Push the dielectric half way into the capacitor

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Homework Help Overview

The discussion revolves around the work required to insert a dielectric material into a parallel plate capacitor and the implications of this action on the system's energy. Participants explore the physical principles governing capacitors and dielectrics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the dielectric and the relationship between electric fields and stored energy. Questions arise regarding the calculation of work done and the initial and final energy states of the capacitor system.

Discussion Status

The conversation includes various interpretations of the energy changes associated with inserting a dielectric. Some participants provide insights into the nature of the forces at play, while others seek clarification on the underlying principles and calculations involved.

Contextual Notes

There are ongoing questions about the assumptions made regarding the capacitor's connection to a battery and the implications of energy changes when a dielectric is introduced. Participants are considering the effects of fixed charge and varying capacitance on energy storage.

indigojoker
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This is a general question because I cannot find any example problems to base this question off of.

Suppose we have two parallel plate capacitors, what work is needed to push the dielectric half way into the capacitor and why does the slab feel a pull into the capacitor?
 
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The force due to the electric field acts in a direction that decreases stored energy... the capacitor with the dielectric has less energy than the capacitor without the dielectric... so the force acts in a way that pulls the slab in (decreasing stored energy).

what happens is that the dielectric gets polarized and is attracted by the capacitor charges, pulling it in.
 
why does the field act in a direction that decreases stored energy?

And how does one calculate the work done to push the dielectric into the capacitor?

W=Fd, what would be the F=Eq? Not sure where to go from here
 
Work done is final energy - initial energy.

What is the initial energy in the capacitor before the slab is put in?

What is the final energy? For this part think of it as 2 capacitors... each with half the area, and each with a charge of (1/2)Q. what is the energy on each capacitor? what is the sum?

The work done will be negative.
 
why does the electrostatic energy decrease when a dielectric is inserted? (assuming that the plates are not connected to a battery)
 
indigojoker said:
why does the electrostatic energy decrease when a dielectric is inserted? (assuming that the plates are not connected to a battery)

Energy = Q^2/(2C)

charge is fixed. capacitance goes up, so the energy goes down.
 
but what is the point of the dielectric if the energy decreases?
 
indigojoker said:
but what is the point of the dielectric if the energy decreases?

I think there are many applications of dielectrics, not related to energy.
 

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