Find the electric field a distance z from the centre of spherical

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SUMMARY

The discussion focuses on calculating the electric field at a distance z from the center of a uniformly dense sphere of radius R. The relevant equation is derived from the integral form of Coulomb's law, specifically 1/4∏εo∫ (σ da/ r^2) cos(theta). The solution involves substituting the distance r using the law of cosines, resulting in r=√R^2+z^2-2Rzcos(theta). Participants emphasize the importance of recognizing the charge density as a volume density within the sphere, not just a surface density.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with spherical coordinates and integration techniques
  • Knowledge of the law of cosines in geometry
  • Basic concepts of charge density (volume vs. surface density)
NEXT STEPS
  • Study the derivation of electric fields for different charge distributions
  • Learn about spherical coordinates and their applications in electromagnetism
  • Explore the concept of volume charge density and its implications in electric field calculations
  • Review integral calculus techniques for solving physics problems involving multiple dimensions
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Students in physics or engineering, particularly those studying electromagnetism, as well as educators seeking to clarify concepts related to electric fields and charge distributions.

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Homework Statement



Find the electric field a distance z from the centre of spherical sphere of radius R which carries uniform density B. treat Z<R (inside) and Z>R (outside)………. By using law of cosine how to solve this problem?

Homework Equations



1/4∏εo∫ (σ da/ r^2) cos(theta)

The Attempt at a Solution


substituting r by law of cosin.. r=√R^2+z^2-2Rzcos(theta)
1/4∏εo ∫[σ da/ R^2+z^2-2Rzcos(theta)]cos(theta)
i am stuck here what should i do next
 

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Hi...
Is the charge density given by a surface density, or a volume density? When talking about a sphere and not a spherical shell, one ought to remember that the inside is filled(in this case with charge).
In any case,
Try the following:
[itex] dA = rdr\sin(\theta)d\theta d\phi [/itex]
[itex] dV = r^2dr\sin(\theta)d\theta d\phi[/itex]
Phi goes from 0 to 2pi,
Theta goes from 0 to Pi.
r from zero to R.
Try it,
Daniel
 

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