# Homework Help: Electron's movement due to electric forces

1. Feb 11, 2014

### Rugile

1. The problem statement, all variables and given/known data
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.

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

3. 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

2. Feb 11, 2014

### Staff: Mentor

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

3. Feb 11, 2014

### 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...

4. Feb 12, 2014

### Rugile

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