- #1
issacnewton
- 1,000
- 29
Hi
Here is a problem I am trying to do. A point charge q is located at a distance r
from the center O of an uncharged conducting spherical layer whose inside and
outside radii are equal to [itex]R_1[/itex] and [itex]R_2[/itex] respectively.
Find the potential at the point O if [itex]R_1 < R_2[/itex].
Now I was thinking of method of images. But since we have a spherical conducting
sphere with some thickness, that would mean we have several concentric equipotential
surfaces surrounding the chrge q. So one image charge would not suffice. We will
need infinitely many of the image charges. So method of images is not practical here. Griffiths
says in his book that the leftover charge on the outer surface in case of a charge
placed in a metal cavity is uniformly distributed. He doesn't give any satisfactory
reasoning. But let's assume what he says. Then the outside of the metal sphere, world
will see that charge q is at the center of the sphere, which means potential outside
is given by
[tex]V(r)=\frac{1}{4\pi\epsilon_o}\frac{q}{r}[/tex]
where r is the distance of any point outside the sphere from the center of
the sphere. So on the outer wall of the sphere, the potential is
[tex]V(R_2)=\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}[/tex]
since sphere is metal , its equipotential, so the potential on the inner wall of the
sphere is same. So what else can we say so that we can get the potential at the
center of the sphere ?
thanks
Here is a problem I am trying to do. A point charge q is located at a distance r
from the center O of an uncharged conducting spherical layer whose inside and
outside radii are equal to [itex]R_1[/itex] and [itex]R_2[/itex] respectively.
Find the potential at the point O if [itex]R_1 < R_2[/itex].
Now I was thinking of method of images. But since we have a spherical conducting
sphere with some thickness, that would mean we have several concentric equipotential
surfaces surrounding the chrge q. So one image charge would not suffice. We will
need infinitely many of the image charges. So method of images is not practical here. Griffiths
says in his book that the leftover charge on the outer surface in case of a charge
placed in a metal cavity is uniformly distributed. He doesn't give any satisfactory
reasoning. But let's assume what he says. Then the outside of the metal sphere, world
will see that charge q is at the center of the sphere, which means potential outside
is given by
[tex]V(r)=\frac{1}{4\pi\epsilon_o}\frac{q}{r}[/tex]
where r is the distance of any point outside the sphere from the center of
the sphere. So on the outer wall of the sphere, the potential is
[tex]V(R_2)=\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}[/tex]
since sphere is metal , its equipotential, so the potential on the inner wall of the
sphere is same. So what else can we say so that we can get the potential at the
center of the sphere ?
thanks