Parallel plate capacitor with a dielectric?

AI Thread Summary
When a battery is connected to a parallel plate capacitor with a dielectric, it pushes charge onto one plate and pulls charge from the other until the capacitor's voltage matches the battery's EMF. If the capacitor's voltage is less than the EMF, the excess voltage can lead to breakdown if it exceeds the capacitor's breakdown voltage, resulting in sparks. If the dielectric does not fill the space between the plates, Gauss's law can be used to calculate the electric field strength, with the formula E = Q/(ε*A). To find the charge Q, the voltage across the capacitor, which equals the EMF of the battery, should be used in the equation Q = CV. Understanding these principles clarifies how capacitors function in circuits with dielectrics.
basenne
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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.
 
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hi basenne! :smile:
basenne said:
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?

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
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?

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)
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?

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 :smile:
 
That really cleared up my questions.

Thanks a lot!
 
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