Is the Acceleration of a Charge in a Uniform Electric Field Constant?

[Rodgerd]

Consider a test charge is moving in and electric field of a parallel plate capacitor then :

Force F = ma

In an electric field, F = qE

For an electron in field E,

qE = ma

=> a = qE/m

As the charge is in uniform electric field "E" will be constant . As "E" is the force per unit charge so "qE" will be constant and mass is always constant. Therefore, acceleration is constant??

I'm very confused in it. what does an uniform electric field actually means.
is the uniform e.field of a single charge same as the electric .field of a parallel plate capacitor?

 

[jedishrfu]

This may help:

http://en.wikipedia.org/wiki/Electric_field

there's some discussion on uniform electric fields

A uniform field is one in which the electric field is constant at every point. It can be approximated by placing two conducting plates parallel to each other and maintaining a voltage (potential difference) between them; it is only an approximation because of edge effects.

 

[sophiecentaur]

. . . . .
I'm very confused in it. what does an uniform electric field actually means.
is the uniform e.field of a single charge same as the electric .field of a parallel plate capacitor?

This, to me, seems like you want to consider the force on a pair of parallel plates (with their distributed charges), due to a single charge.
When we talk of a Field, it just describes the force on a unit charge (/mass etc), if it were placed in that field. It is not necessary to think in terms of the field that would be around that charge if it were out on its own somewhere.
The concept of a field is really quite artificial - we are just so used to the idea that we tend to avoid questioning to too deeply. But you seem to be questioning it - reasonably enough.
For instance, we can think of the force between two current conducting wires by looking at the field of either wire and then working out the Lorenz force on the other wire, due to that field. OR we could do it the other way round. The same would apply for two point charges - or, of course, in that case, the Coulomb force can be calculated directly.