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

In summary, the problem involves a metal sphere with a thick concentric metal shell, where a point charge +Q is located inside an irregularly shaped cavity in the sphere. The induced charges on the relevant surfaces are a negative charge on the cavity's surface and a positive charge on the sphere's surface. The surface density of the charge on the outer surface of the sphere (r=c) is Q/(4πc^2). The electric field where a<r<b is given by 1/(4πε0) * Q/r^2, and where b<r<c is zero due to the net charge being zero. The language in the problem may be inconsistent, but it is based on a similar problem in Griffith's E+M
  • #1
jinksys
123
0

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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  • #2
'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:
  • #3
collinsmark said:
'Looks correct to me. :approve:

:smile: Awesome, thanks!
 
  • #4
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?
 
  • #5
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)?
 

What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric field at a closed surface to the net electric charge enclosed within the surface.

What is a metal sphere of radius 'a' surrounded by a shell?

A metal sphere of radius 'a' surrounded by a shell is a system consisting of a solid metal sphere with a radius of 'a' at its center, and a conducting spherical shell with the same center but a larger radius, enclosing the metal sphere.

How does Gauss's Law apply to a metal sphere of radius 'a' surrounded by a shell?

Gauss's Law states that the net electric flux through a closed surface is equal to the enclosed electric charge divided by the permittivity of free space. In the case of a metal sphere of radius 'a' surrounded by a shell, the enclosed charge is simply the charge on the metal sphere, and the electric field at any point outside the sphere is zero due to the shielding effect of the conducting shell.

What is the electric field inside the metal sphere of radius 'a' surrounded by a shell?

The electric field inside the metal sphere is also zero, due to the principle of electrostatic equilibrium. Any excess charge on the metal sphere will distribute itself on the outer surface of the sphere, creating an equal and opposite electric field inside the sphere that cancels out the field from the enclosed charge.

How does the electric field change if the metal sphere is not at the center of the shell?

If the metal sphere is not at the center of the shell, the electric field inside the sphere will no longer be zero. This is because the electric field from the enclosed charge will not be perfectly cancelled out by the induced electric field from the excess charge on the outer surface of the sphere. However, the electric field outside the sphere will still be zero due to the conducting shell.

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