How Do Electric Fields Behave Around Concentric Charged Spherical Shells?

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SUMMARY

The discussion focuses on calculating the electric field around two concentric charged spherical shells, with charge Q on the inner shell and -Q on the outer shell. The electric field inside the smaller shell is zero, while between the shells, it is given by E = kQ/(πr²). For the region outside the larger shell, the electric field behaves as if the total charge were concentrated at the center. The Gaussian surface method is employed to derive these results, emphasizing the importance of symmetry in electric field calculations.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and charge distributions
  • Knowledge of spherical symmetry in electrostatics
  • Basic algebra for manipulating equations
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  • Study Gauss's Law applications in electrostatics
  • Explore electric field calculations for different charge configurations
  • Learn about the concept of electric field lines and their significance
  • Investigate the effects of dielectric materials on electric fields
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Students of physics, particularly those studying electrostatics, educators teaching electric field concepts, and anyone seeking to understand the behavior of electric fields around charged objects.

ibaraku
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Homework Statement


Two concentric plastic spherical shells carry uniformly distributed charges, Q on the inner shell and -Q on the outer shell. Find the electric field (a)Inside the smaller shell, (b)between the shells, and (c) outside the larger shell

Homework Equations



integration(E * n)dA = (4pi) (k) (Qenclosed)

The Attempt at a Solution



In order to get hteh electric field between the shells, we can say that

[abs(E) * abs(n) cos delta](4pi) r^2 = (4pi) (k) (Qenclosed)

and working out the algebra it comes out to be

E = kQ/pi r^2

but what about for the charge outside the larger shell and the charge inside the smaller charge?
Is it safe to say that the we need to divide by 2 for the smaller shell and multiply by 2 for the large shell?
Thanks
 
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ibaraku said:
[abs(E) * abs(n) cos delta](4pi) r^2 = (4pi) (k) (Qenclosed)

and working out the algebra it comes out to be

E = kQ/pi r^2
Yes that's right.
ibaraku said:
but what about for the charge outside the larger shell and the charge inside the smaller charge?
Is it safe to say that the we need to divide by 2 for the smaller shell and multiply by 2 for the large shell?
Thanks
You do this the same way as you did the above, except now the Gaussian surface for (a) cuts through the smaller sphere, and in (c) the Gaussian surface encloses both spheres. There isn't any way you can use the result obtained above to compute for the other two cases. The method used is still valid, though because of symmetry.
 

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