I would say that it is polarization,not induction.

Induction is for motion in a magnetic field (induced EMF) which creates currents.

If you take everything into acount including the earth I do

not see any problem. It would only be a local circumstance.

You may be refering to conservation of charge and not conservation of energy?

My first question after browsing your paper is:

Did you take the energy of the field of the central body into account?

By adding a sphere, the charge on the sphere is separated but at the same time the field of the separated charge is superimposed on the original field resulting in no field at all, inside the wall of the sphere. So, the energy of the field of the combination of central body and surrounding spheres is less than the energy of the field of the central body alone. (Which is balanced, of course, by the mechanical energy gained during placing the spheres.)

The same should be true when grounding the outside of the outmost sphere: Energy is gained by the charge flowing from the sphere to infinity, and the same energy is lost by the disappearance of the field outside the sphere.

Everybody has to make his/her own experiences, of course, but please bear with me if I add a memory of an old man :-) : Many decades ago, I used to invent electrostatic paradoxes myself. This came to an abrupt end when I fully realized that a superposition of electrostatic (as well as dynamic) fields results just in the (vector) sum of the fields. In other words: While notions as "shielding", "penetration depth", "rotation of the plane of polarization" (Faraday effect) etc. are nice shortcuts for engineering purposes, the physicist should always be aware of the fact that static fields reach out to and radiation fields propagate to infinity. If they seemingly become weaker (i. e., their integral over a sphere), rotated, or distorted, this has to be an effect of superposition of further fields. The consequence is: Either we succeed in finding an electrostatic paradox involving only two charged particles (which seems to be a hard task) or we cannot find an electrostatic paradox at all.

Hope this helps.

A look at the "patent" of the publication's author helps to understand the setup:  https://www.researchgate.net/publication/303342836_WIRELESS_TRANSFER_OF_ELECTRICITY_USING_CHARGE_INDUCTION

The whole setup can be considered a giant - though low capacitance - multilayer capacitor.And - honestly - I've never heard of any magic from multilayee capacitors. Nor about any violations of the law of energy.

I only heard about some lack of capabilities of students. Which could be helped by teachers during studies.

Patent WIRELESS TRANSFER OF ELECTRICITY USING CHARGE INDUCTION

Dreher,

that's a different kettle of fish I don't remember attaching that as resources in this context.

Juan,

https://en.m.wikipedia.org/wiki/Electrostatic_induction

Thank you Joerg,

Yes an electrostatic equilibrium will not have been reached unless inside the walls have no electric fields otherwise the charges will flow.

If you draw lines emanating from the charges outward to infinity. You will realise that the disappear in the walls and appear at the surfaces.

Therefore the electric field lines on the outer most sphere is a vector sum of the electric fields on the inner spheres. Therefore the energy of the electric field of the outermost sphere is more than that of the central sphere.

O. k., here it goes: I skip your abstract, and start with the introduction. The part on page 1 looks good but contradicts the statement in your comment above. If the amount of charge on the outermost sphere equals the amount of charge on the central body (right), then - given that the surface of this sphere has to be necessarily larger than the surface of the central body - the charge density on the sphere is lower than that on the central body. Second, the field between the spheres and outside the outermost sphere is the same as in the absence of any spheres; the field inside the walls of the spheres is zero.

Introduction, page 2: "electron shielding effect": As I wrote in my first comment, you can use notions like "shielding" as long as you are aware that any shielding is accomplished by superposition of a secondary field over the (unchanged) primary field.

Second sentence in bold font: In the rest of your text, I cannot find the calculations mentioned here. Let's consider the case that a charged central body is already present and resting (relative to an arbitrary frame); now two half-spheres move in opposite directions from infinity toward the central body until they form a sphere concentric to the central body. I claim that the sum of mechanical energy (caused by the force acting on the moving half-spheres) and heat (produced by the polarization current) equals the field energy lost inside the wall of the sphere. Please prove my claim wrong either by giving the general equations and/or by calculating a concrete example.

P. S. I saw your last comment only after finishing mine. The answer to your last paragraph is: No, because the amount of charge on the inside of any sphere equals the amount of charge on the outside of the sphere but the charges are of opposite polarity. Therefore, the vector sum of the fields of one shell alone yields zero everywhere outside the shell. So, the vector sum of all fields outside the outermost shell equals exactly the field of the central body.

Thank you Jeorg,

In your last paragraph I meant Fig 4 after earthing took place and the spheres have net charges. Therefore the electric field lines on the outer most sphere is a vector sum of the electric fields on the inner spheres. Therefore the energy of the electric field of the outermost sphere is more than that on each inner sphere including the one that was held by central sphere.

To make it simple lets assume that this conductors and charged body have the same capacitance. And we are lucky that the induced charge is 100%. If we momentarily earth each conductor it means they will gain equal but opposite charges to the charged body. Meaning they each store the same amount of energy as the initial charged body.

