Push the dielectric half way into the capacitor

Click For Summary
The discussion centers on the mechanics of inserting a dielectric slab into a parallel plate capacitor and the associated energy changes. When the dielectric is pushed in, it becomes polarized and experiences an attractive force due to the electric field, which decreases the stored energy in the capacitor. The work done to insert the dielectric can be calculated by comparing the initial and final energy states of the capacitor, with the final energy being lower due to increased capacitance. The decrease in electrostatic energy occurs because the charge remains constant while capacitance increases, leading to a reduction in energy. Despite the decrease in energy, dielectrics have various applications beyond energy considerations.
indigojoker
Messages
240
Reaction score
0
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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
2K
Replies
5
Views
587
Replies
14
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
6
Views
3K
Replies
10
Views
4K
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K