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Electric field above a plane?

  1. Oct 28, 2014 #1
    I'm a bit confused about the electric field above a plate. First, I have come across the equation [itex]E=kQ[/itex] http://www.sparknotes.com/testprep/books/sat2/physics/chapter13section3.rhtml which in many ways makes sense to me because around a point the field would be proportional to [itex]1/{r^2}[/itex], and then for a line it would be proportional to [itex]1/{r}[/itex] etc, however I have not come across this formula anywhere else and am finding it hard to understand why moving away from the plate would not have any impact at all on the strength of the force on a changed partcle. Also, if this were the case then I would expect the field between two parallel plates to be [itex]E=2kQ[/itex] due to the principle of superposition. However I know from elsewhere that the E field between two parallel plates is [itex]E=V/d[/itex] and for a capacitor [itex]C=Q/V[/itex], and in the case of just air between the plates, [itex]C=\epsilon A/d[/itex], so putting that together I get that [itex]E=kQ/r^2[/itex]...

    Have I made a wrong assumption somewhere?

    Thank you!

    EDIT: Sorry, for the last part I used [itex]A=4\pi r^2[/itex] which probably will not apply here. In that case, I am not sure what to do for the last part...
  2. jcsd
  3. Oct 29, 2014 #2


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    Staff: Mentor

    The plate is a plane, extending out to infinity so there are no edge effects. The plane carries charge uniformly distributed. Field lines from these charges space themselves apart and never cross each other, so the only possible field lines arrangement is all perpendicular to the plane, like the hairs on the back of a frightened cat. These field lines extend out to infinity and remain parallel to each other. So no matter how far away from the plane, the field strength is uniform and undiminished.

    Your equations were looking good until the last where r leapt in unheralded and uninvited!
  4. Oct 29, 2014 #3
    Thank you for your reply! The thing is I do not know how to get rid of the r. If my equations were to work out, A would have to be [itex]2\pi[/itex]... And I have no idea why this would be the case. The only thing I can think of is that I have approached the question wrong and talking about the charge and area of the plate does not make sense. Instead I should be using the charge per unit area... Could that be it?
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