Coulomb s rate problem, find the formula of Coulomb s law and prove it

This is a problem that you can solve with a strong balance. Point charges are subjected to forces in all directions (note that they are in all directions of space, not just planes), and these forces are all the same (the charges are evenly distributed) and are always reversed, macroscopically the force is zero. When a hole is excavated, the object is subjected to a force that is so much opposite to the force generated by the hole (the charge at the point).

The force of a circular hole on a point charge can be calculated as follows: k times q times q times r divided by r squared

It's that R is much smaller than R.

The areal density of the charge on the spherical shell is:

v=q/s=q/(4∏r^2)

On a disc with radius r:

q'=vs'=qr^2/(4r^2)

Due to the cancellation of the Coulomb force at other locations, only on the area corresponding to the opposite side of the hole.

Charge q'Coulomb repulsion to q.

f=kqq'r 2=kqqr 2 (4r 4) direction points to the small round hole.

First of all, v 4 3 ( r 3), v = 4 3 ( r 3) is uniform so v v=q q1, so q1 = ((r r) 3)*q, f = kq1*q (r-r) 2, bring the corresponding amount into k*q 2*(r r) 3*(r-r) 2, the direction is pointing to the center of the small hole. It can be assumed that at the beginning there is only a charge at the small hole, and the size is the same, but in the opposite direction.

f=k*q*q*r*r (r*r*r*r) in the direction of pointing through the center of the sphere towards the small round hole.

Coulomb's law. It should be two point charges.

The formula between the forces acting in between.

Suppose there is a Q-point charge, and there is a Q-point charge at the distance from it r. In order to calculate the force of q on q, first calculate the strength of the electric field at the distance r.

E size. For a sphere with q as the center and radius r as the center, use the high parallel grasp theorem.

e·4 r =q [0] (the magnitude of the electric field should all be the same, it is e, because the point charge is symmetrical in all directions, and the flux is e multiplied by the area 4 r; On the right is the enveloping charge q divided by the dielectric constant.

Thus e=q 4 r [0]=kq r (k=1 4 [0] is Coulomb's law.

The proportional coefficient inside).

Verification method. If the torsion angle of the Coulomb torsion scale is x, then the moment due to the repulsion force can be obtained as mx (m is the torsion coefficient and is the measurable value).

The moment of repulsion can also be expressed as FL (F is the magnitude of the repulsive force, L is the distance between the two balls connected to the torsion scale, and F is perpendicular to L).

In this way, it is possible to introduce the relationship between F and X, and then Coulomb's law.

It depends on how the landlord thinks ......(By the way, the Coulomb's law mentioned by the landlord should be a formula for the force between two point charges, right?) ）

First of all, the physical laws with the word "law" were originally derived from experiments, that is, experiments calculate the data, and then fit the formula (mentioned upstairs). For example, "Newton's three laws", "Archimedes' law of buoyancy", etc., are all like this. There is also a coincidental physical law, called the "theorem", which was originally proved by deducing the key derivation like mathematics, such as the "kinetic energy theorem", "momentum theorem" and so on.

From this point of view, the most direct derivation of Coulomb's law is experiment, and you can think that this is what the experiment sees, just remember it.

However, what I said above is "initially", not forever, theorems can be verified by experiments (and this is one of the criteria for evaluating the correctness of theorems), and the same laws can be proved by logical derivation, and the same is true of Coulomb's law. But logic needs a starting point (assuming that something is right and we don't doubt it, then we can prove something else), so to prove Coulomb's law, we also need to assume a starting point. One proof that I know of is to use Gauss's theorem as a starting point (Gauss's theorem states that the electric field strength flux of an arbitrarily closed surface is equal to the charge enclosed within it divided by the dielectric constant, see the encyclopedia).

Originally, Gauss's theorem was a "theorem", which was derived from Coulomb's law. Now let's assume that this is correct, and the experiment will prove Coulomb's law. The simple proof process is as follows.

