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Added- What I do know is this, your calculation uses M1 and M2 My formula uses a+b and a+b . I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.
Quote from: Thebox on 23/03/2018 12:00:33Added- What I do know is this, your calculation uses M1 and M2 My formula uses a+b and a+b . I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass. Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.
Quote from: Kryptid on 23/03/2018 16:26:22Quote from: Thebox on 23/03/2018 12:00:33Added- What I do know is this, your calculation uses M1 and M2 My formula uses a+b and a+b . I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass. Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.I wish it was that simple, I wish I had a Maxwell to help me, it is not that I can't do equations, it is a matter of knowing what all the symbols mean and how to present them, additionally what the numbers mean when they have -10 but the - is high or low etc, additionally when the number ends with an E, I have no idea what that suppose to mean. Coulombs constant , might as well be Mandarin , I can't read it.
Quote from: Thebox on 23/03/2018 16:53:20Quote from: Kryptid on 23/03/2018 16:26:22Quote from: Thebox on 23/03/2018 12:00:33Added- What I do know is this, your calculation uses M1 and M2 My formula uses a+b and a+b . I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass. Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.I wish it was that simple, I wish I had a Maxwell to help me, it is not that I can't do equations, it is a matter of knowing what all the symbols mean and how to present them, additionally what the numbers mean when they have -10 but the - is high or low etc, additionally when the number ends with an E, I have no idea what that suppose to mean. Coulombs constant , might as well be Mandarin , I can't read it. Then I'll try to do it for you. I'll just need some information first. How far apart are the proton and electron in your model of the hydrogen atom? Is it in agreement with the current model (i.e. approximately the Bohr radius) or is it some different value?
In my model the Proton and Electron are merged and have a void in the center which is a dense field. Sort of like a maltesers.
Quote from: Thebox on 23/03/2018 19:51:44In my model the Proton and Electron are merged and have a void in the center which is a dense field. Sort of like a maltesers. ...okay then.I can already tell you that the attractive force will be exactly zero in that case. If the proton in Atom 1 is the exact same distance from the proton in Atom 2 as it is from the electron in Atom 2, then the attractive and repulsive forces will be exactly equal to each other and cancel out. If you insist on it, I can do the calculation anyway if you don't believe me. It really shouldn't be necessary though...
then the attractive and repulsive forces will be exactly equal to each other and cancel out
No, the attractive force is not 0, the electron and the proton while merged retain their attractive force to retain form, if there was 0 attractive force they would not sustain being merged.
When r=0 between two atoms , the attractive force remains between electrons and protons of each atom, the fields form bonds.
They do , an atom cannot pass through an atom because it is equally repulsive as attractive.
I'm talking about zero attractive force between the two different atoms, not between the proton and the electron in the same atom. I'm also talking about net attraction, not gross attractions. Sure, the proton in Atom 1 will be attracted to the electron in Atom 2, but it is also being repelled by the proton in Atom 2. Since protons have electric charge equally strong to that of electrons, and since you have defined that protons and electrons are at the same place in the same atom, then the attraction and repulsion between the atoms have to be equal to each other. They are pulling on each other just as strongly as they are pushing away from each other. The net force has to be zero.
I was talking about two different atoms. The force between the two atoms is not 0, both atoms want to merge but can't because the repulsive force stops it merging.
