Electric Fields Created by Capacitor

[justadream]

Correct me if I'm wrong but if you put a charge (e.g., an electron) in the electric field created by an infinitely wide capacitor, then the magnitude of the acceleration that the charge feels is constant anywhere in that field.

Why is that? I would think that as the electron gets closer to the positively charged end of the capacitor, it would accelerate (smaller radius according to Coulomb's law).

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[MedHopeful20134]

Electric field is constant between the 2 plates of a capacitor.
E = V/d

Now, F = q E = m a
Acceleration is equal to qE/m (and is constant as q, E and m are not changing).

 

[Cawolf]

The reason the electric field can be approximated as constant is due to the interaction between the oppositely charged plates.

The field lines point in the same way, but when you are close to the positive plate, you are far from the negative plate - and vice versa.

Since the fields originate at different points at the surface of the plates, these average out to make a constant field (with no field outside). This is an approximation of course, that assumes you are not near the edge of the plate (or an infinite plate as you said).