# Document 10541081 ```291196624
5/28/2016
Assignment HW
26
Calculate
v  r  by convolution
methods for the
following test problem.
Consider a charged
charge density 1. Gauss’s
law tells us that
4r2 E(r)=-(4/3)r3 for r&lt;R
and 4r2 E(r)=-(4/3)R3 for r&gt;R
so that
v(r)=1000/(3r3) r&gt;R
v(r)=50-r2/6 r&lt;R
We also have
V (r )   d 3 r '  (r ') / | r  r ' |
which is a convolution integral. for\CONVOL.for
Interestingly the transform of 1/r is not trivial
g  r   1/ r

G  f    exp j 2 f  r
 1r d r
3
Noting that the z axis of the r coordinate system can be located
along f.
G  f   2 

0
r2  1
exp  j 2 fr   d   dr



1


r
The integral over mu is easy
G  f   2 

0

1
jf


0
r 2 exp  j 2 fr   exp   j 2 fr 
dr
r
j 2 fr
exp  j 2 fr   exp   j 2 fr   dr
This nice integral does not converge. Introduce an epsilon as
a convergence factor.
G f  
1
jf


0
exp  j 2  f  j  r  

 dr
  exp   j 2  f  j  r 
So that
G f  


1 
1
1



jf  j 2  f  j  j 2  f  j  
1
2 f  f  
2
2

f
 j  f  j   
For more detail than desired, see test_problem
1
  f 2 
2
1/1
```