Can you merge these equations and get acceleration

[locika]

m*a=k*Q1*Q2/r
equalizing Newton's first law and coulomb's force to get acceleration of the specific charged body.

 

[brainpushups]

I see one equation. And yes, solving for acceleration is as simple as dividing one factor to the other side (note: you are missing the square on your factor 'r').

 

[locika]

I see one equation. And yes, solving for acceleration is as simple as dividing one factor to the other side (note: you are missing the square on your factor 'r').

Thanks, and because force changes across distance then acceleration wouldn't be constant so is this a valid formula

 

[Noctisdark]

m*a=k*Q1*Q2/r
equalizing Newton's first law and coulomb's force to get acceleration of the specific charged body.

I trield very hard to solve this equation, i came up with numerical solution, but the exact solution is a bessel x'' = kQq/(x^2 + y^2), a second order differential equation, good luck !

 

[rcgldr]

Are you looking for a 1D, 2D, or 3D solution? If looking for a 1D solution, then this previous thread might help:

https://www.physicsforums.com/threads/time-of-approach-of-two-oppositely-charged-particle.761290

 

[hackYou]

m*a=k*Q1*Q2/r
equalizing Newton's first law and coulomb's force to get acceleration of the specific charged body.

For sure there are more competent members here to answer this but I don't think it is possible to merge these two equations since ther nature of the two forces involved is different.

 

[brainpushups]

Thanks, and because force changes across distance then acceleration wouldn't be constant so is this a valid formula

Answer: kind of.

The acceleration is valid for the 'initial' acceleration of the charges. The problem is not that the acceleration varies with position (of course, what you have is a differential equation), the problem is that, once the charges are accelerating, Coulomb's force law is no longer valid.