# Electrons movement due to electric forces

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

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