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In my class we derived an expression for an electric field due to an infinitely long plane of charge and it given as:

$$\bf{E} = \frac{\sigma}{\epsilon_0}\hat{\bf{r}}$$ where $\sigma$ it is the surface charge density on the plane.

But the same was directly applied for the parallel plate capacitors and capacitors are made of plates of finite length. How is it that the relation holds?

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The equation for magnitude of the electric field from a single infinite sheet of charge is not the one you gave, it is

$$E=\frac{\sigma}{2\epsilon_o}$$

Then the field between two infinite parallel sheets of charge is

$$E=\frac{\sigma}{\epsilon_o}$$

But the same was directly applied for the parallel plate capacitors and capacitors are made of plates of finite length. How is it that the relation holds?

The second equation holds for a parallel plate capacitor of finite dimensions provided that the distance $d$ between the plates is much less than the dimensions of the plates. Then the field is uniform except at the ends of the plate (edge effect). So the dimensions of the plates, in actuality, don't have to be "infinite", just very large compared to the plate separation.

Then for the capacitor we have a uniform field of magnitude $E$ that is related to the plate separation $d$ and the voltage $V$ across the plates by

$$E=\frac{V}{d}$$

Hope this helps.

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  • $\begingroup$ So ya mean to say that there is no way the formula derived for infinite sheet or plane of charge can be directly used for parallel plate capacitors. $\endgroup$ – Orpheus May 13 at 15:54
  • $\begingroup$ @Orpheus I don't understand what you are asking. Why would you think the field for a single sheet would apply to the field in a parallel plate capacitor capacitor which has two sheets? The second equation applies to the capacitor, not the first. $\endgroup$ – Bob D May 13 at 16:30
  • $\begingroup$ Got it @Bob D...Thank you! $\endgroup$ – Orpheus May 13 at 17:07

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