Parallel plate capacitor with a dielectric?

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SUMMARY

When a battery is connected to a parallel plate capacitor with a dielectric, charge is transferred to one plate while the other plate loses charge until the voltage across the capacitor matches the electromotive force (EMF) of the battery. If the capacitor's voltage is lower than the EMF, the capacitor will eventually reach the same voltage as the EMF, provided the breakdown voltage is not exceeded. In cases where the dielectric does not fully occupy the space between the plates, the electric field strength can be calculated using Gauss's law, with the electric displacement field (D) being independent of the dielectric material.

PREREQUISITES
  • Understanding of parallel plate capacitors
  • Familiarity with electromotive force (EMF) and breakdown voltage
  • Knowledge of Gauss's law in electrostatics
  • Concept of electric displacement field (D)
NEXT STEPS
  • Study the relationship between charge (Q), capacitance (C), and voltage (V) using the formula Q=CV
  • Explore the implications of dielectric materials on capacitor performance
  • Learn about breakdown voltage and its effects on capacitor safety
  • Investigate the calculation of electric fields in non-uniform dielectric configurations
USEFUL FOR

Students and professionals in electrical engineering, physicists, and anyone interested in understanding the behavior of capacitors in circuits, particularly with dielectrics involved.

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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