7/28/2019 Jmoore Electrodynamics FinalReview

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7/28/2019 Jmoore Electrodynamics FinalReview

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separation of vars in cylindrical. find potential inside wedge such that all sides are zero potential. start with laplace eq inlindrical:

()

+ 12

22

= 0(no charge) Break up = R()()R

() R

= 2

2 2

= 2

eneral solutions: R() = C + D and () = A cos + B sin oundaries: = 0 at = a, = 0, = - plug in reduce eq to:

(, ) =

nAnsin()(

a2 ) = n

ant E and E and (, 0), (, ), (a, )

= E A1 sin

(1 + a2 1 )

=1

A1

cos

(1 a2 1 )

efine region 1 where youre at the top piece of the wedge ( = ) region 2 is on the circle, = a and region 3 is one bottom piece of the wedge( = 0)

egion 1:

(, = ) = 0 E = 0, 0E = A10

(1 a2 1 )

egion 2:

( = a, ) = 0 E = 2A1 0 sin( )a1, E = 0egion 3:

(, = 0) = 0E = 0, 0E = A10

(1 a2 1)

= then you have a flat line with a bump of radius a and by resolving components we get a constnat in the zrection.

= A21

sin()2 + A21

cos()2 E =

A21

z To find charge on bulb:

=

0 ad so Qbulb = 4A1 0 a, Qslab = 20aA1

Problem 3.7

otential with 3 charges: +q at az -2q at origin, and +q at -a z. Potential is:

= 140

q

|ra| 2q

|r0| +q

|r+a|

Simplifying...

=q

40

1

|raz| +1

|r+az| 2r

ow we can expand the Green function in terms of spherical harmonics: 1|rr

0

rlr

l+1

Pl(cos )

hen for l is even we get =q

20(

0

rl

Pl (cos ) 1r )

0 so r

= a and r

= r Expand for 2 terms.. l=0 and l=2

=q

20

a0

r0+1P0 (cos ) +

a2

r2+1P2 (cos ) 1r

s easy to see that the first term is just 1

rand this will cancel with the last term and well be left with:

=q

20

a2

r3P2 (cos )

hich reduces to =

Q

40 r3

(3cos 2 1) after plugging in P2(cos )nclose this thing with a spherical shell use image charges...

=q

40

1

|ra=z| +1

|r+az| 2r

+

q40

1

|rz|+ 1

|r+z| 2

r

ub in q = b

aq and z = b2

a Ive already done some reducing... then were going to again replace in terms of

herical harmonics...Also for the 2nd term Ive put in r lesser as r and r greater as b2

a

=q

20

1

r+ 1

b+ (

0

rl

1b

ra

b2

lPl(cos )

gain only survives for even l, so we take two terms: l = 0 and l = 2 to get:

=q

20

a2

r3 r2a2

b5

P2 (cos )

hich we can simplify for final answer: =Q

20r3

(1 r5b5

)P2(cos )

Problem 3.9 - Tube with = 0 on top and bottom and = V(, z) on side

ssume separable solutions: = R()Q()Z(z)

or Z, 2Z

z2= k2 Z Z(z) = A sin(kz) + B cos(kz)

ut with our BC, z = 0, = 0 so B must be zero.Z(z) = A sin(kz) where k = nL

because z = L, = 0.

or Q,2Q

2= m2 Q Q() = C sin(m) + D cos(m) But m = 0 because of azimuthal symmetry so

e C term is gone.

or R, 2 R

2+ 1

R

(k2 + m22

)R = 0 Substitute x = k And the results for R are modified Bessel

nctions:R(x) = EIm(x) + F Km(x) but we have a finite potential at s = 0 and we have an interior problem somust be zero, because Km(x) doesnt play nice. Therefore, R(x) = EIm(x)ow we have before using last BC for = b:

( = b , , z) = RQZ =

nmEIm(

n

)A sin( nL

z)

( , , z) =

nm

Im(n

)A sin( n

Lz) 2V

LIm

nb

L

0sin( n

Lz)dz

Problem 4.1

lculate multipole moments for all l and 2 sets of nonvanishers +q at ax and ay and -q at ax and aygure out which is just the sum of all the charges at their respective locations in and because all of them have= a

m = qal Yml ( 2 , 0) + Yml ( 2 , 2 ) Yml ( 2 , ) Yml ( 2 , 32 )nce all have =

2and all Ylm have cosine dependence, they will be Pl(0) We can use this to simplify... we learn

at only m=odd survive.

m = qal

2l+1

4(lm)!(l+m)!

Pml

(0)2[1 im] plug and chug l = 1,l = 3 should get qlm = qlm and

sults specifically as q11, q31 , q33you do the same problem we did in 3.7 with 3 charges all on z, only terms are q20, q40 because azimuthal symmetryd m=0.

o get multipole expansion, take answer q20 and plug into multiple expnasion reduce Pl(cos ) to Pl(0) againcause were on xy plane and l = 0 because weve got Q = 0 Answer is:(r , , ) = 1

20

qa2

r3( 1

2(3(0)2 1)) 1

40

qa2

r3

Problem 4.6

aceless quadrupole tensor - use Q11 = Q22 = 12 Q33 andQ33

e= Q to get W = e

4Q

Ezz

from

W =

(x)(x)dx but for quadrupole term...

