Electric field in asymmetric conductor configuration

I was trying to figure out an electrostatics exercise. I am confortable solving these type of problems, when there is an easy application of symmetrical properties and Gauss's Law. But this one took me by surprise. As the picture shows, there is a spherical surface with charge +q, as well as the ellipsoidal surface. Both are conductors and hollow, without any type of contact between them. The figure shows a cross section of the configuration, showing qualitatively that they are not centered at the same point, but still the ellipsoidal axis passes right through the center of the sphere shell. I don't have to calculate the field in all space. All I must do is order the points A,B,C and D in increasing order of electric field. I am stucked because it seems to me that I don't have the tools to contemplate the charge redistribution problem.

If we were dealing with static charges it would be quite easy to realize that the superposition principle applies inmediately. Maybe it is easier if I try to imagine different steps to achieve this distribution. First I imagined an empty spherical shell, with an homogeneus charge distribution throughout it's surface. Using Gauss we would get that the field inside is zero. Later we would add the ellipsoidal. And here is when I ask myself, the charge distribution of the spherical shell at a first instant doesn't affect the ellipsoidal one. The latter creates an electric field that would reaccomodate the spherical charges. After this I am not sure of what could be said about the field inside the sphere.

It is not zero between the ellipsoidal and the sphere. But it is indeed zero inside the ellipsoidal. Inmediately outside the ellipsoidal the field is normal to the surface.

I must say also, that the exercise asks to calculate the external field to the sphere, which I suspect is the same as if there was a sphere charged with $+2q$. This hunch is product of the way the question is being asked. If this is true it would be very nice to know why since it is inconclusive what happens after the "redistribution time".

Hope the doubt is clear enough. Thanks for reading.

I would start with an ellipsoid as if it was alone. The field inside the ellipsoid would be obviously zero. That takes care of point C.

The field at points A and D would be stronger than the field at point B, because, given the same potential, the field is stronger where the radius is smaller.

Next, I would introduce an uncharged sphere. When we add the sphere, the charges on the ellipsoid will induce a negative charge, -q, on the inner surface of the sphere and a positive charge, +q, on the outer surface of the sphere.

Since point A is closer to the sphere, the concentration of the induced charges in that area of the sphere will be greater and, as a result, charges on the surface of the ellipsoid would shift upwards, which will make the field near point A stronger than the field near point D.

So, the answer to the first question would be: C, B, D, A.

The induced charge +q on the outer surface of the sphere will be evenly distributed due to the symmetry of the sphere. Adding +q charge to it, will just double the charge density, as you've suggested. So, the field strength around the sphere will be uniform and could be calculated using the standard formula for a charged sphere.

why there must be a inner charge surface

If we put an uncharged spherical shell around a charged (+q) ellipsoid, the flux over the surface of the sphere will correspond to +q (according to Gauss) and the field will be the same as if the outer surface of the sphere was evenly charged by +q.

For that to hold, the outer surface of the sphere has to be actually contain charge +q, since the field outside the sphere is not affected by anything else: because the field inside the skin of the spherical shell, which is a conductor, is zero.

For the spherical shell to remain neutral, there has to be a charge -q on the inner surface.

And also you answered the ordering question with a total charge of q on the configuration, and this is not quite so since the conservation of charge implies that there is +2q. I mean, the resultant order has to contemplate the initial configuration.

Putting additional charge +q on the spherical shell does not affect the field inside the spherical shell and therefore does not affect the ABCD order. Think superposition.

"The induced charge +q on the outer surface of the sphere will be evenly distributed due to the symmetry of the sphere." why this?

Since, as mentioned earlier, there is no field inside the skin of the spherical shell (a conductor), the distribution of charges over the outer surface of the spherical shell is affected only by the charges on that outer surface, i.e., these charges "see" each other but not any charges inside the shell. Because of that, the charges on the outer surface will be distributed the same way as if it was just a charged sphere, which is a known case: the charges should be distributed uniformly due to the symmetry of the sphere.

I am unaware of a condition that establishes that since a surface is equipotential, the charge distribution must be homogeneous. If we took the ellipsoidal surface, the charge distribution as you said will not be homogeneous but still there is an equipotential surface.

You are right: an equipotential surface does not imply a homogeneous distribution of charge. My statement was and is: "The induced charge +q on the outer surface of the sphere will be evenly distributed due to the symmetry of the sphere". A sphere has a 3D symmetry, ellipsoid - does not. We can also say that a sphere has the same radius everywhere and therefore the same charge density.

• Thanks a lot but I dont quite get why there must be a inner charge surface, this is specifically a charged surface. I thought that it either has charge density or not, I dont think you can make that distinction here. If you can, tell me please why so?. And also you answered the ordering question with a total charge of q on the configuration, and this is not quite so since the conservation of charge implies that there is +2q. I mean, the resultant order has to contemplate the initial configuration. . – ivanculet May 11 '18 at 5:41
• Lets assume that is true that there is an inner surface induced charge as well as an outer one, regardless that the sphere is a widthless shell- What is the condition that establishes that we can take this as if there was a homogeneous distribution of charges during the electrostatic phase of this metamorphosis (adding the sphere, then the charge and finally waiting until they stop moving). "The induced charge +q on the outer surface of the sphere will be evenly distributed due to the symmetry of the sphere." why this? – ivanculet May 11 '18 at 5:51
• Sorry that I am writing the third comment, but I thought that maybe it could be useful if I made myself clearer, and also update you on my chain of reasoning. I am unaware of a condition that establishes that since a surface is equipotential, the charge distribution must be homogeneous. If we took the ellipsoidal surface, the charge distribution as you said will not be homogeneous but still there is an equipotential surface. – ivanculet May 11 '18 at 6:00
• Ok thanks a lot, I was convinced when I first saw this exercise that since the conductors were surfaces , you couldn't use the fact that the field inside the conductor spherical shell is zero, since the shell is widthless. But approaching the way you did it is much more realistic. You have been a big help. Thanks. – ivanculet May 11 '18 at 14:59