### How Is Coulomb’S Law Similar To Newtons?

Abstract Assuming a Winterberg model for space where the vacuum consists of a very stiff two-component superfluid made up of positive and negative mass planckions, Q theory is the hypothesis, that Planck charge, q pl, was created at the same time as Planck mass.

Moreover, the repulsive force that like-mass planckions experience is, in reality, due to the electrostatic force of repulsion between like charges. These forces also give rise to what appears to be a gravitational force of attraction between two like planckions, but this is an illusion. In reality, gravity is electrostatic in origin if our model is correct.

In short they strive to maintain a fixed distance of separation from one another. Unlike mass planckions really do not interact directly, Instead they interact indirectly by being forced close to one another by their respective fluid forces. Consider a string of positively charged planckions, all in a row, along the x -axis, and label them, #1, #2, #3, etc.

Due to the symmetry, forces in the, y, and, z, direction cancel, and we are concerned only with forces in the x -direction. Focus on particle, #3, and sum up the forces acting on that particle, when that 3 rd particle is displaced a distance, x, to the right., and so we obtain, (3-1) In the last line, we used Equation (2-1). The force that particle, #4, exerts on particle, #3, which is,, is pointing to the left, and hence the negative sign in Equation (3-1). The force that particle, #2, exerts on particle, #3, which is,, is pointing to the right, and hence the positive sign associated with this force in Equation (3-1). The positive and negative signs for all subsequent forces are found in this fashion. In Equation (3-1),, is the unperturbed or average nearest neighbor equilibrium distance of separation between two planckions of the same species. The,, is related to the positive mass planckion number density,, through the equation, (3-2) Next, we recognize that, for any arbitrary, ” b “, and, ” β ” values, the following identity holds, (3−3) This can be proven algebraically. We set,, and substitute this together with Equation (3-3), for various, ” b “, values into Equation (3-1). We find, (3-4) In Equation (3-4), we first factor out the,, term on both left and right hand sides. Then we bring the,, term from the left hand side over to the right hand side, and make use of Equation (3-2), to simplify. The result, after pulling out a factor of 4, is, (3-5) Equation (3-5), holds for any value of, including, or, what is equivalent,, Therefore it follows that, (3-6) The infinite series within Equation (3-6), is a known series, (3-7) Here,, is the Riemann Zeta function, and equals Apery’s constant, an irrational number. Numerically,, Making use of this value, we now have an expression for, κ, the planckion spring constant. It reads, (3-8) More generally, irrespective of the value of,, in Equation (3-5), the infinite series within that expression on the right hand side, must equal,, The arguments above for a positive mass/charge planckion particle also hold for a negative mass/charge planckion particle. Some of the signs change, but the outcome is the same. The negative mass planckions are also held in check by what are, essentially, electrostatic forces. We still have to determine the planckion number density,, for an undisturbed vacuum. To fix a value for,, and thus determine a value for κ, by Equation (3-8), we use the fact that our planckions are, more or less, spatially anchored or locked in position. Hence they are confined to a region of space where box quantization must apply. They are also oscillating, i.e.

, continuously accelerating, which produces gravitational radiation. Because they are “boxed in”, and radiating, the energy of the radiation emitted must be related to the energy level jumps, or transitions, permissible within that box. Much like the Bohr atom for Hydrogen, energy level transitions account for the radiation frequency emitted.

The situation here is totally analogous. To see this more clearly, we consider the energy levels for a particle trapped in a 3-dimensional cubic box, having volume, L 3, These are well known to be quantized, and given by the expression, (3-9) The, are quantum numbers, which can take on the values,, The lowest energy level, or ground state, is specified by,, In this instance, Equation (3-9), reduces to, (3-10) For the mass of the confined particle, we have chosen the Planck mass, which in the present epoch, equals,, The next highest energy level has three-fold degeneracy, because,, The next highest energy level also has three-fold degeneracy since,, The fourth level has no degeneracy,, Continuing in this fashion, we can account for all the energy levels. If the positive mass planckion is excited, due to collisions with CBR photons, for example, transitions between energy levels are possible. We have quantum jumps where the energy emitted is, (3-11) The unprimed quantum numbers refer to the situation before, and the primed quantum numbers correspond to the situation after the transition. This is completely analogous to the situation in the Hydrogen atom, where we have the Lyman series, the Balmer series, the Paschen series, etc. The most probable transition is the most frequent one, and there,, This is already obvious from the few examples of energy levels given. Transitioning from higher to lower levels gives exactly this amount. For example, transitioning from,, gives exactly this amount. The energy being emitted is determined by the Planck radiator formula, (3-12) Here, ν, is the frequency of the photon being emitted, and T is the blackbody CBR temperature. The, k B, is Boltzmann’s constant. Nowhere in this formula, is mass, or charge, explicitly stated. We interpret this to mean that any quantum radiator will emit this amount of energy irrespective of whether it is mass or charge which is oscillating., Equation (3-12), the becomes, (3-13) In Equation (3-13), we have chosen the most probable frequency, for the frequency for this particular CBR temperature. In general the peak frequency is specified by, (3-14) The oscillating and continuously accelerating positive mass/charge planckion acts as a radiator, and at a CBR temperature of 2.725 Kelvin emits this specific frequency as its peak frequency. In reality a whole spectrum of frequencies are being emitted as an infinite number of quantum transitions are possible.

