fluidistic said:
So it's like a 3 dimensional hollow sphere. In its center there's a small charge. So if we set the place where the charge is as to be the origin, in order to reach the surface of the sphere I would have to pass through 65 cm of vacuum and then 20 cm of metal (conductor). Am I right?
And they ask what is the charge over the inner surface of the sphere. Wow, I've no idea how to do this. I don't really see how Gauss can help me. Ok so the total charge is 0. Hence the whole sphere (without the charge in its middle) must have a charge of -1.16 \mu C. I don't really know how to find the charge distribution.
In the original question, the metal shell was neutral.
"A positive point charge of magnitude 2.4 micro Coulombs is at the center of an uncharged spherical conducting shell of inner radius 65 cm and outer radius 85 cm."
The geometry has spherical symmetry, so Gauss Law is easy to apply. The electric field lines are radial and the electric field intensity depends only on r, distance from the centre.
Inside the empty sphere, from r=0 to r=Rin, a sphere of radius r encloses the charge at the centre. So the surface integral of E for the sphere is
E*4r^2\pi=\frac{Q}{\epsilon_0}\rightarrow E=\frac{Q}{4\pi\epsilon_0 r^2}
The electric field lines always start out at positive charges and end at negative ones, except those lines which go to infinity. There are Q/epsilon
0 field lines emerging from the central charge. The same number of field lines hit the surface of the inner sphere: so there must be -Q charge there. This means -Q/(4piR
in2 surface charge density. But it is true for any charge Q that Q/epsilon
0 field lines emerge from it or end in it. The electric field at the surface of the inner sphere is
E=\frac{Q}{4\pi\epsilon_0 R_in^2}.
There are no field lines inside the metal. All the field lines ending in a surface charge come from the empty sphere. So the electric field at the inner surface corresponds to surface charge density of magnitude
\sigma=E*\epsilon_0= \frac{Q}{4\pi R_{in}^2}
For R
in<r<R
out, the enclosed charge is 0, and so is the electric field inside the metal shell.
Outside the shell and on the outer surface, the enclosed charge is Q as the shell itself is neutral. The electric field is
E=\frac{Q}{4\pi\epsilon_0 r^2}
and the corresponding surface charge density is
\sigma=E*\epsilon_0= \frac{Q}{4\pi R_{out}^2} .
If the total arrangement is neutral (it is so when the metal shell is grounded) the field in the empty sphere is the same as before, the surface charge density at the inner surface of the shell is the same again, but the field outside the shell is zero, as the enclosed charge is zero.
ehild
