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ELECTRIC NETFORCE IN REAL MATTER 
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NETFORCE 
My quest for comprising gravity as a netforce started in real matter.
Assemble two positive electric charges and two negative electric charges in a square as shown. At first glance one might say "There are two positive charges plus two negative charges, so the total charge is zero and so the total force is zero too". But when counting the number of attractions and repulsions, one sees there are four attractions and only two repulsions. The repulsions act over a little longer distance than the attractions and thus are weaker. The net force isn't zero at all. Under the electric force alone this square should implode! At least at first moment. One can see the forces perpendicular to the plane of the charges are zero. And the forces on each charge are equal and symmetrical. So one can expect the square to remain a square while imploding. Which means the contraction will continue. Lets go one dimension higher. Now we have: 12 attractions of strength 1 in the first figure, minus 12 repulsions of strength 1/2 in the second figure, plus 4 attractions of strength 1/3 in the last figure. equals 12  6 + 4/3 = 6.75. (The distances of the repulsions in the second cube are the square root 2 times the distances of attraction in the first cube, according to Pythagoras. The strength of the electric force falls with the inverse square of the distance. The distances of the attractions in the third cube are square root 3 times the distance of an attraction in the first cube.) We have effective nearly 7 attractions of strength one more than repulsions. So this cube will contract too. For reasons of symmetry one may expect the cube staying a cube while contracting. At first glance it looks stable too. So also this contraction will go on and on. Does this hold for more charges? Suppose we start from a large but finite cloud of an even number of positive and negative charges, homogeneous distributed over space and in a homogeneous distribution of charge. This is the most stable distribution. Every excess of one kind of charges anywhere tends to spread until it is maximum smeared out over the cloud of charges. For simplicity we assume their masses all to be equal and at one certain moment of time t all charges are standing still relative to each other. Select from the cloud a sphere with n positive and n negative charges in it and neglect all other charges as if they weren't there. ("click them off") The total number of attractions in the remaining sphere is # attractions = n (the number of negative charges times the number of positive charges)The total number of repulsions is # repulsions = 2 * = n  n (all possible pairs in the group of negative charges plus all possible pairs in the group of positive charges)The net force is the total number of attractive interactions minus the total number of repulsive interactions netforce = n  (n  n) = n So in a cloud of n positive and n negative charges there are n attractions more than repulsions.The electric force falls with increasing distance. It is only for this reason we try to maintain a maximum homogeneity of spatial and charge distribution. Then the distances of the repulsive interactions are assumed to be distributed ”in the same way” and to have ”the same weight” as the distances of the attractive interactions. As a result a contracting force is working all over the cloud at moment t. The cloud is going to contract, at least at first moment t. Does the spatial distribution of the 2n charges in the cloud keep its shape while contracting? For symmetrical reasons the contracting force is directed to the sphere's center all over the cloud and one may expect this stays like that. But the strength of the force, isn't it stronger at the center and weaker the more one goes towards the surface? Or is it the other way around? I should my simulation keep running again and then I can see. Let's suppose for the moment the force is the same all over the cloud. Suppose we go back to moment t and we take an extra (2n+1)th charge and place it in an appropriate way in the frayed edge of the sphere. We do this simply by slowly enlarging the sphere's radius and, if necessary, displace the center of the sphere a little, until precisely one of the "clickedoff" charges is enclosed in the sphere too and "click it on" again. This ads 2n interactions to the cloud, n attractive and n repulsive ones. But if someone who has never seen the cloud before, is coming to look at the cloud after we added our (2n+1)th particle, that person will not be able to judge which particle it was that we had been adding! In the same way the cloud itself will not be able to tell the new particle from the rest. One could argue that ”the cloud consists of an even amount of wellmixed positive and negative particles of equal mass” and will be electrically neutral with respect to an extra particle outside. But at this moment this person might definitely be pointed to the fact that the cloud is in fact a multipole, an electric field that is the sum of all monopole electric fields of the individual charges. The result may look neutral from large distance but at close detail it should reveal a very complicated pattern of outgoing and ingoing fieldlines that are nearly straight lines originating in the center of the sphere. ”So if the particle accidentally happens to lie on an outgoing fieldline it would recede from the cloud. It would be repelled”, our questioner insists. ”Well, here the homogeneity of the spatial and charge distribution should come to the rescue. It would provide every charge is always a little bit more surrounded by it opposites than by its equals which should make it always lying on an inward directed fieldline  anyway, I think the repelling field line will go into space, turn around there and speed back again into the cloud of charges. The particle will always return. That holds as an attraction!” ”Well, this sounds a little bit fuzzy, but all right, usually this indeed might be the case. But then still the (2n+1)th particle added n repulsive interactions and n attractive ones, which still are supposed to cancel each other out nearly completely. So no matter what fieldline it is on in the resulting multipole field, this is of no significance. Our extra particle will fall behind while the rest of the cloud starts to contract.” But then, which one was our particle? Which particle will remain behind? Because the cloud doesn't distinguish between its 2n+1 particles no one will stay behind. There is just no other way. In other words: a negative (and likewise a positive) charge will be attracted by a ”neutral” cloud of n positive and n negative particles. The contraction only will proceed a little slower. The estimation calculated above now is: # attractions = n (n+1) = n + n
The next step is to remember which one was our (2n+1)th particle and then again slowly enlarge the sphere of charges (and if necessary displace it a little too) so that it encompasses another particle, of opposite charge this time, just next to our (2n+1)th particle. For educative reasons we assume this succeeded this time. What do we have now? At first we made the cloud electrically neutral again. It now consists of n+1 positive and n+1 negative charges. So according to preceding reasoning this cloud is going to contract. But we might interpret the two new particles (number 2n+1 and number 2n+2) as a new minicloud. That cloud contracts because we made the particles being each others opposite. The rest of the cloud, being the n positive and n negative charges with which we started, contracts too. And since the whole cloud of n+1 positive and n+1 negative charge will start to contract anyway, we may interpret this result as follows. The electrical neutral cloud of n+1 positive and n+1 negative charges and the electrical neutral cloud of our two new particles are attracting each other. although a little too fast maybe, we draw from this the general conclusion that two neutral clouds of electrical charges with a homogeneous spatial and charge distribution (which is an equilibriumcondition), will attract each other. NEXT PAGE Up CONTACT 