See Equation (3-11). We singled out one particular frequency, the most probable. Equation (3-9), is explicitly mass dependent. If the mass of the positive mass planckion changes, due to a change in G value, this will affect the energy levels, and the transitions which are possible. Lower mass means higher energy levels for the same quantum numbers according to Equation (3-9).

### comparison between coulomb’s law and gravitational law (Basic Physics)

This means that, in general, larger frequencies will be emitted. But this is exactly what we expect at higher CBR temperature. The frequencies being emitted, in general, will also change as a consequence, being shifted towards higher values. They are still quantized, but will take on different values.

1. What about the negative mass planckions? For a negative mass, such as a negative mass planckion, the energy levels are inherently negative, by Equation (3-9), as measured from the top down.
2. We have inverted box quantization, where instead of a potential energy square well, we have the mirror image, an upright potential energy square well.

This upright well is populated with energy levels, according to Equation (3-9), taken from the top down. The largest energy level jumps are near the top. The highest level in this inverted potential energy well is the “ground state”. And negative mass particles will want to transition to this highest energy level.

1. Equation (3-11), is still valid.
2. But instead of going from less negative energy to more negative energy, which is the case for a positive mass particle, we will be transitioning from positive energy levels to higher positive energy levels for a negative mass particle.
3. By transitioning upwards, a negative mass planckion actually lowers its binding energy.

Due to the symmetry of the energy levels between positive and negative planckions, the most probable transition here, also, releases,, of energy. It is worth mentioning that the frequency of emission, is exactly equal to the difference in oscillating frequencies of the planckions. Since the positive and negative mass planckions are continuously accelerating, with a specific frequency, the frequency of emission should equal the difference,, in oscillating frequencies, where,, is the angular frequency of the planckion before the transition, and,, is the angular frequency after the transition. Since both the positive and negative mass planckions contribute their fair share of energy to the radiator at a particular temperature, we set two times the energy value, indicated by Equation (3-10), equal to one times the value of Equation (3-13), which is the resulting energy output. Thus, (3-15) This is easily solved for length, and we obtain, (3-16) The dimensions of the cubic box, just so happen to equal the distance of separation between nearest neighbor positive, and negative, mass planckions. The only difference is that the geometric center of the box is centered around the positive or negative mass planckion.

1. One will notice that our nearest neighbor inter-planckion separation distance, within a specific species, is very close to the limits what modern day accelerators are able to probe.
2. The diameter of a quark is about, 8.60E−19 meters, and Equation (3-16), is just a hair below that.
3. The LHC at CERN produces 7 TeV protons whose Compton wavelength is 1.78E−19 meters.

All these distances are comparable to the nearest neighbor inter-planckion separation distance. If our estimates are correct, we may be on the verge of establishing an inherent “graininess” for the vacuum. Having determined the nearest neighbor separation distance, we proceed to find the average number density for both the positive and the negative mass planckions, when the vacuum is in the undisturbed state. (3-17a) (3-17b) The number densities are, needless to say, very high. But now, we are attempting to find a graininess to the vacuum, or space, which many believe is smooth and continuous. Finally, let us come back to the spring constant for planckions, Equation (3-8). Using Equations (3-8), and (3-17a), we can now evaluate its value. We obtain for this calculation, (3-18) This is a current epoch value because the spring constant will scale upon expansion of the universe, as shown in reference, This large value for a spring constant justifies our assumption that we are dealing with a very stiff superfluid/ supersolid when we are considering the vacuum. was associated with the restoring force if either the gravitational force or the electrostatic force got the upper hand, For equilibrium, the electrostatic force of repulsion counteracted the gravitational force of attraction between two like mass planckion particle.

1. Realizing that we are dealing with the same κ, it should be recognized that the gravitational force is really electrostatic in nature.
2. We can look more carefully at the sum given in Equation (3-1).
3. All the negative sign contributions pull the #3 particle to the left.
4. All the positive sign contributions pull planckion particle #3 to the right.

If,, then there is no displacement and particle #3 is in equilibrium. In other words, both the individual forces pulling to the left and the individual forces pushing to the right add up to zero. We can identify the gravitational force with the sum of either one of these net forces, pushing or pulling.

1. The electrostatic force would then be identified with the counteracting force.
2. In short, the force of gravity is really electrostatic in origin within the Q theory.
3. When the planckions were first created two force magnitudes were simultaneously created.
4. But one counteracted the other, and being an attractive force between two masses was treated as a gravitational force.

The gravitational force has evolved with cosmological time through a varying G value, whereas the electrostatic force has not. That is our hypothesis.4. Gravitational Displacements, the Gravitational Potential, Latent Gravitational Field Energy, and Vacuum Resilience We saw that the inter-planckion distance of separation, between like mass planckions, is of the order, ~5E−19 meters.

This must be close to the maximum displacement that a planckion particle can experience. Gravitational displacement within the vacuum will occur whenever we have an external applied field, such as that what might be found just outside the surface of a black hole. It is now time to look at such situations.

