# Just understanding capacitors and e-fields

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

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:
$$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$$ so $\sigma = \frac{Q}{A}$.

Maybe in this case $$E = \frac{q}{\epsilon_0 L} = \frac{\lambda}{\epsilon_0}$$ where $\lambda = \frac{q}{L}$.

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