The Electric Field Above a Plane: Have I Made a Wrong Assumption?

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SUMMARY

The discussion centers on the electric field above an infinite plane charged uniformly. The participant initially references the equation E=kQ, which is applicable for point charges, but struggles with its application to a plane. They correctly note that the electric field between two parallel plates is given by E=V/d, and they express confusion over the uniformity of the electric field strength above the plane, which remains constant regardless of distance from the plate. The participant concludes that they may need to consider charge per unit area instead of total charge to resolve their confusion.

PREREQUISITES
  • Understanding of electric fields and their properties
  • Familiarity with the concept of charge density
  • Knowledge of capacitor equations, specifically E=V/d
  • Basic grasp of the principle of superposition in electrostatics
NEXT STEPS
  • Study the derivation of the electric field due to an infinite plane sheet of charge
  • Learn about charge density and its role in calculating electric fields
  • Explore the concept of electric field lines and their implications
  • Investigate the differences between point charges and continuous charge distributions
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as anyone seeking to deepen their understanding of electric fields and charge distributions.

21joanna12
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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...
 
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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!
 
NascentOxygen said:
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!
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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