We first start, however, with another example, the cosmos as a whole. It was shown in a previous work, that we can define a cosmic gravitational field, due to all the ordinary, and polarized bound mass contained within the cosmos. Polarized bound mass was identified as dark matter. (4-1) The subscript, “0”, refers to the present epoch, and the bar indicates an average or smeared quantity. At distance scales in excess of about, 100 Mpc, the universe is fairly homogeneous and isotropic. The cosmic gravitational field,, in Equation (4-1), is the value obtained at the surface of the Hubble bubble, our Gaussian surface, and takes into account all matter, ordinary, and bound, contained within this sphere. Even though,, technically, is a surface gravitational field, we argued that it holds point for point within the cosmos, because one observer’s surface is another observer’s point of reference. Any observer within the universe would measure this same value, irrespective of location., with dark energy. It permeates all of space, takes into account source matter, made up of quarks and leptons, as well as bound, or polarized matter, due to macroscopic ordering of gravitational dipole moments within the vacuum. Using the ΛCDM parameters in Friedmann’s equation, it was found that numerically, the relative cosmic gravitational permittivity,, in the present epoch. Moreover, the gravitational permittivity, defined by,, equals, in the current era. The, G, is Newton’s constant, which may or may not be a cosmological constant. Since,, represents dark energy, it follows that (4-2) Solving this for,, gives us the value as is indicated in Equation (4-1). We are using the latest cosmological parameters as found by the Planck collaboration (final release), Let us find the average displacement associated with this cosmic gravitational field. We can use either Equation (2-20), to find this displacement, or, since, Equation (2-22). We’ll use Equation (2-20), because we cannot be sure that, We first rewrite Equation (2-20) in the form, (4-3) Next we evaluate the right hand side using Equations (4-2), (3-17a), and, (4-4) Evaluating the right hand side of Equation (4-3), gives (4-5) This is clearly a very small value and we are justified in using the approximation,, We next substitute our value for, y, Equation (2-19), for the left hand side of Equation (4-5). This renders, (4-6) This equation can be solved for,, For the planckion spring constant, κ, we’ll use Equation (3-18). The Planck mass is specified in Equation (4-4). And the speed of light is also known. Evaluating gives, (average cosmic displacement) (4-7) This is an incredibly small displacement. However, we keep in mind that space is very, very dilute, only about 6 hydrogen atoms per cubic meter. Due to this dilution, there is hardly any displacement. In the true voids, where there is no source matter, we would expect zero gravitational field displacement.

Also keep in mind that the maximum displacement seems to be in the neighborhood of about, 5E−19 meters, as is indicated by Equation (3-16). Another example might be the gravitational field of the earth. On the surface of the earth, the source gravitational field is, 9.81 m/s 2, The relative permittivity, K, is essentially one because, as far as we are able to determine, there is virtually no vacuum susceptibility on the surface of the earth, if we take this surface to equal our Gaussian surface.

There is virtually no polarized mass enclosed within this surface. For that we have to have a substantial vacuum or “empty” space, which doesn’t exist close to the earth. We calculate,, using Equation (2-16), and find (4-8) The gravitational permittivity equals,, Following the same steps as before, but now using the new gravitational field mass density, indicated by Equation (4-8), we find for the new gravitational displacement, (surface of earth) (4-9) Compared to the value in Equation (2-6), this is about 10 orders of magnitude larger. However, it is still extremely small in value. We next look at the gravitational potential and its relation to the gravitational pressure. We designate the gravitational potential by,, because it is really a difference in gravitational voltage that we are considering. The subscripts indicate that this difference in voltage is due to gravitational fields. We start with Equation (2-8). If we bring the infinitesimal displacement, d x, over to the left hand side, we really have,, where,, is the infinitesimal change in gravitational potential energy. However, we know that gravitational potential is related to gravitational potential energy by the relation,, Once the, d x, has been brought over to the left hand side in Equation (2-8), we divide both sides by, to obtain the relatively simple expression, (4-10) We integrate this from,, to, x, and find, (4-11) To take the two species of planckion particles into account, the positive and the negative mass species, we multiply this expression by a factor of two. Thus, (4-12) Finally, using Equation (2-9), this can be shown to equal, (4-13) This is a very simple and straightforward expression. Once we calculate a gravitational displacement, due to a specific gravitational field, we can find the corresponding increase in gravitational voltage, or gravitational potential, within the vacuum.

We simply use Equation (4-13), with our κ value, specified in Equation (3-18). The planckion mass in the present epoch, is indicated in Equation (4-4). Some numerical examples are as follows. For the cosmos as a whole, the average displacement of planckions within the vacuum, due to the presence of source and bound mass, is specified by Equation (4-7).

