Speed of transfer of disturbance in E,M field

[sadhu]

what i have studied is that any change in the electric field travels in space with the speed of light...right
suppose any electric field generated by an infinite plane sheet of charge
E=\sigma/2\epsilon...where \sigma is surface charge density

now suppose \sigma is doubled instantaneously.
this results in a disturbance in electric field traveling linearly with the velocity of light
now consider a gaussian cube of edge e at a distance of (c i.e 3*10^8 m) away from plate
such that one of its face is // to the plate.
suppose its one half is outside the plane ( c distance away from charged plate)and other half on other side of it.
now E field disturbane must have reached only till one half of cube
1 second after the instant when we doubled charge density

now calculating net flux

flux through four surfaces // to EF =0

we are left with two surfaces.

one at a distance < c...A
other > c.....B
at time one sec past the instant when we doubled charge density

flux through A will double but negative of B anyhow net flux will not come out to be 0
but isn't it against Gauss law as the cube has no charge in it

 

[Gear300]

I'm not sure whether Gauss's Law applies to changing electric fields over time or not...but I'm thinking it does not...

 

[sadhu]

well my teacher while prooving the transverse nature of EM wave used it on time varying field

 

[Gear300]

I'm not saying that they're completely invalid for such cases; you could actually reformat/derive/retrieve equations from the equations set for static fields so that they apply to non-static fields. Is it possible that I can see the mathematics your teacher did?

 

[granpa]

now suppose \sigma is doubled instantaneously.

how?

 

[sadhu]

hey gearo here is the working

1)imagine a 3-d cordinate system

2) now suppose a longitudinal wave is traveling along X-axis (left To Right)

3)mark a pont on the X axis

4)draw a parallelopiped at that point of dimension(dx,dy,dz)

closedintegral(E)=0...no charge withen the parallelopiped

as electric waves are supposed to be longitudinal four faces of parallelopiped will be // to Electric field hence flux=0

the face at point X and X+dx are now scrutinised

at X electrin field=E
at X+dx electric field =E+dE

net flux= E(dzdy)+dE(dzdy)-E(dzdy)=0...gauss law(note that electric field should be time varying )
dE(dxdy)=0
dE=0

i.e electric field is static which obviously means no wave

 

[Gear300]

wait...aren't electromagnetic waves transverse? For // vectors, the scalar product shouldn't come out to 0 if the vectors aren't 0? Can you explain a bit further?

 

[sadhu]

well in that working we first assumed that electromagnetic waves are longitudinal and then prooved that such waves are not consistent with gauss law and don't exist

but just mentioning that in a traverse wave the front and back face of parrellelopiprd will be// to electric field henceflux=0
and the four faces will have flux passing through them but flux passing through opp faces will cancel out as Electric field vector only changes with x
E=E(x,t)
hence field lines entering parallelopiped fron one face will leave from the opp one
with the result that net flux=0...fully consistent with gauss law

 

[Gear300]

In the case your teacher presents, the electromagnetic wave seems to have already been passing through the parallelopiped. In the case you presented in your opening post, what's happening is that when increasing the charge, the very front of the "additional" wave is still propagating through the structure. That seems to be the difference between the two situations.