Max,

your design has the big advantage of spherical symmetry. To keep things simple, let's define earth as well as spherical and concentric. Then earth is a spherical shell whose radius is very large, tending to infinity, and earthing means to provide charged particles the freedom to move to or from infinity from or to the earthed body, still in spherical symmetry.

Earthing the outermost sphere results in removing the charge from the outside of the sphere to infinity. During this process, the charge gains and/or transfers mechanical energy, and the outer field vanishes because the sum of the remaining charge is zero.

Earthing the second sphere (counting from the outside) results in removing the charge from the outside of the second sphere to the inside of the first sphere. There the process stops (apart from possible oscillations) because if the charge would move further, the remaining charge would not add up to zero and would attract the freely movable charge under consideration. During this process, the field between sphere 1 and sphere 2 vanishes.

And so it goes on: By earthing all spheres, a certain amount of mechanical energy becomes available, and the same amount of energy is lost by the vanishing of the fields.

I have a bad conscience while explaining this, because the effect on you would be stronger and accompanied by much more surprise or joy if you arrived at these conclusions by your own effort.

Thank you Jeorg once again for your time.

Yes your right the energy lost when each individual electric  field collapses is equal to the energy gained as heat, mechanical work, sound etc.

Here is the real problem.

Let's figureout how much energy we input into the system and how much we are getting out. Let's assume we are standing initially on the earth with zero electrostatic potential energy. We then use the resources of the  earth to build our setup. We expend energy say 1Joule to create the field on the central sphere and then bring in the different spheres. After earthing the system, new electric fields appear that have more energy than what we expended on the central sphere.

The fields of this spheres carry more energy because there are a vector sums of the inner fields.Therefore the single low energy field you did work to create is giving rise to many high energy fields.

Dear Max,

sorry to state: this "kettle of fish" is still a number of capacitors in series connection. Discharging any of these caps results in a decrease of the potential of the most-inner plate vs. "earth". There is no energy to be gained.

Max, I feel we will have no progress with pure prose. So, let's take the 1 J you mentioned in your last comment, and restrict this to the process of forming the central charged ball (no spheres as yet). Now you have the freedom to choose the final radius of the central ball, and then please calculate the charge on the ball.

You don't have to solve tedious integrals manually if you don't like it, you might as well use a math program. But please don't restrict your answer to the numerical results but post the used equations as well.

Consider 3 hollow conductor spheres A, B and C together with the charged hollow conductor sphere X of charge Q.

Let the spheres A, B, C and X be of a similar capacitance C. i.e.The difference in their capacitances is so small to be significant.

Bring the spheres A, B and C towards X so that the enclose it.

Momentarily earth the spheres A, B, C and X so that A, B and C each gain a net charge Q. Let the volumes of the spheres A, B, C and X each be V. i.e. The difference in their volumes is so small to be significant.

The charge density Z on X was,

Z = Q/V

The charge densities Z on A, B and C are,

Z = Q/V+Q/V+Q/V

Z = 3Q/V

The results show that the charge density of A, B and C charge system is higher than that which was on X. This is possible because the electrostatic field on the outer most sphere is the vector sum of the electrostatic fields emanating from the charged inner spheres. So the charge density on the outermost sphere is more than that which was on the central sphere if we add the net and induced charges on the surface of the outermost sphere.

Therefore a higher charge density is being created at a certain point in space without as expending energy.

I'm sorry, Max, but I give up. The good point of your style of communication is: You do not become aggressive when contradicted; try to remain so. The bad point: You are not responding to the answers. This gives the impression that you do not really have a question but simply intend to circulate your point of view.

My last advice: Build a prototype, and measure the excess energy. And then respond at least to Mother Nature's answer.

Max

In your article you make a mistake in stating that the charge distribution will change if you earth the internal spheres after having earthed the external sphere.  When you earth the external sphere, the positive charge will go, but all internal charge distributions will be unchanged.  When after that you earth each internal sphere in turn the charge distribution will not change because the charges will be kept in place by the opposite charges on the surface facing them.  Or to put it differently, the assembly of spheres is now neutral and will not discharge to earth, as this would leave it charged, i.e. require work to separate the discharge from the charge left behind.  So the whole calculation of different charges on different spheres is wrong.

Also the energy gained when the hemispheres are allowed to drop into place around the sphere is lost when the hemispheres come to rest - it must be absorbed either by the apparatus stopping the hemispheres from falling as fast as they want, or be turned into heat when the hemispheres meet.  Some of the work will be lost by heat due to movement of the electrons, but that is very small, and most will need to be absorbed some other way, or the hemispheres would bounce apart almost as far as they started after being allowed to collide with each other.