

mxa

sinny

bsin

nyb

= 16ab

(z z )m,n

sinmx

asin

mxa

sinny

bsin

nyb


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However, since sin kx forms an orthogonal basis, each term in this sum mustvanish by itself. This results in the differential equation

d2

dz 2 2mn gmn (z, z ) =

16ab

(z z ) (1)where mn = (m/a )

2 + ( n/b )2 is given in part c). Note that the Fourier sineexpansion automatically satises Dirichlet boundary conditions for x and y. Theremaining boundary condition is that gmn (z, z ) vanishes whenever z or z is equalto 0 or c.

b) Solve the Greens function equation for gmn (z, z ) subject to Dirichlet boundaryconditions and write down the result for G(x,y,z ; x , y , z ).

We may solve the Greens function equation (1) by rst noting that the homoge-neous equation is of the form

g (z )

2

mng(z ) = 0

This is a second-order linear equation with constant coefficients admitting thefamiliar solution

g(z ) = Ae mn z + Be mn z

However, we want g(z ) = 0 when z = 0 or z = c. This motivates us to writeout the solutions

u(z ) = sinh mn z 0 < z < zv(z ) = sinh[ mn (c z )] z < z < c

The Greens function solution is then given by

gmn (z, z ) =Au(z ) z < zBv (z ) z > z

The matching conditions

g< = g> ,d

dzg< =

ddz

g> +16ab

then give the system

A sinh mn z = B sinh[ mn (c z)],A cosh mn z = B cosh[ mn (c z)] +

16ab mn

which may be solved to yield

A =16 sinh[ mn (c z)]

ab mn (cosh mn z sinh[ mn (c z)] + sinh mn z cosh[ mn (c z)])=

16 sinh[ mn (c z)]ab mn sinh mn c

=16

ab mn sinh mn cv(z)


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and

B =16 sinh mn z

ab mn (cosh mn z sinh[ mn (c z)] + sinh mn z cosh[ mn (c z)])=

16 sinh mn zab mn sinh mn c =

16ab mn sinh mn cu(z)

As a result, the full Greens function solution is then given by

gmn (z, z ) =16

ab mn sinh mn cu(z< )v(z> )

=16

ab mn sinh mn csinh mn z< sinh mn [(c z> )]

Hence

G( x, x ) =16ab

m,n

1 mn sinh mn c

sinmx

asin

mxa

sinny

bsin

nyb

sinh mn z< sinh mn [(c z> )](2)

c) Consider the boundary value problem where the potential on top of the box is(x,y,c ) = V (x, y ) while the potential on the other ve sides vanish. Using theGreens function obtained above, show that the potential may be written as

(x,y,z ) =

m =1

n =1

Amn sinmx

asin

nyb

sinh mn z

where mn = (m/a )2 + ( n/b )2 andAmn =

4ab sinh mn c

a

0dx

b

0dy V (x, y )sin

mxa

sinny

b

Since we only have to worry about the potential on the top of the box (and sincewe assume there is no charge inside the box), the Greens function solution maybe written

( x ) = 1

4 z = c (x ) G ( x, x )n da=

14 z = c V (x , y ) G ( x, x )n dz dy

(3)


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Noting that the outward-pointing normal n on the top of the box is in the + zdirection, we compute the normal derivative of (2)

G ( x, x )n z = c

=G ( x, x )

z z = c

= 16ab

m,n

1 mn sinh mn c

sin mxa

sin mxa

sin nyb

sin nyb

sinh mn z mn cosh mn [(c z )] z = c= 4

m,n

4ab sinh mn c

sinmx

asin

nyb

sinh mn z

sinmx

asin

nyb

Inserting this into (3) then straightforwardly gives the desired result. Note thatthe primes may be dropped from the double integral Amn once it has been isolatedfrom the rest of the expression.

2. The potential on the surface of a sphere of radius a is specied by

V 0

0

0

V (, ) =0, 0 < V 0 ,

0, <

There are no other charges in this problem.

a) Show that the potential outside the sphere may be expressed as

(r,, ) =l =0 ,2 ,4 ,6 ,...

