Finding equilibrum point between two charges

[dock]

[SOLVED] finding equilibrum point between two charges

this is pure physics;ain't no theory development;so don't move me around.

assume you have two charges q₁ and q₂ on distance R.the problem i'll solve here is how to find the equilibrum point between those charges.
to do that you have to introduce new charge q₃ in the system and place it somewhere between the first two on the conecting line.assume that distance between q₁ and q₃ is x then:
F₁₃x^2=-kq₁q₃
and
F₂₃(R-x)^2=-kq₂q₃
but it also has to be
F₁₃=-F₂₃
this system of three equations gives the following:
-x^2q₂=(R-x)^2q₁
the solution for x is:
x₁=(q₁+sqrt(-q₁q₂))R/(q₂+q₁)
x₂=(q₁-sqrt(-q₁q₂))R/(q₂+q₁)

i wonder why if the charges are same by sing then x is some imaginary num and why if it is q₁=-q₂ then x is infinite/undefined point?

am gratefull for your answers!

 

[cala]

If q1=-q2, the condition F13+F23=0 can't exist, so there is no equilibrium point on space.

If the charges are of the same kind, the equations are:

F13x^2=-kq1q3 or F13x^2=kq1q3

F23(R-x)^2=kq2q3 or F23(R-x)^2=-kq2q3

F13+F23=0 or F13=-F23

so x^2q2=(R-x)^2q1

There is no - on one term, so it seems to me that you made the ^2, but better than that, make the sqrt of each term.

Then, the solution is:

x=sqrt(q1)R/(sqrt(q2)+sqrt(q1))

when q1=q2, it gives: x=R/2

 

[chroot]

Actually, you don't have to introduce any test charges. The electric field at any point in space can be represented as a superposition of the fields generated by all the sources present in the system. Thus you can immediately come up with an equation that represents the total field magnitude at any point in space -- it's just two copies of Coulomb's Law added together. Then you solve it for zeros.

- Warren

 

[dock]

i'd like to tell you why i was looking for this equilibrum point.
begin simulation
choose Q1,Q2,R;
find the equilibrum point (EP)=R/(1+(+/-)sqrt(abs(Q1/Q2)));
make X1 the equilibrum vector starting in (EP) ending in Q1;
make X2 the equilibrum vector starting in (EP) ending in Q2;
make F1 the force vector akting upon Q1 and normal with X1;
make F2 the force vector akting upon Q2 and normal with X2;
make E1=F1xX1=const the energy vector of Q1;
make E2=F2xX2=const the energy vector of Q2;
begin loop
choose dX1 the displacement vector for Q1 such that F1xdX1>0 and
||X1+dX1||=||X1||;
choose dX2 the displacement vector for Q2 such that F2xdX2>0 and
||X2+dX2||=||X2|| but also ||X1||*||dX2||=||X2||*||dX1||;
make the new X1 equal to the old X1 plus dX1;
make the new X2 equal to the old X2 plus dX2;
onscreen(Q1,Q2)
end loop
end simulation

this is how i could find the trajectories of Q1 and Q2 without ever using F=ma i.e. Newton the 2nd.the values are relative to (EP). using standard 3d vector procedures like adding, rotating etc i can get all the possible trajecoties.

i consider this to be the greatest discovery of mine so far.

 

[chroot]

From what I can tell, nothing that you're doing even resembles real physics.

If you put two free, opposite charges near each other, they will come together and "disappear." If the charges have no mass, then they will travel at the speed of light towards each other. If they have finite mass, you will eventually have to use Newtonian or special relativistic relationships to determine their trajectories.

I have no idea why you think your little program does anything physical. After all, the only vector dX1 that satisfies the condition "||X1+dX1||=||X1||" is the zero vector. Perhaps you are running across machine rounding errors and thinking you've discovered the Holy Land.

Furthermore, without a specified mass, you really can't say anything about how the charges will move -- at least not in this reality.

I know you're a total kook, and I'm not even sure why I'm discussing anything with you.

- Warren

 

[Claude Bile]

Couloumbs law is a specific case for a more general Gauss' law. Gauss' law can be rewritten in terms of electric potential using Poisson's equation -

(Laplacian)V = (Charge density)/(Epsilon0)

For regions of no charge density (If we neglect the discontinuous point charges), it reduces to Laplaces equation.

(Laplacian)V = 0

Solving for Laplace's equation, it is possible to show that there can be no maxima of minima in the potential field, hence no equilibrium points.

Charges cannot be confined by electric fields, however they can be confined by magnetic fields.

 

[chroot]

We weren't talking about the potential (scalar) field -- we were talking about the electric (vector) field. There are equilibrium positions (i.e. zeros) in the electric field. They are unstable equilibrium points, however -- which is what [nab]²V is telling you.

- Warren

 

[dock]

Originally posted by chroot
From what I can tell, nothing that you're doing even resembles real physics.

If you put two free, opposite charges near each other, they will come together and "disappear." If the charges have no mass, then they will travel at the speed of light towards each other. If they have finite mass, you will eventually have to use Newtonian or special relativistic relationships to determine their trajectories.

I have no idea why you think your little program does anything physical. After all, the only vector dX1 that satisfies the condition "||X1+dX1||=||X1||" is the zero vector. Perhaps you are running across machine rounding errors and thinking you've discovered the Holy Land.

Furthermore, without a specified mass, you really can't say anything about how the charges will move -- at least not in this reality.

I know you're a total kook, and I'm not even sure why I'm discussing anything with you.

- Warren

only one question for you since you know that much:
Energy is a vector product of force vector and the equilibrium distance vector.of this I'm so sure and i can prove it to you even if you wake me in 3AM in the morrning.

what is the energy vector acording to your understanding and scenario and how can it be conserved since the system has no energy input/output from/to outside?

"scratch my back and i'll scratch yours" - says Konfuchie

 

[STAii]

sorry if this seems dumb, but energy is a scalar quantity, not a vector one

 

[chroot]

Last time I checked, 'energy' does not seem to have a directional quality.

- Warren

 

[dock]

Originally posted by chroot
Last time I checked, 'energy' does not seem to have a directional quality.

- Warren

this is due to the fact that the revolution of physics directed by dock has a maximum warp speed.

just kidding!

there is new topic on PF
https://www.physicsforums.com/showthread.php?s=&threadid=1000