W = 16

ij

QijEixj

Problem 4.7

calized charge distribution - right away we make our lives easier by realizing that theres a sin()2 term in the volumearge so we should probably rewrite in terms of spherical harmonics.

(r) = 164r2

er

sin()2

, r , (r) = 0n()2 = 1 cos()2 = 2

3(P0 (cos()) P2 (cos())) - from this we know only l = 0 and l = 2 term

survive from qlm because of orthogonality. should get q00 =1

2

and q20 = 6

5

4

The essence of multipole expansion is only valid in the far field case . = 140

1r 6

r3P2 (cos()

Now to find at any point, we have to use the freespace green function but wait ! this can ALSO be expandedof spherical harmonics, and solved just like part aOR we can rewrite it in terms of no dependence (ie, m = 0) and use this form: G(x,

l

rlr

l+1

Pl (cos )Pl (cos ) Then we have orthogonality again with l = 0 and l = 2 but

to be careful because now our

0

dr integral must be split to

r

0dr and

r

dr

(r, , ) = 140

124

1r

(... + ...)

and for far field, er = 0 and reduces back to part a) also

you have to taylor expand perhaps up to r5

If you see a function of it is desirable to express in terms of Spherical Harmonics might greatly reduce the pMultipole expansion is only valid for far field case not local field.

Problem 4.8concentric cylinder filled with dielectric material.

Solve potential in three regions - inner circle, in between a and b and outside b (inner radius a, outer radius b)

I =

m=0Fmm cos m (includes origin - so 2nd term is not allowed)

II =

m=0

Bmm + Dm

m II I =

m=0Am

m + Cmm

Far field case, this should just look like the regular electric field Eo cos All m > 2 Am will have to be zero, otherwise first term blows up.Also, all the initial constants are not impfinding E because well be taking derivative of

I = F cos , II = (B +D

) cos , I II = (C E0 ) cos Take E as negative gr

potential: E = EI = F cos , EII = (D2 B) cos , EII I = (C2 + E0) cos E = EI = F sin , EII = (B + D

2) sin , EII I = (C2 E0 ) sin D-Boundary and E-Boundary must be continuos across these jumps.0EII I (b) = EII (b), EIII (b) = EII (b) 0 EI (a) = EII (b), EI (a) = EII (a)Four constants, four unknown - solve. this is gross.

Problem 4.10concentric cylinder 1/2 dielectric material, 1/2 with empty space

If E is in the radial dierction, then the normal component is automatically zero and both boundary conditions are conD nda = Qenclosed

Define the Gaussian surface justoutside of a (inner radius) and get E =Q

2r2(+0 )

To get on inner surface, use boundary condition ( D2 D1 ) n12 = with D1 zero in both cases becaof a there is no charge.and we get:

R() Q

2r2 (1+0 )

( E) L(0 ) Q

2r2 (1+)

0

(0E)

Calculate prob. density induced on the surface at r=a, P = 0 eE

Noting that all charge comes from the dielectric side - and that the normal is inward r then we get P = (or P = Q

2r20+0

and the charge on this would be + if it were asking at the other side (b) lots of little

cancel out in the middle, net effect on edges.

Problem 5.3

solenoid of N turns - find B in terms of 1 and 2 For one loop of wire, above z axis all contributions for B fie

except the z component. Just use amperes law. B(z) =04

I

dl|x|2 cos

B(z) =0

2a2

a2+z2

and we can integrate this to figure out solenoid. to get in

B(z) =0 IN

2(cos 1 + cos 2)

Problem 5.6

magnetic flux density, or more familiar, Find B.Make one of radius a going one direction and current going other direction for radius b and add them togethyou add in the vector d which separates the centers of the circles, it turns out not to matter where theBhole(r) =

12

0J(z d)

Problem 5.7

so for a part we get the same thing as before on Bz for 1 loop when we add a second loop a distance b awa

fer more interesting esp. if we put the origin at the halfway point. Restricting to the center is crying for an e0

an(x c)n c is usually zero.

Problem 5.22

based on prob 5.19a) Be =0

2

z+L

(z+L)2 +a2

zz2 +a2

z We need to know this at

B(L) because the force on this cylinder will be nonzero there. Otherwhere it will be zero because n is p erpendicand the M is uniform.In the limit L >> a we add these two up and get that the 1st one cancels with the 2nd term of the 2nd one

F 02

M2A relate this to W = 2ax20

for electrostatics.W = Fx and

W = 12

v

1 M B0 dx analogous to W = 12

v1 P E0dx

M(x) = 14

v

M(x)|xx| dx

+ 14

S

n M(x)da|vecxvecx|

First term goes to zero with uniform M!

Problem 5.13

random tidbit: C for concentric cylinders C =20 L

ln( ba

and F = + Wt

instead of normal -.