Parallel plate capacitor with a dielectric?

basenne

If you have a simple circuit with only a battery and a parallel plate capacitor with a dielectric, what exactly happens when the battery is turned on?

Also, if the voltage of a capacitor with a dielectric is less than the EMF of the battery, what happens to the rest of the voltage?

One last question... if the dielectric doesn't fill the space between the parallel plates, how does one calculate the electric field strength in the gap in between a plate and the dielectric? I know that you'd use Gauss's law and find that E = Q/(ε*A)... but how would you find Q? Would you use Q=CV? If so, would you use the EMF of the battery or the voltage of the capacitor?

Thanks, as always for all your help.

tiny-tim

hi basenne!
the battery starts "pushing" charge onto one plate of the capacitor, and "pulling" charge off the other plate

eventually, the voltage across the capacitor is the same as the emf
do you mean the breakdown voltage of the capacitor?

if the emf exceeds the breakdown voltage, then sparks will pass across the middle of the capacitor

if the emf doesn't exceed the breakdown, then the capacitor voltage is always the same as the emf (eventually)
the capacitor voltage is the same as the emf

so use that to find D (the electric displacement field) …

D is independent of the dielectric …

then find E from D, for each part of the space

basenne

That really cleared up my questions.

Thanks a lot!