Electric Field Intensity

Only so long as the plate diameter (or, for a square plate, side length) is much larger than the plate separation.

 

As a general statement it sounds wrong to me !!!
The equation for electric field STRENGTH (I assume that is what your text means as intensity) is
E = V/d where V = the voltage between the plates and d is their separation.
Where did the text come from?

 

Your'e right, technician, that E = V/d, but for the parallel plate capacitor with plate separation << plate diameter, Gauss's law yields another expression for E, namely E = Q/[(epsilon-nought)A]. So if (like David Furlong) we use charge rather than voltage as our electrical variable, E is indeed independent of plate separation, with the proviso I gave above.

 

Assuming a constant charge (capacitor disconnected from battery) the voltage will increase when you increase the distance between the plates so that the ratio V/d stays the same and the electric field will be constant. Of course, as long as the distance is small compared with the width of the plates.

 

Thanks for responses Nasu and Philip !!!!!
I had a battery connected to parallel plates in mind.
Cheers

 

In this case (battery connected) the electric field will be dependent on the distance between plates. The charge will change when d changes.

 

It's because of symmetry. If the width and height of the plates are much larger than their separation, they are effectively flat and infinite planes. For a flat infinite plane of uniform charge distribution, symmetry dictates that the only possible field produced is uniform and constant. And you don't need two plates. One plate will have the same result. If you are very close to a single plate with uniform charge, such that it appears to be infinite and height and width, the field will be constant and uniform.