jackson 3 3 homework solution

8/12/2019 Jackson 3 3 Homework Solution

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Jackson 3.3 Homework Problem SolutionDr. Christopher S. Baird

University of Massachusetts Lowell

PROBLEM:

A thin, flat, conducting, circular disc of radiusRis located in thex-yplane with its center at the origin,and is aintained at a fi!ed potential V. "ith the inforation that the charge density on a disc at fi!ed

potential is proportional to #R$%$&-'($, whereis the distance out fro the center of the disc,

#a& show that for r )Rthe potential is

r ,,=$V

R

r

l=*

'l

$ l'Rr$ l

P$ lcos

#+& find the potential for rR.

#c& "hat is the capacitance of the disc

SOLUTO!:

"arnin#$he solution that /ac0son gives is wrong. Let us solve the pro+le the wrong way #the way

/ac0son e!pects&, then show why this solution is wrong. hen let us solve the pro+le the right way

and figure out where /ac0son went wrong.

T%e "ron# "a&:

#a& he surface charge density was stated to +e1

= S

R$r$for rRand * otherwise

he three-diensional charge density is then1

=S'

R$r$

/ $r

for rRand * otherwise

o find out what Sis in ters of the potential V, use Coulo+2s law which integrates over all the

charge density to find the potential at the origin and set it e3ual to V1

= '

4* ' 2'' 2d' 2

V='

4 * ' 2' 2 d' 2


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V= '

4*

*

$

*

*

R

S '

R$r2$

( 2/$)r2

'

r2r2

$sin 2d r2d 2d 2

V=S '

$**

R'

R$r2$d r2

V=S'

$ * [sin' r2R ]*

R

S=4* V

5ow 0nowing S, plugging it +ac0 in, we have the final for of the charge density1

=4*V

'

R$

r$

(/$)r

for rRand * otherwise

5ow apply Coulo+2s Law to find the potential at any point on theza!is1

='

4 * ' 2'' 2d' 2

Use '

rr*=

l=*

r,l

r)

l+'Pl(cos 2) to e!pand this #which is allowed +ecause we are on theza!is&.

= '4 * *

$

*

*

R

4 * V 'R$r2$ 2 /$r2 l=*

rl

r)l'Plcos 2 r2$ sin 2 d r2 d 2 d 2

=$V

l=*

*

R'

R$r2$r

l

r)l'Pl* r2 d r2

"e have to treat the two regions separately. Let us loo0 at the r)R2 region +ecause its integral iseasier. 6or r)Rwe also have r) r2 so that1

=$V

l=*

Pl*

rl'

*

R'

R$r2$

r2l'

d r2

Ma0e a change of varia+les u=R$r2$ and udu=r2 d r2

=$V

l=*

Pl*

rl'

*

R

R$u$l/ $ du

5owPl#*& is 7ero for lodd, so only even lters contri+ute. Let us rela+el to ta0e account for this fact1


3/8

=$V

l=*

'

r$ l'

*

R

P$ l*R$u$ldu

8f we do the integral case +y case for l9 *, ', $... the integration is trivial and we soon see a pattern1

r ,,= $V

R

r

l=*

'l

$ l'R

r$ l

5ow we ust ree+er that this is only valid on theza!is. "e can a0e use of the handy theore

that for pro+les with a7iuthal syetry, the general solution is :ust the solution on the 7-a!is,

ultiplied +yP#cos &1

( r , ,)=$V

R

r

l=*

(')l

$ l+'(Rr)$ l

P$ l(cos) for r >R

his is the solution #the wrong one /ac0son e!pects& to the potential in the e!terior region.

#+& o find the potential in the near region #rR&, first note that the pro+le has a7iuthal syetry

and no charge in the near region, so the general solution to the Laplace e3uation holds1

( r , ,)=l=*

(Alrl+Blr

l')Pl(cos)

"e need a finite solution at the origin, so the solution ust have the for1

( r ,, )=l=*

AlrlPl(cos) for r


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=/$ and rR, independent of r. 8t should +e o+vious that the solution a+ove does not reduce to Von the disc. he factorPl#*& is 7ero only for l odd, +ut this solution only has leven. he correct solution

will have only lodd values.

T%e (orrect "a&:5ote that there are really four regions that we need to treat,

separately, as indicted in the diagra. 8n each region, there is no

charge, there is a7iuthal syetry, and the poles are included, sothe solution to the potential has the for1

( r , ,)=l=*

(Alrl+Blr

l')Pl(cos)

he inner regions include the origin, so they ust have allBl7ero

to have a finite solution at the origin. Siilarly, the outer regionsinclude infinity, which we can assue to have 7ero potential,

leading to allAl+eing 7ero in these regions. ;ur solutions in all

regions therefore +ecoe1

out,up(r , , )=l=*

Blrl'

Pl(cos ) for r)Rand < =($

out,down(r , , )=l=*

Bl ,down rl'

Pl(cos) for r > Rand < ) =($

in,up(r ,, )=l=*

AlrlPl(cos) for rRand < =($

in,down(r ,,)=l=*

Al , down rlPl(cos) for r < Rand < ) =($

6irst, due to syetry, the potential at any point in an upper region ust e3ual the potential at the

irror point across thex-yplane1

(z)=(z)

in,up(cos)=in,down(cos ) and out,up(cos )=out,down(cos)

l=*

AlrlPl(cos)=

l=*

Al , down rlPl(cos) and

l=*

B lrl'

Pl(cos)=l=*

Bl ,down rl'

Pl(cos)

Al , down=Al(')l

and Bl , down=Bl(')l

"ith these findings, our solutions now +ecoe1

z

xR

>out,up

>out,down

>in,down

>in,up

V


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out,up(r , , )=l=*

Blrl'



Bl(')lr

l'Pl(cos) for r > Rand < ) =($

in,up(r ,, )=l=*


in,down(r ,, )=l=* Al(')l

rl

Pl(cos) for r < Rand < ) =($

5ote that +y forcing the upper and lower region potentials to +e irror iages, we autoatically ade

the atch up at their interface, and have already ta0en care of this +oundary condition.