Quote from: Thebox on 24/03/2018 01:12:25I was talking about two different atoms. The force between the two atoms is not 0, both atoms want to merge but can't because the repulsive force stops it merging. So how do the attractive forces between the two atoms become stronger than the repulsive forces between the same two atoms? The repulsive force between two protons should be exactly equal to the attractive force between a proton and an electron (at the same distance apart).Let me do the equations. I'll consider a pair of hydrogen atoms one nanometer apart. I'll label the proton in the first atom P1, the proton in the second atom P2, the electron in the first atom as E1 and the electron in the second atom as E2. Between these two atoms, the possible interactions are P1P2 (proton-proton repulsion), P1E2 (proton-electron attraction), P2E1 (proton-electron attraction) and E1E2 (electron-electron repulsion). Take note that P1E1 and P2E2 are not considered because they are interactions within the atoms, not between the atoms. Now let's calculate the force involved in each interaction. Positive numbers indicate repulsion, negative numbers indicate attraction:FP1P2 = k((Q1Q2)/r2)FP1P2 = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(1.602 x 10-19 C)/(10-9 m)2)FP1P2 = (8.9875 x 109)((2.566 x 10-38)/(10-9)2)FP1P2 = (8.9875 x 109)((2.566 x 10-38)/10-18)FP1P2 = (8.9875 x 109)(2.566 x 10-20)FP1P2 = 2.306 x 10-10 newtonsFP1E2 = k((Q1Q2)/r2)FP1E2 = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(-1.602 x 10-19 C)/(10-9 m)2)FP1E2 = (8.9875 x 109)((-2.566 x 10-38)/(10-9)2)FP1E2 = (8.9875 x 109)((-2.566 x 10-38)/10-18)FP1E2 = (8.9875 x 109)(-2.566 x 10-20)FP1E2 = -2.306 x 10-10 newtonsFP2E1 = k((Q1Q2)/r2)FP2E1 = (8.9875 x 109 N m2 C−2)((1.602 x 10-19 C)(-1.602 x 10-19 C)/(10-9 m)2)FP2E1 = (8.9875 x 109)((-2.566 x 10-38)/(10-9)2)FP2E1 = (8.9875 x 109)((-2.566 x 10-38)/10-18)FP2E1 = (8.9875 x 109)(-2.566 x 10-20)FP2E1 = -2.306 x 10-10 newtonsFE1E2 = k((Q1Q2)/r2)FE1E2 = (8.9875 x 109 N m2 C−2)((-1.602 x 10-19 C)(-1.602 x 10-19 C)/(10-9 m)2)FE1E2 = (8.9875 x 109)((2.566 x 10-38)/(10-9)2)FE1E2 = (8.9875 x 109)((2.566 x 10-38)/10-18)FE1E2 = (8.9875 x 109)(2.566 x 10-20)FE1E2 = 2.306 x 10-10 newtonsNow let's add up all of the forces and see what the net attraction or repulsion is:Fnet = FP1P2 + FP1E2 + FP2E1 + FE1E2Fnet = (2.306 x 10-10 newtons) + (-2.306 x 10-10 newtons) + (-2.306 x 10-10 newtons) + (2.306 x 10-10 newtons)Fnet = 0 newtonsThere's the math for you. There is no net electrostatic force between two hydrogen atoms when they have the structure that you propose that they have.
That looks complex and the answer should be 0 , but 0 means a balance of 0 not no force.
Imagine weighing two identical mass on pan scales to measure 0.
Quote from: Thebox on 24/03/2018 14:37:06That looks complex and the answer should be 0 , but 0 means a balance of 0 not no force.The net force absolutely is zero. If there is more force on the attractive side than the repulsive side, please explain where it comes from. The numbers certainly don't show it.QuoteImagine weighing two identical mass on pan scales to measure 0.That's a great analogy, actually. Even though both masses are experiencing a force pulling them down towards the Earth, they are keeping each other from tipping the scales in either direction. Their forces are equal so that the scale doesn't move. Same thing with two hydrogen atoms. The attractions and repulsions are equal, so there is no movement.
The fields repulsion is not as great as the atoms attraction
Quote from: Thebox on 24/03/2018 14:54:36The fields repulsion is not as great as the atoms attractionThis is the part that you have not demonstrated. The equations show quite clearly that the attractions and repulsions are equal. Where are you getting this extra attraction from? It certainly isn't from electrostatic attraction.
You are not considering this correctly
consider two rubber balls and squeeze them together, that is the repulsion because of field density N-field particle.
Move them a length apart , the spacial field between them is not dense enough to stop them being drawn together by force to a 0 radius.
So you're saying that Coulomb's law is wrong? That's quite a bold assertion.
What force? It isn't electrostatic. I've already shown mathematically via Coulomb's law that it can't be.