V 0 [P l +1 (cos ) P l 1 (cos )]ar

l +1P l (cos )

where we take P 1 (x) = 0. Note that Legendre polynomials satisfy the relation(2l + 1) P l (x) = P l +1 (x) P l 1 (x).There are several ways of solving this problem. Perhaps the most straightforwardis to realize from azimuthal symmetry that the potential necessarily admits aLegendre expansion

( x ) =l

lar

l+1P l (cos )


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The quadrupole is given by l = 2. Hence the quadrupole moment is related tothe 2 term in the expansion. Since we do not care about normalization (we onlycare to maximize the moment) it is sufficient to write

q2 ,0 2 P 3 (cos ) P 1 (cos )We extremize this quantity by taking a derivative with respect to and settingthe result to zero. Noting that taking a derivative simply undoes the integration,we end up solving

dq2 ,0d

P 2 (cos )sin = 12 (3cos2 1)sin = 0

This is solved by = 0 and = cos 1 (1/ 3). Clearly the quadrupole momentvanishes when = 0 (as the entire sphere is at a constant zero potential). Hencethe maximum is when = cos 1 (1/ 3), provided we restrict 0 / 2.Note that > / 2 does not really make sense, except in a formal manner whereV 0 V 0 (corresponding to interchanging the integration limits). Technically,the problem should have asked to maximize the magnitude of the quadrupolemoment instead of to make q2 ,0 as positive as possible (which would depend onthe sign of V 0 ).

3. A line charge on the z axis extends from z = a to z = + a and has linear chargedensity varying as

x

0

0

| z |

z

a

a

y (z) = 0 z , 0 < z a

0 |z| , a z < 0

where is a positive constant. The total charge on the 0 < z a segment is Q (andthe charge on the a z < 0 segment is Q).a) Calculate all of the multipole moments of the charge distribution. Make sure to

indicate which moments are non-vanishing.

Noting that a uniformly charged line charge on the + z axis has charge density

= 0

2r 2(cos 1)

we see that a varying line charge yields

=(r )2r 2

(cos 1)


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This may be checked by observing

dq = d3 x = r 2 drdd (cos ) = (r )drd2 cos =1

(Note that there is no distinction between r and z for cos = 1.) Hence, for thepositive and negative line charge, we have

= 0 r

2r 2[(cos 1) (cos + 1)]

We may normalize 0 by integrating from 0 to a to obtain the total charge Q

Q = a

0(z) dz =

a

0 0 z dz =

0 a +1

+ 1

Hence

= Q + 12ar 2

ra

[(cos 1) (cos + 1)]

The multipole moments are then

qlm = r l Y lm () r 2 dr d = Q

+ 1a

a

0r l

ra

dr Y lm ()[(cos 1) (cos + 1)] d2

By azimuthal symmetry, only the m = 0 moments are non-vanishing

ql, 0 = Q + 1

aa l +1

+ l + 1 1

1 2l + 14 P l (cos )[(cos 1) (cos + 1)] d(cos )= Qa l

+ 1 + l + 1 2l + 14 [P l (1) P l (1)]

= Qa l + 1

+ l + 1 2l + 14 [1()l ]Hence all moments vanish unless l is odd and m is zero. Then

ql, 0 = Qa l + 1

+ l + 1 2l + 1 l oddb) Write down the multipole expansion for the potential in explicit form up to the

rst two non-vanishing terms.


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The multipole expansion gives

=1

4 0l,m

42l + 1

qlmY lm (, )

r l+1

= 14 0l odd 42l + 1 ql, 0 P

l (cos )r l+1

=2Q

4 0 rl odd

+ 1 + l + 1

ar

lP l (cos )

=2Q

4 0( + 1) a

+ 2P l (cos )

r 2+

( + 1) a3

+ 4P 3 (cos )

r 4+

=2Q

4 0( + 1) a

+ 2cos

r 2+

( + 1) a3

2( + 4)5cos3 3cos

r 4+

(4)

c) What is the dipole moment p in terms of Q, a and ?

Note that the dipole term in (4) has the form

=1

4 02Q( + 1) a

+ 2zr 3

Comparing this with the dipole expression

=1

4 0 p xr 3

gives

p =2Q( + 1) a

+ 2z

Alternatively, one could compute directly

p = x d3 x = a

axQ( + 1) |z|

a +1sgn(z) dz

x = y =0

The z component is the only non-vanishing component

pz = 2 Q( + 1) a

0|z| +1a +1

dz =2Q( + 1) a

+ 2

f07 practice midterm solutions

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