Here, Equation (4-13), gives (average gravitational potential within cosmos) (4-14) On the surface of the earth, we have a difference displacement, indicated by Equation (4-9). This leads to a different gravitational potential in the amount, (surface of earth) (4-15) Note that the units for gravitational potential are the same units as, c 2, The general relation between gravitational pressure, and gravitational potential, is considered next. Start with Equations (2-19), and, (2-20), and recognize that,, We rewrite Equation (2-20), as, (4-16) We next multiply Equation (4-16), through by the factor, c 2, to obtain the gravitational pressure, (4-17) This is our desired relation between gravitational pressure,, and gravitational potential,, In most instances,, and we can use the approximation,, In this approximation, Equation (4-17), simplifies to, (4-18) Here, we see a direct proportion between gravitational pressure and gravitational potential. The mass of the planckion, and planckion number density are also important. Incidentally, planckion pressure and gravitational field energy density,, are equal to one another, since (4-19) We already identified,, with dark energy, See also Equations (2-16), and, (4-2). Gravitational pressure, gravitational field mass density, which is the same as planckion mass density, and dark energy, are all synonymous with one another. Equations (4-17), and (4-18), are another way to find gravitational potential, They will of course give the same results as before. We close this section with some thoughts on the magnitude of the gravitational pressure here on the surface of the earth. As we saw, the mass density equaled the value indicated by Equation (4-8), namely, 6.373E−7 (MKS). (4-20) This is a comparatively large value for pressure, or energy density. Atmospheric pressure, for example, equals, 1.013E5 N/m 2, Why don’t we feel this gravitational pressure? Why can’t the energy in one cubic meter be released? In cosmology, the energy densities can be related to the stress-energy tensor.

• To release the energy trapped in a box, means we would have to alter the stress energy tensor.
• For that to happen, it takes a certain violent gravitational reaction, such as a supernova explosion, or a black hole merger.
• These are not the conditions found here on earth.
• To give you an analogy, there is a lot of energy trapped within the nucleus.

But only under certain circumstances can this be released, such as in a reactor, or in a bomb, or in a star. We believe something similar happens here. This is not energy trapped in matter, but energy stored, or trapped, within the vacuum, i.e., space itself.

It cannot be released without altering the gravitational field itself. If there is no gravitational field, then there is no gravitational pressure, nor is there a gravitational field energy density. We would have to alter the, 9.81 m/s 2, here on earth, in order to tap into this energy, or release the gravitational pressure associated with this slightly gravitationally stressed vacuum.

If we could eliminate the 9.81 m/s 2 in a box, one cubic meter in size, here on earth, we would liberate, 5.736E10 Joules, according to Equation (4-20). We can refer to this energy as latent gravitational energy. This brings us to mechanical resiliency.

Wherever there is a gravitational field, the vacuum is stressed, i.e., the individual planckions making up the vacuum are displaced from equilibrium. Our vacuum is very much mechanical in origin. In material science, resilience is the ability of a material to absorb energy when it is deformed elastically.

Once the stress is released, the elastic energy goes away upon unloading. This describes the vacuum particularly well to our thinking. Resilience, or mechanical energy storage capacity, can be defined for the vacuum, and is measured in units of, N/m 2, or J/m 3,

The maximum resilience of space seems to be in the neighborhood of about, 8.66E33 J/m 3, to about, 1.54E34 J/m 3, The first value holds on the surface of a four solar mass black hole, where the gravitational field is particularly strong, about, 3.81E12 m/s 2, The second value is the gravitational field energy density for a three solar mass black hole, where the gravitational field is even stronger, approximately, 5.08E12 m/s 2,

No black holes have been found in nature having a mass less than three solar masses. The cutoff between neutron stars and black holes seems to lie in the neighborhood between three to four solar masses. Neutron stars can have gravitational fields as high as roughly, ~2E12 m/s 2, once newly formed.

1. So the gravitational fields above seem to fit the scheme.
2. Now it is known, that next to a black hole, three-dimensional space will develop a rip or a tear in the space-time continuum according to the general theory of relativity.
3. In other words, three-dimensional space starts to break down.
4. This would be our version of “gravitic breakdown”.

Just like there is dielectric breakdown when the electric fields get too strong for the medium, we can expect that gravitationally, something similar happens. The gravitational fields listed above must be close to those limits. It should have been mentioned that the smallest mass black holes have the largest gravitational fields, due to the Schwarzschild condition.

• So the above gravitational fields are probably the strongest macroscopic gravitational fields known to science.5.
• Net Planckion Imbalance in the Present Universe Gravitational field energy density, which is the same as gravitational pressure, depends on stressing the vacuum, through the introduction of gravitational fields.

The number density of planckions is, thereby, directly affected. This is seen explicitly in Equation (2-17), where we set the mass density of planckions, equal to the gravitational field mass density. The left hand side of Equation (2-17), is no longer equal to zero, if,, In fact, we will have an imbalance in planckion number density, where now, Before we consider the universe at large, let us consider other situations where we have a gravitational field and a net imbalance in planckion density. We start with the examples of a three, and, four, solar mass black hole. Just outside the surface, we probably will have the largest gravitational fields found in nature. (5-1) Use of the Schwarschild condition,, has been employed to reduce the expression to a relation only involving the black hole mass,, The radius of the event horizon is,, We notice that the least massive black holes have the largest gravitational fields, by Equation (5-1). Just like the earth, we assume negligible vacuum susceptibility right outside the black hole. Little enclosed polarized mass is thought to exist within the black hole itself. We substitute the appropriate black hole masses into Equation (5-1), and obtain, (3 solar mass black hole) (5-2a) (4 solar mass black hole) (5-2b) Next we calculate the gravitational field mass densities, just like we did for the earth. See Equation (4-8). Using our new values for surface gravitational fields, indicated in Equations (5-2a, b), we evaluate,, and find, (3 solar mass black hole) (5-3a) (4 solar mass black hole) (5-3b) If we multiply these,, values by, c 2, we obtain the corresponding gravitational field energy densities on the surface of these black holes. They are the, 1.54E34 J/m 3, and, 8.66E33 J/m 3, respectively, the same values as mentioned previously, at the end of section IV. (3 solar mass black hole) (5-4a) (4 solar mass black hole) (5-4b) These mass densities are only about an order of magnitude larger. The exact number is twelve times larger, which can be easily proven for any mass spherically symmetric black hole. Just set up the ratio,, and work out the details, using the defining relations for the variables within the ratio. We find that,, a very interesting result since it is independent of mass or size. We know from Equation (2-17), that (5-5) Therefore, using Equations (5-3a, b), and, (4-4), it is possible to find the planckion number density imbalance. From Equation (5-5), we have (3 solar mass black hole) (5-6a) (4 solar mass black hole) (5-6b) Because of the intense gravitational fields, and the very extreme gravitational field mass densities, we have an enormous imbalance in planckion number density. In one cubic meter on the surface of these black holes expect about, 10E24 – 10E25, more positive mass planckions than negative mass planckions.

