Dielectric homework problem

  • #1
thereddevils
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When a dielectric is inserted between the plates of a parallel plate capacitor , what would happen to its charge , Q and electric potential , V as compared to when there is no dielectric ?

My thoughts are the presence of dielectric cause the resultant electric field to decrease , and from E=V/d , the electric potential decreases too .

As for the charge , i am not so sure . Could it be that it is constant according to the charge principles , i don see where the charges can go .
 

Answers and Replies

  • #2
ehild
Homework Helper
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You are right, the charge stays the same, the electric field decreases.

ehild
 
  • #3
thereddevils
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You are right, the charge stays the same, the electric field decreases.

ehild

thanks ehild for clarifying , my book is wrong then , it says the charge increases in the dielectric case .
 
  • #4
ehild
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thanks ehild for clarifying , my book is wrong then , it says the charge increases in the dielectric case .

It does if the capacitor is connected to a battery, but the voltage is constant in this case. If the capacitor is isolated, the charge remains the same and the voltage decreases.

ehild
 
  • #5
thereddevils
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It does if the capacitor is connected to a battery, but the voltage is constant in this case. If the capacitor is isolated, the charge remains the same and the voltage decreases.

ehild

oh , yes its connected to the battery so the charges would increase with the presence of the dielectric but why ? Is it because the charges cannot go across to the other plate ?
 
  • #6
ehild
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You know that inserting a dielectric in the capacitor it will decrease the electric field. If E decreases, so does the voltage across the plates. But in case the capacitor is connected to the battery, charges will flow between the battery and capacitor plate until equilibrium is reached again when the voltage across the capacitor plates is the same as the voltage of the battery.

ehild
 
  • #7
thereddevils
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You know that inserting a dielectric in the capacitor it will decrease the electric field. If E decreases, so does the voltage across the plates. But in case the capacitor is connected to the battery, charges will flow between the battery and capacitor plate until equilibrium is reached again when the voltage across the capacitor plates is the same as the voltage of the battery.

ehild

thanks ehild !
 

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