Electric Field Magnitude b/w Capacitor Plates

[ComptonFett]

Electric field magnitude between capacitor plates independent of distance? [SOLVED]

The electric field of a point charge Q is defined as

$E=\frac{F}{q}=\frac{\frac{1}{4\pi \varepsilon _{0}}\frac{Qq}{r^{2}}}{q}=\frac{1}{4\pi \varepsilon _{0}}\frac{Q}{r^{2}}$

where q is the test charge. The magnitude of an electric field is therefore inversely proportional to the distance (squared) from the point charge.

However, my physics textbook declares that the magnitude of an electric field between the plates of a capacitor is independent of the range between the plates. Specifically,

$E=\frac{Q}{\varepsilon _{0}A}$

where A is the surface are of the plates and Q is the charge of each plate. Distance is not part of the equation.

This appears very counterintuitive to me. The plates are charged with a large number of point charges and to me it would make sense that the magnitude of the electric field between the plates would be inversely proportional to their distance.

Obviously I'm missing something but what?

 

[ZapperZ]

It assumes that it is an INFINITE PLATE, and that the edges are not in play. In that case, the E-field has no dependence on the distance from the plate. The E-field geometry remains perpendicular the the plate no matter how far away you go.

Zz.

 

[ComptonFett]

Thanks, you instantly turned counterintuitive to intuitive.

It's a shame that neither of my high school physics series mentioned that the formulas are based on an infinite plane. I guess that's obvious to most readers but to some (sigh) it isn't.