Like we said, this is probably some sort of upper limit for what three dimensional space will tolerate, as far as gravitational stress is concerned. For the earth, the numbers are drastically reduced because the surface gravitational field is so much less. We still use Equation (5-5), but now the mass density is given by Equation (4-8).

Substituting this value into Equation (5-5), and carrying through the calculation, gives (5-7) This is a very interesting result. The imbalance only amounts to approximately 29 more positive mass planckions than negative mass planckions in one cubic meter. Next we look at the cosmos. There we also have a net macroscopic gravitational field due to the ordinary and polarized matter, which is contained within it. (average within the cosmos) (5-8) This is not the lowest value possible. In a void, where there is no source mass present, and thus no polarized mass, the value would be exactly,, But then we would have an undisturbed vacuum. For the very dilute universe we have today, there is an average of one excess positive mass planckion over negative mass planckion, every, Why the universe has a net positive mass is unknown. The result in Equation (5-8), may suggest that, when planckions were first formed, there was an excess of positive mass planckions over negative mass planckions. But this would go counter to the Q theory, because as planckion mass was created, so too was planckion charge.

• And, as far as we can tell, the universe has zero net charge.
• Somehow between then, when planckions were first created, and now, the imbalance must have formed.
• And it probably had something to do with the formation of quarks and leptons.
• Why and how negative mass planckions were used up in this process is unclear.

We close with a quick calculation for the imbalance in terms of absolute planckion numbers. If the Hubble radius in the present epoch is,, then, (5-9) For,, we have used Equation (5-8). And for the radius of the observable universe,, we took this to equal,, a value found in reference,, The,, is the present day difference in planckion numbers, with the positive mass planckions being more plentiful than negative mass planckions. This excess, while it appears large, is only an insignificant amount when compared to total planckion numbers. Using Equations (3-17a), or, (3-17b), and Equation (5-8), we see that (5-10) This fraction is very minute.6. Summary and Conclusions We introduced a model where Planck mass and Planck charge were frozen out of the vacuum simultaneously. We treated mass and charge as two components of a more fundamental particle, the planckion.

• Based on previous and extensive work by Winterberg, the vacuum is a vast assembly (sea) of positive and negative mass planckions, which form a two-component superfluid and fill all of space.
• This ether is initially massively, and electrically neutral, has zero net mass density, zero net gravitational pressure, and zero net entropy in the undisturbed state.

Within the Winterberg model, we introduce Q theory, the notion that Planck mass and Planck charge were created simultaneously, as well as two force laws, one electrostatic, and one seemingly gravitational in nature. The electrostatic force keeps two planckions of the same species, whether they have positive or negative mass, apart.

1. The gravistatic force brings them together.
2. Thus equilibrium is achieved, where individual planckions maintain a fixed distance of separation from each other within their species.
3. Along with the simultaneous creation of mass and charge, we also posited the simultaneous creation of two force laws.
4. Equation (2-2), shows us how they are connected.

Planckions are anchored, or locked, in position spatially via fluid forces. Equation (2-7), holds for positive mass planckions, and Equation (2-10), is valid for negative mass planckions. Both lead to number density functions, which will tend to bring the planckions back to equilibrium, if displaced.

• See Equations (2-9), and (2-12).
• The total gravitational pressure is zero, and so is the energy density, if planckions are undisturbed.
• See Equations (2-9), and (2-15).
• If the vacuum is disturbed through the introduction of an applied gravitational field, we obtain Equations (2-17), and, (2-20).
• We identified the gravitational field energy density, or dark energy, with planckion mass density.

This leads to Equation (2-22), relating planckion displacement to gravitational field intensity. In Section III, we looked at the restoring force acting on individual planckions more carefully and discovered it was entirely electrostatic in origin. In other words, the gravitational force introduced in Section II, within the Q theory, is, in reality, electrostatic in nature, see Equation (3-1).

• The planckion spring constant could thus be evaluated, and the result is Equation (3-8).
• Moreover, by appealing to box quantization, and Planck’s radiator formula, we could estimate the average number density for the positive and negative mass planckions in the undisturbed vacuum.
• Equation (3-17), is the result.

