The electric field produced by oscillating charges contained in a slab

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1 The electric field produced by oscillating charges contained in a slab Ref: Feynman Lectures Vol-1, Chapter 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

2 The electric field produced by oscillating charges contained in a slab 1. An accelerated charge produces electromagnetic fields Y q Line of sight r E y E y () = - c 2 X (1) Electric field at point at the time t Component of the acceleration of the charge perpendicular to the line of sight The accelerations a y 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 q 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. E () = - 4π ε o c 2 (2) c 2 Normal component of the charge's acceleration

3 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 - lane section full of charges, all oscillating synchronously with the same amplitude x o 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= x o COS (ωt) = Real ( x o e iωt ) (3)

4 a = - x o ω 2 e iωt Acceleration of the charge q (located at point Q) at the time t (4) q Electric field at point, 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 at due to the E ( ) - x o ω 2 e iω (t - r/c) charge q located at point Q ~ Replacing (4) in (2) at due to the E ( ) = 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 glass. consider that the oscillating charges on the glass plane are excited by a light beam incident on a given The cross total section electric A of field the at glass "" is the result of a Sheet of contribution from all the charges in this Sheet of The total electric field at will be the charges cross section area charges result of a contribution from all the charges in this cross section. λ

5 Since the cross section area A is assumed to be far away from the observation point, 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.) da A r r E (at ) = ( (6) A r E (at ) = ( Sheet of charges = = E (at ) = ρ = 0 ( (7) r= (8)

6 E (at ) = ( = r= (9) Evaluation of the integral: r= (10) Sheet of charges η 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= = ( dr) e -i (w/c) + ( dr) e -i (w/c)(+ dr) + (11) the sum of many small complex numbers, each of magnitude dr and with an angle that increases by starting from θ o = - (w/c). Im Δr Δr Δr Δr Real

7 Real Let's calculate R (12) (13)

8 (14) Im Real But may have a definite value if we assume that η (the charge density) taper off as r increases. Let's see why. The integral above becomes, Integral= η ο ( dr) e -i (w/c) + η 1 ( dr) e -i (w/c)(+ dr) + (15)

9 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)

10 (17) η ο Replacing (17) in expression (9), Sheet of charges E (at, t) = (18) where ηη ο o is isthe surface charge at density the center of the sheet. E (at, t) = (19) Can we say that the field at is independent of the position of the plate? (See AENDIX-1 of these Lecture Notes).

11 3. The field produced by charges in a slab of finite thickness In the Appendix 1 we show that the electric field produced by an infinitesimally thin in plate, at point along the central axis and away from plane, is independent of the position of the plate. Accordingly, if we had not at infinitesimal thin plate, but a thick plate of thickness d, then the total electric field at will be a superposition of the fields produced by many infinitesimal sheets. The only difference will come from the term. For a sheet of charges: A) was the total number of charges in the region of interest ( being the surface charge density) For a slab of thickness d N (d A) will be the total charges from the region of interest (N being the volumetric charge density.) Thus, in the expression for the field produced by a sheet of charges we have to replace by N o d. (We put N 0, since we again assume that the volumetric charge density N(r) tapper off, decreasing away from the center of the plate; otherwise the integral would not converge.) E (0, 0, ) = Nod x 2 q 0 0 c ( i ) ei ( t /c) (20) (7) X (0, 0, ) Y d