Gauss's Law:Metal sphere of radius 'a' surrounded by a shell

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SUMMARY

The discussion centers on the application of Gauss's Law to a metal sphere of radius 'a' surrounded by a thick concentric metal shell with inner radius 'b' and outer radius 'c'. A point charge +Q located within an irregular cavity of the sphere induces charges on the surfaces of both the sphere and the shell. The surface charge density on the outer surface of the sphere at radius r=c is calculated as σ=Q/(4πc^2). The electric field is determined to be zero in the region between the shell and the sphere (b PREREQUISITES

  • Understanding of Gauss's Law
  • Knowledge of electric field concepts
  • Familiarity with charge induction principles
  • Basic proficiency in electrostatics
NEXT STEPS
  • Study the implications of charge induction in conductors
  • Learn about electric field calculations using Gauss's Law
  • Explore the concept of surface charge density in electrostatics
  • Review Griffith's "Introduction to Electrodynamics" for related problems
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jinksys
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Homework Statement



A metal sphere of radius a is surrounded by a thick concentric metal shell (inner radius b, outer radius c). Neither the shell nor the sphere carries any charge, but there is a point charge +Q located inside an irregularly shaped cavity in the otherwise solid sphere as shown in the figure. The irregular cavity is not concentric with the sphere.

a) Sketch the induced charges on all the relevant surfaces.
b) What is the surface density of the charge on the outer surface of the sphere r=c?
c) What is the electric field where a<r<b?
c) What is the electric field where b<r<c?


The Attempt at a Solution



a) The positive charge within the irregular cavity will induce a negative charge on the surface of the cavity. This will in turn induce a positive charge on the surface of the sphere. A negative charge will be induced on the surface at b and a positive charge will be induced on the surface at c. This is illustrated in this image.

b) Q=σA where A=area of surface.
σ=Q/A
σ=Q/(4πc^2)

c) E=1/(4πε0) * Q/r^2
The total enclosed charge is Q.

d) This point is inside the metal shell so the E-field should be zero.
The enclosed charges are +Q from the sphere and -Q from the 'b' surface. The net charge is zero, so E= 1/(4πε0) * 0/r^2 == 0.

I'm using Griffith's E+M text.
 
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'Looks correct to me. :approve:

[Just out of curiosity though, part b) uses the word "sphere" but then has "r=c" which would imply outer radius of the "shell." Maybe it's nothing, sometimes shells are called spheres. But it seems a little inconsistent for Griffiths. Are you sure that part b) isn't asking you for charge density at r = a?]
 
Last edited:
collinsmark said:
'Looks correct to me. :approve:

:smile: Awesome, thanks!
 
jinksys said:
:smile: Awesome, thanks!
In case you missed my edit (which I didn't update until after your last post), I'll repeat it again here:

Just out of curiosity though, part b) uses the word "sphere" but then has "r=c" which would imply outer radius of the "shell." Maybe it's nothing, sometimes shells are called spheres. But it seems a little inconsistent for Griffiths. Are you sure that part b) isn't asking you for charge density at r = a?
 
collinsmark said:
In case you missed my edit (which I didn't update until after your last post), I'll repeat it again here:

Just out of curiosity though, part b) uses the word "sphere" but then has "r=c" which would imply outer radius of the "shell." Maybe it's nothing, sometimes shells are called spheres. But it seems a little inconsistent for Griffiths. Are you sure that part b) isn't asking you for charge density at r = a?

My homework problem is based off a similar problem in griffith's, so its the teacher's language and not Griffith's. Part-b says (verbatim) What is the surface density of charge on the outer surface of the sphere (r=c)?
 

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