How Does Inserting a Copper Slab Affect a Capacitor's Performance?

Click For Summary

Homework Help Overview

The discussion revolves around the effects of inserting a copper slab into a parallel plate capacitor, focusing on changes in capacitance, energy storage, and the work done during the insertion process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the capacitance changes due to the copper slab, discussing energy ratios before and after insertion, and questioning the nature of work done on the slab. Some participants seek clarification on the implications of energy changes and external work.

Discussion Status

There is ongoing inquiry into the correctness of reasoning regarding the work done on the slab and the energy changes in the system. Some participants suggest that the energy increase indicates external work, while others are seeking further clarification on related concepts such as potential difference and electric fields.

Contextual Notes

Participants are working under the assumption that the voltage across the capacitor remains constant during the process. There are also references to external work done by a battery or other source, which may influence the discussion.

stunner5000pt
Messages
1,447
Reaction score
5
A slab of copper is thrust into parallel plate capacitor as shown in the figure.
a) What is the capacitance after the slab is introduced?

This is like two capacitors in series so
[tex]C_{1} = \frac{\epsilon_{0} A}{d-b-x}[/tex]
[tex]C_{2} = \frac{\epsilon_{0} A}{x}[/tex]
when added it yields [tex]\frac{epsilon_{0} A}{d-b}[/tex]
Find the ratio of the stored energy before and after the slab is inserted if the voltage is kept constant
dividing U1 and U2 which are
[tex]U_{1} = \frac{1}{2} \frac{\epsilon_{0} A}{d-b} \Delta V^2[/tex]
[tex]U_{2} = \frac{1}{2} \frac{\epsilon_{0} A}{d} \Delta V^2[/tex]
[tex]\frac{U_{1}}{U_{2}} = \frac{d}{d-b}[/tex]
Find the work done on the slab as it is inserted. Is it pulled in or pushed in?
Well find the difference in the energy U1 - U2 right?
[tex]U_{f} - U_{i} = \epsilon_{0} A (\frac{1}{d-b} - \frac{1}{d}[/tex]
this change is positive value. So the slab must be pushed in. Is this right?
Please do advise on any mistakes i may have made.
Thank you for your help!
 

Attachments

  • charge.JPG
    charge.JPG
    9.4 KB · Views: 726
Last edited:
Physics news on Phys.org
can someone advise me on whether i am right or not on the last part of this problem? I know the first and second parts are correct but what about the last part ? Since the nergy of the system goes up there must be external work done... so the slab mustb e pushed in, yes?
 
stunner5000pt said:
can someone advise me on whether i am right or not on the last part of this problem? I know the first and second parts are correct but what about the last part ? Since the nergy of the system goes up there must be external work done... so the slab mustb e pushed in, yes?
Some of the external work is done by the battery or other source that keeps the voltage constant across the capacitor.
As the final capacitance is higher than the initial one and the voltage stays the same, the charge increases. The work done by the battery is the constant voltage multiplied by the increment of charge. The change of energy is the summ of your work and the work done by the battery
[tex]\Delta E = \Delta Q U + W[/tex]
ehild
 
I have the same problem but how do i find the potential difference as a function of x. Also what is the electrostatic energy of the system and force on the dielectric.
 
Do you know the electric field in the dielectric and in the free space region? What is the relationship between V and E ?
 
Electromagnetic Theory by Reitz is a good book to learn more about dielectric slabs from...
 

Similar threads

Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
17
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 20 ·
Replies
20
Views
3K