Line charge inside a conducting sphere?

[ebru]

 

[vanhees71]

Solve for the Green's function of the spherical conducting shell (it can only be a shell, because otherwise there cannot be any non-vanishing charge distribution inside in the static case) and then do the integral to get the em. field inside and outside the shell.

 

[ebru]

Solve for the Green's function of the spherical conducting shell (it can only be a shell, because otherwise there cannot be any non-vanishing charge distribution inside in the static case) and then do the integral to get the em. field inside and outside the shell.

I don't know how to do that.

 

[vanhees71]

The Green function $D(\vec{x},\vec{x}')$ is the solution of
$$\Delta D(\vec{x},\vec{x}')=-\delta^{(3)}(\vec{x}-\vec{x}').$$
It's the solution for a point source of charge 1 subject to the boundary conditions of the problem, i.e., for you case you get it by putting a unit charge somewhere inside the sphere at $\vec{x}'$ with $|\vec{x}'|<R$.

Now I suppose the shell is grounded. Then you have to simply find the solution subject to the boundary condition
$$D(\vec{x},\vec{x}')|_{|\vec{x}|=R}=0.$$
For the field inside the sphere, i.e., for $|\vec{x}|<R$ you can put an image charge $q''$ at $\vec{x}''$ with $|\vec{x}''|>R$. You have to adjust $\vec{x}''$ and $q''$ such that the boundary condition is fulfilled.

For the field outside the sphere, i.e., for $|\vec{x}|>R$ everywhere $\Delta D=0$ must hold with the sphere as equipotential surface. This can only be the Coulomb field with the unit charge in the origin.

Now you get the field for the line charge by
$$\vec{E}(\vec{x})=\int_0^{R/2} \mathrm{d} z' \lambda D(\vec{x},z' \vec{e}_z).$$

 

[BvU]

Point is the exercise does not ask for the field, but for the locus of the line charge. When I first saw this OP I was inclined to try and solve it along the lines of example 2 by Errede -- only inverted with the charge inside the spherical cavity and the mirror charge outside. Didn't work it out but still find it a good path to explore. Anyone agree ?

 

[hutchphd]

That should work. In fact if they only want the "locus" of the line charge I think you need only specify the two end points (it must be a line by symmetry)? It will not be uniform I believe but that is not the question..

Edit: I seem to have misread the question... apologies..

 

[vanhees71]

Ok, if they don't want the field and only the (net?) force, just calculate the image charge distribution, which you can of course do first for a point particle and then "smear" it along a line as a line-charge density. Then you have the original line-charge density (I assume it's along the 3-axis)
$$\rho(\vec{x})=\lambda \delta(x) \delta(y) \Theta(0\leq z \leq R/2)$$
and the mirror-charge distribution $\rho'(\vec{x})$, which has support completely outside the fear. The net force then is
$$\vec{F}=\int_{\mathbb{R}^3} \mathrm{d^3} x \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(\vec{x}) \rho'(\vec{x}')}{4 \pi \epsilon_0} \, \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$

 

[BvU]

Can @ebru follow all this ?