Just understanding capacitors and e-fields

  • #1
It's a problem were we are given the usual - 2 plates separated in air, distance D apart, each of length L, one excess +ve charge the other exces -ve charge

The question then states:
the magnitude between the e-field plates is: E = q/EoL, calculate the potential (blah blah..)

My question is why have they used L?
Wouldn't this give the electric field units of: C / (C^2.N^-1.m^-2)*(m)
(apologies for no latex)
= Newton-metres per Coulomb?

when it should be Newtons per Coulomb (or Volts per metre)?

Is the question assuming 1D plates?
Just looking to get this cleared up, thanks
 

Answers and Replies

  • #2
555
0
I don't know why they used L either... However, you have not specified what q is. Is it a charge, thus units Coulomb? Could it be that q is a charge per length with units C/m ?

I think it's weird using the length of a capacitor plate though... It's the area that usually matters. Usually it's:
[tex]E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}[/tex] so [itex]\sigma = \frac{Q}{A}[/itex].

Maybe in this case [tex]E = \frac{q}{\epsilon_0 L} = \frac{\lambda}{\epsilon_0}[/tex] where [itex]\lambda = \frac{q}{L}[/itex].
 
Last edited:

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