Suppose there is a Q-point charge, and there is a Q-point charge at the distance from it r. In order to calculate the force of q on the clear q, first calculate the electric field strength e at the distance r. For a sphere with q as the center and radius r as the center, use Gauss's theorem:

e·4 r =q [0] (the magnitude of the electric field should all be the same, it is e, because the point charge is symmetrical in all directions, and the flux is e multiplied by the area 4 r; On the right is the enclosed charge q divided by the dielectric constant [0])).

Thus e=q 4 r [0]=kq r (k=1 4 [0] is the proportionality coefficient in Coulomb's law).

Finally, the above is just a proof, and there are other ways to prove Coulomb's law. I've heard that the inverse square of Coulomb's law can be proved from the static mass of a quantum mechanical photon of 0, but I don't know how to prove it.

e m = yes, if the charge of these two electrons is more than the electron cluster of a ball made up entirely of electrons, then it is fine, but there are also protons and neutrons in the conductor, and their charge-to-mass ratio is much smaller.

is the specific charge of electrons.

The specific charge of different charged bodies is not the same.

The specific charge value you use here is the specific charge of the electrons, and of course it will not work for the charged ball.

Neutrons, before decaying, are in a critical stable state, therefore, should be in a state of equilibrium of forces, and generally speaking, there should be two forces:

f(electric field force) = k qp qe rn rn).

f (centrifugal force) = (mn mp) c c rn

These two forces should be equal, where k is the electromagnetic constant; qp is the amount of power of the proton; qe is the amount of electricity of the electron; MN is the mass of the neutron; mp is the mass of the proton; rn is the radius of the neutron; c is the speed of light.

Based on this formula, we calculated:

The radius of the neutron is e-15 m

The density of neutrons is e+17 kg m3

The radius of the proton is e-15 m

The radius of the electron is e-17 m

The radius of the nucleus ra = ru a (1 3).

The radius of the atomic mass unit ru = e-15 m

where a is the mass of the nucleus.

The experimental results of scientists show that the radius of the helium-4 nucleus is femtometers, and the radius of the helium-6 nucleus is femtometers. Calculated according to our formula:

The radius of the helium-4 nucleus is femtometers, and the ratio of the experimental value to the theoretical value is ; The radius of the helium-6 nucleus is femtometers, and the ratio of the experimental value to the theoretical value is . The error between the experiment and the theory is 5%.

In the semi-empirical theoretical research of scientists, the well-known formula for the radius of the nucleus is:

ra = ro× a ^（1/3） （ro ≈ 1. 23 × 10^(-15) m ）

where a is the mass of the nucleus, and ro is called the nucleus radius constant. However, a large number of experimental data show that RO is not a constant, and it systematically decreases with the increase of a, from light to heavy nucleus by about 10 %, and this deviation indicates that important factors are ignored.

ru / ro =

where ru is exactly 10 % smaller than ro Therefore, the above formula proves that our nucleus radius formula is correct.

In addition, we found that the charge radius of the proton is different from the material radius of the proton, the charge radius of the proton is exactly on the material bisector of the proton, and the charge radius of the proton is femtometers, which is the data given in our **. Thanks to these data, we calculated the mass of the photon from the equilibrium relationship of the forces as: e-35 kg; The distance between the center of the electropositive and negative particles of the photon is:

Femtomi. All light and particles can be applied to our data to give precise calculations, and that's our value.

When charged with a different kind, the two balls attract each other.

Therefore, the isolation B is buried and hidden, and the Coulomb force mg is offset by the upward direction of the state hall.

t2=0 The overall total gravity is g=2mg

The Kusan force is an internal force, regardless of it, so t1 = 2mg

To sum up, choose a don't understand and ask again, hopefully.

In a vacuum, it is said that there is no air resistance, and in the medium in the real world, Coulomb's law formula is kq1q2 nr 2, where n refers to a constant, and in a vacuum it is 1, and in air it is greater than 1. In this problem, the surface is insulated and smooth and round, which can be understood as having no resistance, which is equivalent to being in a vacuum.

At the same time, the amount of their charge is very small for their distance, so it can be regarded as a point charge, which is solved using Coulomb's law.