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Bradley j. Nartowt
electromagnetic theory I
Jackson pr 1.3
dr. thorn
due: Monday, august 29, 2011
Charge distributions as Dirac delta-functions: Using Dirac delta functions in the appropriate coordinates, express the
following charge charge distributions distributions as three-dimensional three-dimensional charge charge densities densities (x) . a) in spherical coordinates, coordinates, a charge Q uniformly-distributed uniformly-distributed over a spherical shell of radius R. the charge is symmetric over solid angle, so we must have a charge density of the form (x) k ( r R) ( r) ,
Q ( r ) d 3 x k (r R) r 2dr d 4 k ( r R ) r 2dr k
Q 4 R
2
(r )
Q 4 R2
(r R )
[I.1]
b) in cylindrical coordinates, a charge per unit (z) length uniformly-distributed over a cylindrical surface of radius b. our is unit charge per unit z-length, so
dq dz
dq dz
d dxqdz d 2 x (x)d 2 x . Meanwhile, (x) k ( s b) ( s) , 2
sd , which gives us, ds sd
2
and we integrate d x
2
0
b s
0
k ( s b) ds s d 2 b k
k
2 b
( s)
2 b
( s b)
[I.2]
c) in cylindrical coordinates, a charge Q spread uniformly over a flat circular disk of negligible thickness and radius R. 3
let the disk be located at z = 0. use d x
ds sd dz . The charge density must spike at z = 0 and “shut off” beyond s =
R, so guess ( s, z ) k ( z) ( R s) while using the condition Q (x )d x . we immediately get,
Q
z
R
z
0
2
0
k ( z ) ds sd dz k 2 12 R 2
z
z
( z )dz k
3
Q R
2
( s, z )
Q R2
( R s) ( z ) [I.3]
d) repeat part c using spherical coordinates. we need only make one change to [I.3]: we need ( z ) the property ( x)
1
cos ) . Also, we need a trivial replacement: s r . using ( r co
( x ) and r r . we have the guess (x) RQ 2 ( R r) 1r (cos ) ( r , ) ; seeing if this
works,
(
x
)d x 3
Q R 2
R
2
0
0
1 (cos ) r
1
2
dr d (cos ) d r
RQ2r ( R r ) Q 1 2 R 1 2 Q ( r, ) R2 2 ( c o s )
