Electron's movement due to electric forces

[Rugile]

Homework Statement

4 electrons are moving due to electric forces. Find their velocity when they are very far away from each other, if initially they were on a square's (side length a = 20 cm) vertexes.

Homework Equations

$F = \frac{kq_1q_2}{r^2}$
$k = \frac{1}{4\pi\epsilon_0}$

The Attempt at a Solution

Firstly, each electron is pushed by the other three electrons. Let the forces be $\vec{F_1}, \vec{F_2}$ and $\vec{F_3}$, and $\vec{F_1}, \vec{F_2}$ are the forces exerted by adjacent electrons. Since those two vectors are right-angled, the scalar sum of those vectors will equal to $F_{12} = \sqrt{F_1^2 + F_2^2}$. Both those forces are $F_1=F_2=\frac{kq^2}{x^2}$, where q is electron's charge and x is distance between two electrons at some point of time. So $F_{12} = \sqrt{2\frac{k^2 q^4}{x^4}} = \frac{k q^2}{x^2}\sqrt{2}$ The opposite electron's (to the first one) force vector is in the same direction as the force $F_{12}$, so $F_{123} = F = F_3 + F_{12}$. Since we have a square here: $F_3 = \frac{kq^2}{ (\sqrt{2}x )^2 } = \frac{kq^2}{2x^2}$. Now we know that $F=ma$, so $ma = \frac{kq^2}{x^2}(\sqrt{2} + \frac{2}{2})$. Now what's really confusing me is that the accelerations seems to be a variable here, since x is changing. I have found that $v = \frac{d^2 x}{dt^2}t + \frac{d^3x}{dt^3}t^2$, but that doesn't really help me a lot ( I don't know how to solve such differential equations).

Any help appreciated!
Rugile

 

[gneill]

Have you considered an energy conservation approach? What's the total PE in the initial configuration?

 

[BvU]

Would it be an idea to think of this in terms of energies? Because of the symmetry I would expect each electron to fly off into infinity, thereby converting electrostatic potential energy into kinetic energy.

Ah, again, even quick replies cross. Well, the hints are alike, so there you go...

 

[Rugile]

Oh, that's right! Not so difficult as I thought :) thank you!

 

[HalfLostAndSearching]

I would like to clarify a few things about the scenario described in this problem. Firstly, it is important to note that the movement of electrons is not solely due to electric forces. Electrons also have their own inherent motion due to their mass and energy. Additionally, the electric forces between electrons are constantly changing as they move, so it is not accurate to assume a constant force between them.

That being said, let's assume that the problem is asking for the velocity of the electrons when they are very far away from each other, such that the electric forces between them can be considered negligible. In this case, we can use the equations provided to solve for the velocity.

Firstly, we can find the total force acting on each electron by summing up the forces exerted by the other three electrons. This can be done using vector addition, as described in the problem attempt. However, it is important to note that the distance between the electrons will be changing as they move, so we cannot use a fixed distance value in our calculations.

Next, we can use the equation F=ma to find the acceleration of the electrons. Again, we must use the total force and the mass of a single electron in this calculation.

Finally, to find the velocity, we can use the equation v=u+at, where u is the initial velocity (which we can assume to be zero) and t is the time it takes for the electrons to reach the point where the electric forces can be considered negligible. We can find this time by using the equation d=ut+0.5at^2, where d is the distance between the electrons at the starting point (20 cm) and at the point where the forces are negligible (very far away).

In summary, the velocity of the electrons when they are very far away from each other will depend on the initial conditions (such as their initial velocities and positions) and the time it takes for them to reach a point where electric forces can be considered negligible. This problem can be solved using the equations provided, but it is important to consider the limitations and assumptions made in the solution.