Calculating Power Dissipation on Cylinder Surface

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SUMMARY

The discussion focuses on calculating power dissipation over a cylindrical surface using the Poynting vector, specifically the integral \(\oint \mathbf{E} \times \mathbf{H} \, ds\). It clarifies that for a closed cylinder, the surface area elements differ: for the end-caps, \(dS = s \, ds \, d\phi\), and for the curved surface, \(dS = s \, d\phi \, dz\). The derivation of these area elements is referenced from Griffiths' "Introduction to Electrodynamics," particularly section 1.4.2 and page 40 of the 3rd edition.

PREREQUISITES
  • Understanding of Poynting vector and its application in electromagnetism
  • Familiarity with cylindrical coordinate systems
  • Basic knowledge of vector calculus
  • Access to Griffiths' "Introduction to Electrodynamics" (3rd edition)
NEXT STEPS
  • Study the derivation of area elements in cylindrical coordinates from Griffiths' "Introduction to Electrodynamics"
  • Explore applications of the Poynting vector in electromagnetic theory
  • Learn about power dissipation in different geometrical configurations
  • Investigate further into vector calculus techniques relevant to electromagnetism
USEFUL FOR

Students and professionals in physics, particularly those focusing on electromagnetism, electrical engineers, and anyone involved in calculating power dissipation in cylindrical geometries.

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Homework Statement


I am trying to calculate power dissipated over a cylindrical surface using poynting vector -
[tex]\oint[/tex] ExH ds

I know ds for a sphere is r^2 sin [tex]\theta[/tex] d[tex]\theta[/tex] d[tex]\phi[/tex]

But now sure what ds is for a cylinder?

Homework Equations





The Attempt at a Solution


 
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likephysics said:
I know ds for a sphere is r^2 sin [tex]\theta[/tex] d[tex]\theta[/tex] d[tex]\phi[/tex]

But now sure what ds is for a cylinder?

It depends on which surface you are talking about. A closed cylinder has 3 surfaces; one curved surface and two flat circular end-caps. For the end-caps, [itex]dS=s ds d\phi[/itex]. While, for the curved surface, [itex]dS=s d\phi dz[/itex]. (Using [itex]\{s,\phi,z\}[/itex] for the cylindrical coordinates)

Griffiths' Introduction to Electrodynamics derives the infinitesimal displacements ([itex]dl_s=ds[/itex], [itex]dl_\phi=s d\phi[/itex], [itex]dl_z=dz[/itex]) in cylindrical coordinates in section 1.4.2. And the author gives a brief discussion of how to obtain area elements from these infinitesimal displacements at the end of page 40 (3rd edition).
 
Great. I am going to go take a look at Griffith's right now.
 

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