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

AI Thread Summary
A metal sphere of radius 'a' with a point charge +Q in an irregular cavity induces charges on its surfaces, resulting in negative charges on the cavity surface and positive charges on the sphere's outer surface. The surface charge density on the outer surface at radius 'c' is calculated as σ = Q/(4πc^2). The electric field in the region where a < r < b is determined to be E = 1/(4πε0) * Q/r^2, while in the region where b < r < c, the electric field is zero due to the net enclosed charge being zero. There is some confusion regarding the terminology in part b of the problem, as it refers to the "sphere" at r = c, which may imply the outer shell instead. Clarification on the language used in the problem statement may be necessary for accurate interpretation.
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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