Displacment equation of a charged particle

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SUMMARY

The discussion focuses on the displacement equation of a charged particle influenced by another fixed charged particle along a single axis. The force acting on the charged particle is defined by the equation F = (const)(q1*q2)/(r^2), with simplifications leading to acceleration expressed as a = 1/(x^2). The challenge lies in determining displacement as a function of time, given that displacement (X), velocity (v), and acceleration (a) are interdependent. The differential equation to solve is d²x/dt² = 1/x².

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thetrice
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In a case of a charged particle attracted to another fixed chrged particle both on one axis.
force on charged particle will be F= (const)(q1*q2)/(r^2)
for simplification = q1*q2*const =1 , and also mass of particle will be equal to one .
so acceleration = 1/(r^2)
r is distance between 2 particles, Another simplification is to assume r = X which is displacemnt of chrged particle(assuming fixed particle at fixed cordinates of(0,0,0) .
so equation is accelration=1/(x^2)
where x= velocity*time , velocity = initial velocity + acceleration *time.
so how can i get displacement in variable of time ?
Problem is that X is affected by v , v is affected by a , and a is affected by X. so its a loop.
 
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thetrice said:
In a case of a charged particle attracted to another fixed chrged particle both on one axis.
force on charged particle will be F= (const)(q1*q2)/(r^2)
for simplification = q1*q2*const =1 , and also mass of particle will be equal to one .
so acceleration = 1/(r^2) r is distance between 2 particles, Another simplification is to assume r = X which is displacemnt of chrged particle(assuming fixed particle at fixed cordinates of(0,0,0) .
so equation is accelration=1/(x^2)
where x= velocity*time , velocity = initial velocity + acceleration *time.
so how can i get displacement in variable of time ?
Problem is that X is affected by v , v is affected by a , and a is affected by X. so its a loop.

Putting both charges on the x axis...
You need to solve:

[tex]\frac{d^2x}{dt^2} = \frac{1}{x^2}[/tex]
 

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