Download The field produced by charges in a slab of finite thickness “d”

The field produced by charges in a slab of finite thickness “d”

This section shows that the electric field produced by an infinitesimally-thin
plate, at point along the central axis and far away from plane, is independent
of the position of the plate.
[Basically, the justification argument is that if the sheet is shifted to the right,
then the separation between the sheet at the observation point decreases,
hence given rise to a change in phase of the field arriving at P. But, the field
at the sheet would have also been shifted. These two phase shift compensate.
This argument is detailed below].
Consider two plates, one located at z = 0 and having 1 charges per unit area and
another at z =  having 2 charges per unit area. Charges on these plates will be excited
by the incident external field. We wish to calculate the electric field produced by the
plate’s oscillating charges at a point P.
Y
2
1
External
incident
electric field
P
Einc ( z,t )  Eo ei(ωt - kz )
z
Z

z
At z = 0
the electric field
changes with time as
E(0,t)  Eoeiωt
At z = 
the electric field changes with
time as
E(  ,t)  e - ik  E eiωt
o
Since the local charges at each plate respond to the local electric fields, the acceleration
of the charges at each plate will have, then, similar phase differences
At z = 0
the acceleration
of the charges vary
with time as
a(0,t )  a eiωt
o
At z = 
the acceleration
of the charges vary
with time as
a( ,t )  e- ik  a eiωt
o
The charges at the different plates will create an electric field at P(0, 0, z)
(1)
We will assume that the charges on the plate at the right are excited by the incident
radiation that did not interacted with the charges on the left. We also arbitrarily
discount that the fields produced by the charges on the left plate do not excite the
charges on the right plate.)
The external fields put in motion the charges on the plates
Motion of the charges
at z = 0
x(t)  xo (ω) eiωt
v(t)  ix (ω) eiωt
o
Motion of the charges
at z = 
x(t)  xo (ω) e - ikeiωt
v(t)  ix (ω) e - ikeiωt
o
The charges in motion create an electric field at point P(0, 0, z)
Contribution from the Plate-1 to the field at P
E1 ( z, t )  -1
q
2 c
i xoei (t - z/c )
(2)
o
which was basically obtained from the following calculation
~ [acceleration of one charge ] x [ a value resulting when
]
at plate-1
integrating the contribution
from many charges
~ [ acceleration at plate -1 ] But evaluated
at a time before t
[ i (c/w) ]
How long before?
Answer: a time t = distance/c
where “distance” is the separation between the
plate and the point pf observation
For the case of sheet-1: “distance” =z
~[
eit ]at t- z/c
~[
eit-z/c)
[ i (c/w) ]
] [ i (c/w) ]
Plate-1
Here z is the distance from the point P to the plate-1.
Contribution to the field at P from the Plate-2
E2(z,t)
~ [acceleration of one charge ] x [ a value resulting when
]
at plate-2
integrating the contribution
from many charges
(3)
~ [ acceleration at plate -2 ] But evaluated
some time before t
[ i (c/w) ]
How long before?
Answer: a time t = distance/c
where “distance” is the separation between the
plate and the point pf observation.
For the case of sheet-2: “distance” = z-
~ [
e - ik eiωt ] at t - (z - ) /c [
~ [
e - ik eiω [t -
( z -  )/c]
i (c/w) ]
] [ i (c/w) ]
Using k= /c
~ [
eiωt e
-i z/c
] [ i (c/w) ]
plate-2
(4)
Notice, one obtains the same result expressed in (3) for the plate-1.
E2 ( z,t )  - 2
q
2 c
[ i xo (ω) eiω( t - z/c )]
Independent of 
(5)
o
Contribution to the field at P from the sheet-2
That is, from (4) and (5), the electric field produced by a plate is independent of the
position of the plate along the z-axis.


The main conclusion we draw from the above analysis is that,
the net effect of a sheet of charges on the point of observation “P” is to
produce a field component that is /2 lagging the field produced by the
source.
No matter where we place the sheet, the net effect is to produce a field
that is  /2 lagging the field produced by the source.

The above result implies that if we had not at infinitesimal-thin plate, but a thick
plate of thickness “d”, the total electric field at P will have the form obtained for an
infinitesimally-thin plate, except that instead of we would have to input Nod
(where No is the number of charges per unit volume, at the center of the plate. No
decreases away from the center.)
E due to a plate ( z,t )  - N o d
of thickness d
External
incident
electric field
Einc(z,t)  Eoei(ωt - kz )
q
2 c
[ i xo (ω) eiω( t - z/c )]
o
Y

xo()
d
P
z
Z