The electric field produced by oscillating
charges contained in a slab
Ref: Feynman Lectures Vol-1, Chapter 31
1. An accelerated charge produces electromagnetic fields
2. Calculation of the field produced by charges oscillating
in a thin plane sheet
3. The field produced by charges in a slab of finite thickness
The electric field produced by oscillating charges
contained in a slab
1. An accelerated charge produces electromagnetic fields
Y
q
Ey
Line of sight
(1)
X
P
r
Ey (P) = Electric field at point
“P” at the time “t”
c2
Component of the
acceleration of
the charge
perpendicular to
the line of sight
The accelerations ay is evaluated at a retarded
time t- r/c because the signal takes a time r/c to
arrive from the charge location to the point P
More general, the electric field
produced by the accelerated charge q is
- oriented perpendicular to the line-ofsight, and
- has a magnitude proportional to the
component of the acceleration along
the normal direction of the line-ofsight.
q
P
(2)
E (P) = - 4π ε
22
o cc
Normal component of the
charge's acceleration
Electric fields produced
by a charge whose
acceleration points in the
vertical direction (on the
plane of the paper).
2 Calculation of the field produced by charges oscillating
in a thin plane sheet
η
P
- Plane section full of charges, all oscillating synchronously
with the same amplitude xo and the same phase.
- There are η charges per unit area of the plane
- Each charge is equal to q , and moves up and down (around its
average position) according to,
x= xo COS (ωt) = Real ( xo eiωt)
(3)
a = - xo ω2 eiωt
(4)
Acceleration of the charge q
(located at point Q) at the
time t
q
P
Electric field at point P, at the time t,
due to a charge q located at Q is
proportional to the acceleration that
the charge q had at the earlier time
t - r/c
E (
at P due to the
charge q located
at point Q
)
~
- xo ω2 e iω (t - r/c)
Replacing (4) in (2)
(
E (
at P due to the
charge q located
at point Q
) =
(5)
Let's consider that the oscillating charges on the glass plane sheet are
excited by a light beam incident on, for example, a cross section of the
Let's
consider that the oscillating charges on the glass plane are excited by a light
glass.
beam incident on a given
A of
theat
glass
Thecross
totalsection
electric
field
"P" is the result of a
contribution from all the charges in this
Sheet of
The total electric field at P will be the
Sheet of
cross section area
charges
charges
λ
result of a contribution from all the
charges in this cross section.
z
P
Since the cross section area A is assumed to be far away from the observation point P,
considering
the omission of having to consider the component of the acceleration normal to the
line of sight is justified (a more detailed justification of this step is provided in the
appendix-2).
(A more detailed justification is provided in the attached
Appendix-1 file.)
zz
dA
A
P
rr
(
(6)
E (at P) =
A
r
P
z
(
E (at P) =
(7)
ρ=0
Sheet of
charges
=
=
(
E (at P) =
r=z
(8)
E (at P)
(
=
(9)
=
r=z
(10)
Evaluation of the integral:
r=z
Sheet of
charges
η
z
P
r
r+dr
Let's see what happens if we assume, first, that the
charge density on the glass plate η is constant.
Then the integral
can be interpreted as
r=z
= ( dr) e -i (w/c)z + ( dr) e -i (w/c)(z+ dr) + …
(11)
the sum of many small complex numbers, each of
magnitude dr and with an angle that increases
by
starting from θo= - (w/c)z .
Im
Δr
Real
Δr
Δr
Δr
Real
Let's calculate R
(12)
(13)
(14)
Im
Real
But
η
may have a definite value if we assume that
(the charge density) taper off as r increases. Let's see why.
The integral above becomes,
Integral= ηο( dr) e -i (w/c)z + η1 ( dr) e -i (w/c)(z+ dr) +
(15)
Im
Real
Δr made big just for
illustration purposes
The dashed line correspond to the case where the adding complex numbers have the
same magnitude Δr ; i.e. when η (r) is constant (independent of r).
The solid line correspond to the addition of complex numbers whose magnitude
decreases progressively; this happens when η (r) decreases as r increases
For very small
(infinitesimal)
values of Δr we
should have,
Im
Real
(16)
(17)
ηο
Replacing (17) in expression (9),
Sheet of
charges
E (at P, t) =
P
z
(18)
whereηη
thesurface
surfacecharge
charge
where
ο o isisthe
] the sheet.
atdensityDWU
the center of
E (at P, t) =
(19)
Can we say that the field at P is independent of the position of the plate?
(See APPENDIX-1 of these Lecture Notes).
3. The field produced by charges in a slab of finite
thickness
IntheAppendix1weshowthattheelectricfieldproducedbyan
infinitesimallythininplate,atpointalongthecentralaxisandawayfrom
plane,isindependentofthepositionoftheplate.
Accordingly,ifwehadnotatinfinitesimalthinplate,butathickplateof
thickness“d”,thenthetotalelectricfieldatPwillbeasuperpositionofthe
fieldsproducedbymanyinfinitesimalsheets.Theonlydifferencewillcome
fromtheterm KR.
Forasheetofcharges:
K'A) wasthetotalnumberofchargesintheregionofinterest
(Kbeingthesurfacechargedensity)
Foraslabofthickness“d”
N(d'A)willbethetotalchargesfromtheregionofinterest
(Nbeingthevolumetricchargedensity.)
Thus, in the expression for the field produced by a sheet of charges we have
to replace KR byNod.(WeputN0,sinceweagainassumethatthevolumetric
chargedensityN(r)tapperoff,decreasingawayfromthecenterofthe
plate;otherwisetheintegralwouldnotconverge.)
E(0,0,z)=
q N o d x0 Z iZ (t z/c)
(i )e
c
2H 0
(20)
(7)
X
(0, 0, z)
P
Y
d