How to Derive Displacement Current and Energy Density in a Capacitor?

  • Thread starter Thread starter hitman0097
  • Start date Start date
  • Tags Tags
    Intro Optics
Click For Summary

Homework Help Overview

The discussion revolves around deriving the displacement current density in a parallel plate capacitor and understanding the energy density stored within it. The problem involves concepts from electromagnetism, specifically relating to capacitors and their behavior under changing electric fields.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses confusion about deriving the displacement current density and seeks clarification on related equations and concepts.
  • Some participants question the definitions of terms used, such as displacement current and conductivity.
  • Others suggest using the relationship between charge, capacitance, and voltage to derive current.
  • There are references to energy density equations and their derivations, indicating attempts to connect various physical concepts.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some have made progress in understanding the relationships between the variables involved, while others continue to seek clarification. There is no explicit consensus, but the conversation appears to be productive in terms of exploring various interpretations and approaches.

Contextual Notes

Participants mention consulting notes and external resources, indicating that they are working within the constraints of their current understanding and available materials. The original poster also plans to discuss the topic with their professor, suggesting a desire for further clarification.

hitman0097
Messages
30
Reaction score
0

Homework Statement


Show that the displacement current density which flows when a parallel plate capacitor gets charged is ε(dE/dt) where ε is the permittivity of the medium between the capacitor. What is the energy density stored in the capacitor and the force per unit area (pressure) on the plates?

Homework Equations


D=εoE + p= εoE +εoχE=εE
p=[itex]\alpha[/itex]E (mircoscopic)
p=χE (marcoscopic)
(Constitute of Relations) p is the polarization of the medium

The Attempt at a Solution


(a) I am really confused as to how to show the first part, which I think is needed for the whole problem.
(b) For the work, from my notes I have[itex]\tau\tau[/itex]=[itex]\sigma[/itex]E which means it behaves equivallently to Ohms Law V=IR so the Voltage which is the work done per unit charge is equal to that expression.
(c) For the energy density of the capacitor Ue=1/2 εo E^2
And the Total energy density is Uto/2 *E^2 +B^2/2μo which reduced to 2εoE^2
(d) P=1/2 εoE^2 +B^2/μ
P=εo E^2
P=s/c (s is the Poynting vector)

I am just really confused but this is what my notes and external sites have led me to. Any help to make sense of this is appreciated (I'll be talking to my professor today too as well so hopefully I can understand this better)
 
Physics news on Phys.org
tau is the displacement current, and sigma should be the conductivity? I think
 
I didn't need the equation for D in the first part, All I needed was C=eA/d for capactiance and q=CV
 
So If i take the time derivative of q=cv I get the current I = e A/d *dV/dt
 
Nvm, I figured most of it now.
 
Oh, I thought you had figured it out five hours ago.
 

Similar threads

Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 11 ·
Replies
11
Views
7K
Replies
5
Views
3K