Displacment equation of a charged particle

AI Thread Summary
The discussion focuses on deriving the displacement equation for a charged particle attracted to a fixed charged particle along one axis. The force acting on the charged particle is defined by F = (const)(q1*q2)/(r^2), with simplifications leading to acceleration = 1/(x^2). The challenge arises in expressing displacement as a function of time, given that displacement (X), velocity (v), and acceleration (a) are interdependent. The key equation to solve is d²x/dt² = 1/x², highlighting the complexity of the relationships between these variables. Ultimately, the discussion seeks a method to express displacement in terms of time despite the circular dependency.
thetrice
Messages
19
Reaction score
0
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.
 
Physics news on Phys.org
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:

\frac{d^2x}{dt^2} = \frac{1}{x^2}
 
Thread 'Motional EMF in Faraday disc, co-rotating magnet axial mean flux'
So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday...
Back
Top