This planckion number density holds only in the present cosmological epoch. Given the relation between number density, and mean distance of separation, Equation (3-2), we could determine the nearest neighbor inter-spatial distance between planckions of the same species., as indicated in Equation (3-18). The vacuum is very stiff, and this spring constant value is also epoch dependent. In section IV, we considered various examples of planckion displacements. For the cosmos as a whole, there is a net gravitational field due to all the matter contained within it, ordinary and bound, i.e., which is Equation (4-7). For the gravitational field of the earth, on the surface we find a shift in the amount,, Equation (4-9). We derived relations between gravitational potential, and gravitational pressure. Those relationships are presented in Equations (4-17), and, (4-18). An easy way to calculate gravitational potential is through Equation (4-13). It was also recognized that dark energy is synonymous with gravitational field mass density, which is the same as planckion mass density.

• See Equations (2-17), (4-2), (4-19), and, (4-20).
• Whenever we have a gravitational field we have a gravitational, or planckion, pressure imposed upon the vacuum.
• The vacuum is stressed, even here on the surface of the earth, by a minute amount.
• This is a latent form of energy we argued, and it can only be released under extreme conditions where the gravitational field gets wiped out.

See the discussion following Equation (4-20). We also argued that the vacuum has a certain mechanical resiliency, which we estimate lies in the neighborhood of about, 8.66E33 J/m 3, to, 1.54E34 J/m 3, Gravitational fields beyond that may lead to “gravitic breakdown” or “vacuum breakdown”.

1. Finally, in section V, we considered the imbalance in planckion mass density between positive and negative mass planckions.
2. Because there is ordinary mass in the universe, a given, and polarized mass in the cosmos, an assumption, we have a net smeared gravitational field which does not vanish for the cosmos as a whole.

By Equations (5-5), and, (5-8), this forces us to accept that, in the present epoch, the average positive planckion number density exceeds the average negative mass planckion number density, There are more positive mass planckions per unit volume than negative. The exact amount has been calculated in Equation (5-8). The reason for this is unclear, although it may have something to do with ordinary matter, made up of quarks and leptons, being formed in the universe.

When planckions were first created in the very early universe, there were equal numbers according to the Q-theory. There is also the intriguing possibility, although not very likely, that there are regions in the universe where planckion mass density, or dark energy, is negative. This would balance the total number densities between positive and negative mass planckions, then, in the early universe, as well as now, in the present epoch.

There is, however, no direct observational evidence for this, i.e., negative dark energy does not appear to exist. As such, we are left with Equation (5-8), which shows a net imbalance. Q-theory explains charge and mass neutrality in the early universe, but it does not explain quasiparticle formation, collective excitations in the Winterberg model.

Nor does it explain positive planckion density imbalance over negative planckion density, in the present epoch. Q theory can, however, provide a connection between gravistatic and electrostatic force laws. In the very early universe these were formed simultaneously. The gravistatic force of attraction, upon closer inspection, turned out to be electrostatic in origin.

Work is in progress on a microscopic theory of planckions, which would include scaling behavior upon expansion of the universe. Acknowledgements To my family, my parents, my brother, and my children. Conflicts of Interest The author declares no conflicts of interest regarding the publication of this paper.

#### Is Coulomb’s law in Newtons?

Electric Force and Acceleration – Suppose that a rubber balloon and a plastic golf tube are both charged negatively by rubbing them with animal fur. Suppose that the balloon is tossed up into the air and the golf tube is held beneath it in an effort to levitate the balloon in midair.

This goal would be accomplished when the spatial separation between charged objects is adjusted such that the downward gravity force (F grav ) and the upward electric force (F elect ) are balanced. This would present a difficult task of manipulation as the balloon would constantly move from side to side and up and down under the influences of both the gravity force and the electric force.

When the golf tube is held too far below the balloon, the balloon would fall and accelerate downward. This would in turn decrease the separation distance and lead to an increase in the electric force. As the F elect increases, it would likely exceed the F grav and the balloon would suddenly accelerate upward. Suppose that at some instant in the process of trying to levitate the balloon, the following conditions existed: A 0.90-gram balloon with a charge of -75 nC is located a distance of 12 cm above a plastic golf tube that has a charge of -83 nC. How could one apply Newton’s laws to determine the acceleration of the balloon at this instant? Like any problem involving force and acceleration, the problem would begin with the construction of a free-body diagram. There are two forces acting upon the balloon. The force of gravity on the balloon is directed downward. The electric force on the balloon is exerted upward since the balloon and golf tube are like-charged and the golf tube is held below the balloon.

1. These two forces are shown in the free-body diagram at the right.
2. The second step involves determining the magnitude of these two forces.
3. The force of gravity is determined by multiplying the mass (in kilograms) by the acceleration of gravity.
4. F grav = m • g = (0.00090 kg) • (9.8 m/s/s) F grav = 8.82 x 10 -3 N, down The electric force is determined using Coulomb’s law,

As shown below, the appropriate unit on charge is the Coulomb (C) and the appropriate unit on distance is meters (m). Use of these units will result in a force unit of the Newton. The demand for these units emerges from the units on Coulomb’s constant.

