Capacitance and dielectric

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Homework Statement


This is not a problem from any book or any kind of source. I was just thinking on this. Suppose u have a capacitor attached to a battery. And then u insert the dielectric slab. Then do the electric field between the plates changes???


Homework Equations





The Attempt at a Solution


I think it should not.Assume the electric charge before inserting the dielectric is Q.As soon after we insert the dielectric the electric field reduces to E/K. This also causes a decrease in potential between the plates of the capacitor by the factor K. Thus it will be V/K. Since the capacitor should be at the same potential as the battery always the battery wills send an charge equivalent to KQ in order to maker the potential same as before, And thus the Electric field also will again increase from E/K to E as before. Am I correct????
 

Answers and Replies

  • #2
ehild
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There can be two limiting (ideal) cases:

1. The capacitor is connected to an ideal voltage source. The voltage on the capacitor would be constant. The electric field intensity in case of a parallel-plate capacitor is E=V/d independently if there is dielectric between the plates or there is none.
2. The capacitor is charged to some Q and then disconnected from the voltage source. The electric field between the plates is Q/(A K) so it is reduced by inserting the dielectric.

If the capacitor is not disconnected from the battery, but there are some resistances in the circuit, the electric field will change with time, depending how fast the dielectric is inserted, and what are the parameters of the capacitor and the value of resistance. But the voltage will equal to the emf of the battery after very long time.

ehild
 
  • #3
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so suppose there is two dielectric slabs with k1 and k2 placed in parallel in a capacitor. Then can we say that the resultant electric field between the two parallel capacitors now is different.
 
  • #4
ehild
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If you fill the distance D between the plates up to d1 with one dielectric and from d1 to D with an other one, this arrangement is equivalent with two capacitors connected in series. The surface charge density σ is the same on both, but the electric field intensity is different, being E1=σ/K1 and E2=σ/K2.

ehild
 
  • #5
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no I m not talking about capacitors in series I m asking about what if they are parallel?? I mean both of the dielectric are placed side by side nd both of their edges touch the two plates
 
  • #6
Gokul43201
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no I m not talking about capacitors in series I m asking about what if they are parallel?? I mean both of the dielectric are placed side by side nd both of their edges touch the two plates
What do you think would happen?

How would the voltages across each of the parallel capacitors compare (and therefore the electric fields)?
 
  • #7
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the voltage will be the same and the electric field too provided the battery is attached to the circuit. But charge on the plate wont be uniform. the charge on the part of the plate near the k1 will be different from k2. Am I correct??
 
  • #8
ehild
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the voltage will be the same and the electric field too provided the battery is attached to the circuit. But charge on the plate wont be uniform. the charge on the part of the plate near the k1 will be different from k2. Am I correct??
...

and provided that the distance between the plates is uniform. And you are right, the surface charge density will be different. The charge can be the same on both parts, with appropriate choice of the area of the capacitors.

ehild
 
  • #9
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thanks a lot ehild...
 

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