Jan 19, 2001

Notes for Lecture Notes for Lecture #2

Examples of solutions of the one-dimensional Poisson equation

Consider the following one dimensional charge distribution:

r(x) = ì
ï
ï
í
ï
ï
î
0
for x < -a
-r0
for -a < x < 0
+r0
for 0 < x < a
0
for x > a
(1)

We want to find the electrostatic potential such that

d2 F(x)
d x2
 = - r(x)
e0
 ,
(2)
with the boundary condition F(-¥) = 0.

In class, we showed that the solution is given by:

F(x) = ì
ï
ï
í
ï
ï
î
0
for x < -a
(r0/(2e0)) (x + a)2
for -a < x < 0
-(r0/(2e0)) (x - a)2 + (r0 a2)/e0
for 0 < x < a
(r0 a2)/e0
for x > a
.
(3)

The electrostatic field is given by:

E(x) = ì
ï
ï
í
ï
ï
î
0
for x < -a
-(r0/e0) (x + a)
for -a < x < 0
(r0/e0) (x - a)
for 0 < x < a
0
for x > a
.
(4)

The electrostatic potential can be determined by piecewise solution within each of the four regions or by use of the Green's function G(x,x¢) = x < , where,

F(x) = 1
e0
 ó
õ 		¥

-¥ 
 G(x,x¢) r(x¢) d x¢.
(5)

In the expression for G(x,x¢) , x < should be taken as the smaller of x and x¢. It can be shown that Eq. 5 gives the identical result for F(x) as given in Eq. 3.

File translated from TEX by TTH, version 2.20.
On 19 Jan 2001, 09:06.