Image charge of source charge in spherical cavity

[feynman1]

All are used to finding the image charge induced by a source charge outside a conducting sphere. The solution is supposed to also work for the case where the source charge is inside the conducting sphere, in which case the sphere is now a conducting cavity. But the solution suggests the image charge is infinite in charge and infinitely far from the source charge. How does this abnormal image charge produce such a normal electric field inside the void of the spherical cavity?

 

[rude man]

The solution is the same as with the original charge outside the sphere. In this case the two charges are just reversed. The image charge is outside the sphere and the original charge is inside.

What I never liked about this picture is that if an original charge were placed inside a conducting sphere it would immediately go to the surface. Then the potential is constant everywhere within the sphere as we all know it has to be.

But, for what it's worth, try this, immediately after eq. 733:
http://farside.ph.utexas.edu/teaching/em/lectures/node64.html

Also https://en.wikipedia.org/wiki/Method_of_image_charges

 

[feynman1]

Thanks. Note that I mean a source charge in the void of a conducting cavity, not the interior of the conductor, so the source charge is fixed from an external force.

 

[rude man]

Thanks. Note that I mean a source charge in the void of a conducting cavity, not the interior of the conductor, so the source charge is fixed from an external force.

I found your sentence confusing: The solution is supposed to also work for the case where the source charge is inside the conducting sphere, in which case the sphere is now a conducting cavity. .
What is a "conducting cavity"? Seems like an oxymoron.
Did you mean an evacuated cavity (a void) within a solid sphere, or just a solid sphere?
The "opposite situation" with the identical soution is for a solid sphere with the original charge somehow glued within that solid sphere.

I don't know about a void (an evacuated volume) within a solid sphere and the original charge within that void..

 

[feynman1]

I found your sentence confusing: The solution is supposed to also work for the case where the source charge is inside the conducting sphere, in which case the sphere is now a conducting cavity. .
What is a "conducting cavity"? Seems like an oxymoron.
Did you mean an evacuated cavity (a void) within a solid sphere, or just a solid sphere?
The "opposite situation" with the identical soution is for a solid sphere with the original charge somehow glued within that solid sphere.

I don't know about a void (an evacuated volume) within a solid sphere and the original charge within that void..

A conducting cavity is a hollow conductor. The solid region is conducting and the hollow region is vacuum. A source charge is in the void, and the void is of a spherical shape.

 

[vanhees71]

The most simple setup is to consider just a spherical conducting shell of radius $a$ around the origin. Now put a charge inside the shell. We choose the reference frame such that this charge sits at $\vec{r}_0=r_0 \vec{e}_3$. For simplicity assume that the shell is grounded, i.e., we look for a solution of the Poisson equation
$$\Delta \phi=-q \delta^{(3)}(\vec{r}-\vec{r}_0)$$
with the boundary condition
$$\phi(\vec{r})|_{|\vec{r}|=0}=0.$$
It is now immidiately clear that in this situation the unique solution for $|\vec{r}|>a$ is the Coulomb potential fulfilling the boundary conditions since there are no charges outside and the spherical shell $|\vec{r}|=a$ is an equipotential surface. Thus we have
$$\phi(\vec{r})=\frac{q}{4 \pi r}-\frac{q}{4 \pi a}.$$
For $|\vec{r}|<a$ we use the image-charge method. By symmetry the image charge $q'$ must sit on the $x_3$ axis outside of the shell. To find $q'$ and its location $\vec{r}_0'=r_0 \vec{e}_3$ we work in spherical coordinates. Inside we have a field given by the charge $q$ and the image charge $q'$:
$$\phi(\vec{r})=\frac{q}{4 \pi \sqrt{r^2+r_0^2-2 r r_0 \cos \vartheta}} + \frac{q'}{4 \pi \sqrt{r^2+r_0^{\prime 2} -2 r r_0' \cos \vartheta}}.$$
Now for $r=a$ this must be 0, i.e.,
$$q \sqrt{a^2+r_0^{\prime 2} -2 a r_0' \cos \vartheta}=-q' \sqrt{a^2+r_0^2-2 a r_0 \cos \vartheta}.$$
Squaring this gives
$$q^2 (a^2+r_0^{\prime 2} -2 a r_0' \cos \vartheta)=q^{\prime 2} (a^2+r_0^2-2 a r_0 \cos \vartheta).$$
This can only be true if the coefficients in front of $\cos \vartheta$ and the part independent of $\vartheta$ are equal, i.e.,
$$q^2 r_0'=q^{\prime 2} r_0$$
$$q^2 (a^2+r_0^{\prime 2})=q^{\prime 2} (a^2+r_0^2).$$
This leads to a quadratic equation for $r_0'$ which is uniquely solved under the constraint that $r_0'>a$ must hold,
$$r_0'=\frac{a^2}{r_0}$$
Plugging this into the first equation yields
$$q'=\pm \frac{q}{r_0} q.$$
Plugging it into the ansatz for our potential the boundary condition $\phi(\vec{r})|_{|\vec{r}|=a}$ gives then uniquely
$$q'=-\frac{a}{r_0} q.$$
So our potential for $r \leq a$ reads
$$\Phi(\vec{r})=\frac{q}{4 \pi \sqrt{r^2+r_0^2 - 2 r r_0 \cos \vartheta}}-\frac{q a}{4 \pi \sqrt{r^2 r_0^2+a^4 - 2 a^2 r r_0 \cos \vartheta}}.$$

 

[feynman1]

Thank you. The solution is correct. I somehow missed saying in the original question that I was looking for a case when exactly ro=0, source charge being in the centre of the void.

 

[vanhees71]

Then it's trivial, and the solution is
$$\Phi(\vec{r})=\frac{q}{4 \pi r} -\frac{q}{4 \pi a}.$$
This you also get from the general solution by setting $r_0=0$.

 

[feynman1]

Then it's trivial, and the solution is
$$\Phi(\vec{r})=\frac{q}{4 \pi r} -\frac{q}{4 \pi a}.$$
This you also get from the general solution by setting $r_0=0$.

But where's the image charge and how much charge?

 

[vanhees71]

You don't need an image charge in this case, because the boundary condition is fulfilled by symmetry when the charge sits at the center of the sphere.

 

[feynman1]

You don't need an image charge in this case, because the boundary condition is fulfilled by symmetry when the charge sits at the center of the sphere.

Right but it's nicer to understand the image charge in this special case