F elect = k • Q 1 • Q 2 /d 2 F elect = (9 x 10 9 N•m 2 /C 2 ) • (-75 x 10 -9 C) • (-83 x 10 -9 C) / (0.12) 2 F elect = 3.89 x 10 -3 N, up The net force is the vector sum of these two forces. The upward and downward forces are added together as vectors. F net = ·F = F grav (down) + F elect (up) F net = 8.82 x 10 -3 N, down + 3.89 x 10 -3 N, up F net = 4.93 x 10 -3 N, down The final step of this problem involves the use of Newton’s second law to determine the acceleration of the object.

The acceleration is the net force divided by the mass (in kilograms). a = F net / m = ( 4.93 x 10 -3 N, down ) / (0.00090 kg) a = 5.5 m/s/s, down The above analysis illustrates how Newton’s law and Coulomb’s law can be applied to determine an instantaneous acceleration.

### What is Coulomb’s law similar to?

Comparing Electrical and Gravitational Forces – Electrical force and gravitational force are the two non-contact forces discussed in The Physics Classroom tutorial, Coulomb’s law equation for electrical force bears a strong resemblance to Newton’s equation for universal gravitation, The two equations have a very similar form. Both equations show an inverse square relationship between force and separation distance. And both equations show that the force is proportional to the product of the quantity that causes the force – charge in the case of electrical force and mass in the case of gravitational force.

• Yet there are some striking differences between these two forces.
• First, a comparison of the proportionality constants – k versus G – reveals that the Coulomb’s law constant (k) is significantly greater than Newton’s universal gravitation constant (G).
• Subsequently a unit of charge will attract a unit of charge with significantly more force than a unit of mass will attract a unit of mass.

Second, gravitational forces are only attractive; electrical forces can be either attractive or repulsive. The inverse square relationship between force and distance that is woven into the equation is common to both non-contact forces. This relationship highlights the importance of separation distance when it comes to the electrical force between charged objects.

### Is Coulomb’s law consistent with Newton’s third law?

Electric Charges and Fields. Does Coulomb’s law of electric force obey Newton’s third law of motion? Yes Coulomb’s law is in accordance with Newton’s third law of motion. Coulomb forces are equal in magnitude, opposite in direction and act on different bodies.

## How does Coulomb law relate to force?

Home Science Physics Matter & Energy Alternate titles: electronic attraction, law of electrostatic attraction Coulomb’s law, mathematical description of the electric force between charged objects. Formulated by the 18th-century French physicist Charles-Augustin de Coulomb, it is analogous to Isaac Newton ‘s law of gravity,

Both gravitational and electric forces decrease with the square of the distance between the objects, and both forces act along a line between them. In Coulomb’s law, however, the magnitude and sign of the electric force are determined by the electric charge, rather than the mass, of an object. Thus, charge determines how electromagnetism influences the motion of charged objects.

Charge is a basic property of matter, Every constituent of matter has an electric charge with a value that can be positive, negative, or zero. For example, electrons are negatively charged, and atomic nuclei are positively charged. Most bulk matter has an equal amount of positive and negative charge and thus has zero net charge.

Like charges repel each other; unlike charges attract. Thus, two negative charges repel one another, while a positive charge attracts a negative charge. The attraction or repulsion acts along the line between the two charges. The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value, If the charges come 10 times closer, the size of the force increases by a factor of 100. The size of the force is proportional to the value of each charge. The unit used to measure charge is the coulomb (C). If there were two positive charges, one of 0.1 coulomb and the second of 0.2 coulomb, they would repel each other with a force that depends on the product 0.2 × 0.1. Thus, if each of the charges were reduced by one-half, the repulsion would be reduced to one-quarter of its former value.

#### Which best describe Coulomb’s law?

Problems on Coulombs Law – Problem 1: Charges of magnitude 100 microcoulomb each are located in a vacuum at the corners A, B and C of an equilateral triangle measuring 4 meters on each side. If the charge at A and C are positive and the charge B negative, what is the magnitude and direction of the total force on the charge at C? Sol. $$\begin F_ =\frac _ }}. \frac ^ }}\end$$ along AC The on C due to B, i.e., F CB in direction CB, is given by; $$\begin F_ = \frac _ }}. \frac ^ }}\end$$ along CB Thus, the two forces are equal in magnitude. The angle between them is 120º. The resultant force F is given by;

1. $$\begin F =\sqrt ^ _ }}}+ ^ _ }}}+2 _ }\times _ }\cos 120 ^\text }\end$$
2. $$\begin =\frac ^ }} ^ }} \\= \frac ^ }\times ^ })}^ }} ^ }}\end$$
3. This force is parallel to AB.

= 5.625 Newton Problem 2: The negative point charges of unit magnitude and a positive point charge q are placed along the straight line. At what position and for what value of q will the system be in equilibrium? Check whether it is stable, unstable or neutral equilibrium.

• $$\begin F_ =\frac _ }}.\frac _ }^ }\ towards\ A\end$$
• Force on q due to B
• $$\begin F_ =\frac _ }}. \frac _ }^ }\ towards\ B\end$$

, These two acting on q are opposite and collinear. For the equilibrium of q, the two forces must also be equal i.e. |F qA | = |F qB | or $$\begin \frac _ }}. \frac _ }^ } = \frac _ }}. \frac _ }^ }\end$$ Hence, rA = rB. So for the equilibrium of q, it must be equidistant from A & B, i.e., in the middle of AB.

