- #1

Rugile

- 79

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## 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

[itex]F = \frac{kq_1q_2}{r^2}[/itex]

[itex]k = \frac{1}{4\pi\epsilon_0}[/itex]

## The Attempt at a Solution

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

Any help appreciated!

Rugile