5e!t, the potential in the inner regions ust +ecoe Von the disc.

V=l=*

AlrlPl(cos(/$)) and V=

l=*

Al(')lr

lPl(cos(/$))

leading to1

A*=V , l='

AlrlPl(*)=* , and

l='

Al(')l

rlPl(*)=*

5ote thatPl#*& 9 * for all lodd, in which case the last two e3uations are autoatically satisfied. 6or leven,Pl#*& is not 7ero, so1

Al=* for leven, l!*

;ur solution so far is1

out,up(r , , )=l=*Blrl'



Bl(')lr

l'Pl(cos) for r > Rand < ) =($

in,up(r ,, )=V+ l=',?,@. .


in,down(r ,, )=V+ l=',?,@. ..

Al(')lr

lPl(cos ) for r R and /2

5ote that now that lis odd, #-'&lis always :ust -'. he potential in the inner-down region therefore

+ecoes in,down(r ,, )=V l=',?,@. ..

AlrlPl(cos) . he solutions in the two inner regions can now +e

co+ined intoin (r ,, )=V+sgn(cos) l=',?,@...

AlrlPl(cos ) where sgn#cos


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out(r ,, )=B*

r +sgn(cos)

l=',?,@...

B lrl'

Pl(cos)

in(r ,, )=V+sgn(cos) l=',?,@...

AlrlPl(cos )

5e!t, the potentials of the inner and outer regions should atch at r9R1

in(R ,,)=

out(R ,,)

V+sgn(cos ) l=',?,@...

AlRlPl(cos)=

B*

R+sgn(cos)

l=',?,@. .

B lRl'

Pl(cos )

he Legendre polynoials are orthogonal, so we atch up coefficients. Matching up all coefficients,we find1

B*=V R and Bl=AlR$ l+'

he solution so far +ecoes1

out(r ,,)=V R

r+sgn(cos)

l=',?,@...

AlR$ l+'

rl'

Pl(cos) for r)R

in(r ,, )=V+sgn(cos) l=',?,@...

AlrlPl(cos ) for rR

All theAlat this point are ar+itrary, so let us redefineAlasAl(Rlto a0e these e3uations syetric,

leading to1

out(r ,, )=VRr+sgn(cos ) l=',?,@...

Al

(Rr)

l+'

Pl(cos ) for r)R

in(r , ,)=V+sgn(cos) l=',?,@...

Al(rR)l

Pl(cos) for rR

he last set of coefficients can +e found +y relating the electric field across the plate1

(E$E')"n'$=#*

(Ein,downEin,up)"= #* where #= S

R$

r$ for a conducting plate

o find out what Sis in ters of the potential V, use Coulo+2s law to integrate over all the charge

density and find the potential at the origin and set it e3ual to V1

V= '

4* # (' 2)' 2 d a


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V= '

4*

*

$

*

R

S '

R$r2$

'

r2r2sin 2d r2d 2

V=S '

$**

R'

R$r2$d r2

V=S'

$ * [sin' r2R ]*

R

S=4 * V

Using this value, the +oundary condition on the electric field across the plate now +ecoes1

(% in,down+% in,up)" = 4V 'R$r$

'

r

&in,down&

+'

r

&in,up&

=4V

'

R$r$

&in,down

& +

& in,up&

=4V

r

R

'

'(r/R)$

erfor a +inoial e!pansion on the right side, using'

'x$='+

'

$x

$+'"?$"4

x4+

'"?"@$"4"C

xC+...

&in,down&

+ & in,up&

=4V[

r

R+

'

$(r

R)?

+'"?$"4 (

r

R)@

+'"?"@$"4"(

r

R)

+...]&in,down&

+&in,up&

=4V[

l=',?,@. ..

(l$)! !(l')! ! (rR)

l

]

&in,down&

+&in,up&

=4V[

l=',?,@. ..

(')l'$ Pl'(*)(rR)

l

]A Legendre polynoial identity was used in the last step to get the right side in a for that weanticipate will +e on the left side. 5ow evaluate the derivatives1

$ l=',?,@...

Al(rR)l

[ && Pl(cos)]=/$=4V[

l=',?,@...

(')l'$ Pl'(*)(rR)

l

]

$ l=',?,@...

Al(rR)l

[ &&xPl(x)]x=*=4 V[

l=',?,@...

(')l'

$ Pl'(*)(rR)l

]


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$ l=',?,@...

Al(rR)l

[ l x Pl(x)l Pl'(x)x$' ]x=*=4 V[

l=',?,@. ..

(')l'

$ Pl'(*)(rR)l

]

$ l=',?,@...

Al

(

r

R

)

l

[ l Pl'(*)]=4V

[ l=',?,@...

(')l'

$ Pl'(*)

(

r

R

)

l

]Al=

$V

l (')

l+'$

;ur final solution is therefore1

out(r ,,)=V R

r+sgn(cos)

$V

l=',?,@. ..

(')l+'$

l (Rr)l+'

Pl(cos) for r)R

in(r ,, )=V+sgn(cos) $V l=',?,@...

(')

l+'$

l (rR)

l

Pl(cos) for rR

5ote that this solution o+eys all the +oundary conditions that it should. ;n the plate,Pl#cos

jackson 3 3 homework solution

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