1. Now for the equilibrium of the system, A and B must be in equilibrium.
2. For the equilibrium of A
3. Force on A by q
4. $$\begin =\frac _ }}. \frac _ }^ }\end$$
5. Force on A by B
6. $$\begin =\frac _ }}. \frac _ }+ _ })}^ }}\end$$
7. $$\begin =\frac _ }}. \frac _ })}^ }}\end$$

towards q away from q The two forces are opposite and collinear. For equilibrium, the forces must be equal, opposite and collinear. Hence, $$\begin \frac _ }}. \frac _ }^ }= \frac _ }}. \frac _ })}^ }}\end$$ or q = 1/4 in the magnitude of either charge.

• Given
• q 1 = 6×10 -6 C
• q 2 = 4×10 -6 C
• r = 0.040 m
• Sol.
• $$\begin F_ = k \frac q_ } }\end$$

$$\begin F_ = \frac \left ( 6\times 10^ \right )\left (4\times 10^ \right )} \right )}\end$$ $$\begin F_ = \frac \left ( 2.4\times 10^ \right )} }\end$$ $$\begin F_ = \frac }\end$$ F e = 134.85 N Problem 4: Two-point charges, q 1 = +9 μC and q 2 = 4 μC, are separated by a distance r = 12 cm. What is the magnitude of the electric force?

1. given
2. k = 8.988 x 10 9 Nm 2 C −2
3. q 1 = 9 ×10 -6 C
4. q 2 = 4 ×10 -6 C
5. r = 12cm = 0.12 m
6. Sol:
7. $$\begin F_ = k \frac q_ } }\end$$

$$\begin F_ = \frac \left ( 9\times 10^ \right )\left (4\times 10^ \right )} \right )}\end$$ $$\begin F_ = \frac \left ( 3.6\times 10^ \right )} \end$$ $$\begin F_ = \frac \end$$ F e = 22.475 N The electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Coulomb’s law holds good for stationary point charges. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz Visit BYJU’S for all Physics related queries and study materials

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View Quiz Answers and Analysis : Coulomb’s Law – Vector Form, Limitations, Examples, Key Points

## What is Coulomb’s constant equal to?

The constant of proportionality k is called Coulomb’s constant. In SI units, the constant k has the value k = 8.99 × 10 9 N ⋅ m 2 /C 2.

#### What is the difference between Newton’s gravitational force and that of a Coulomb force?

Difference between Coulomb’s Force and Gravitational Force – The main distinction between gravitational and electrostatic forces is that gravitational force is the force that causes the earth to attract other things due to its mass. It is a conservative political party. Coulomb’s force is the force exerted by an object as a result of its charge. It is a conservative force as well.

 Sr. No. Gravitational Force Coulomb’s Force 1. It is the force exerted by an object as a result of its mass. It is the force exerted by an object as a result of its charge. 2. It is a very strong force as compared to the electric force. It is a very weak force as compared to the Electric force. 3. It is only attractive. It is attractive as well as repulsive. 4. It does not depend on the medium. It depends on the medium. 5. Its constant is “G”. Its constant is “K”. 6. It originates from the mass. It originates from the charge. 7. Its formula is: F = G × M × m/r 2 Its formula is: F = k × q 1 × q 2 /r 2

## What are the differences and resemblances between the gravitational force and the Coulomb’s force?

What are the differences between Coulomb force and gravitational force ? Text Solution Solution : 1) The gravitational force between two masses is always attractive but Coulomb force between two charges can be attractive or repulsive, depending on the nature of charges.2) The value of the gravitational constant G = 6.626 xx 10^(-11) Nm^(2) kg^(-2).

The value of the constant k in Coulomb law is k = 9 xx 10^(9) m^(2)C^(-2). (3) The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on nature of the medium in which the two charges are kept at rest. (4) The gravitational force between two point masses is the same whether two masses are at rest or in motion.

If the charges are in motion, yet another force (Lorentz force) comes into play in addition to coulomb force. : What are the differences between Coulomb force and gravitational force ?

### What are four similarities between electrostatic and gravitational forces?

Gravitational v Electrostatic Forces –

The similarities and differences between gravitational and electrostatic forces are listed in the table below:

Table Comparing G and E Fields

The key similarities are:

The magnitude of the gravitational and electrostatic force between two point masses or charges are inverse square law relationships The field lines around a point mass and negative point charge are identical The field lines in a uniform gravitational and electric field are identical The 2 ” data-title=”Gravitational Field Strength” data-toggle=”popover”> gravitational field strength and electric field strength both have a 1 / r 2 relationship in a radial field The gravitational potential and 0 r)” data-title=”Electric Potential” data-toggle=”popover”> electric potential both have a 1 / r relationship Equipotential surfaces for both gravitational and electric fields are spherical around a point mass or charge and equally spaced parallel lines in uniform fields The work done in each field is either the product of the mass and change in potential or charge and change in potential

The key differences are:

The gravitational force acts on particles with mass whilst the electrostatic force acts on particles with charge The gravitational force is always attractive whilst the electrostatic force can be attractive or repulsive The gravitational potential is always negative whilst the electric potential can be either negative or positive