# Distance dependent force

1. Oct 23, 2004

### Homo sapiens sapiens

I’m trying to solve the following problem for a long time:
There are two charged particles, one doesn’t move, the other is being accelerated and is initially at rest, I would use the particle that doesn’t move as the referential, this referential has only one dimension, how do I find the position as a function of time.
I tried this approach, by Coloumb’s law the force in the second particle is F=(k*q*Q)/r^2, for simplification I put a C in place of k*q*Q, now I use Newton’s second law, I get a=C/(m*x^2), then I try to put velocity as a function of time, so the work done on the particle as a function of the distance is found by solving the integral of the force as a function of the distance between x and x0(this is the initial position), I got W=(c/x)-(c/x0), knowing that the particle starts at rest, its kinetic energy must be equal to the work done by the electrical force so W=(mv^2)/2, I put velocity as a function of the position, that is v=√((2c/(xm))-(2c/(x0m))), now the idea of all this stuff was to use the relation a=dv/dt to get the position as a function of time, this is what I don’t know how to do and would appreciate someone’s help in solving the problem this way or a different propose for a solution.

2. Oct 23, 2004

### jcsd

You know that from the defintion of accelartion:

$$\frac{d^2r}{dt^2} = \frac{C}{mr^2}$$

multiply both sides by $2\frac{dr}{dt}$, integrate with respect to r and take the square root:

$$\frac{dr}{dt} = \sqrt{\frac{-2C}{mr} + A}$$

Where A is a constant.

Then it's a case of seperating the variables to solve the equation.

3. Oct 24, 2004

### Homo sapiens sapiens

“Then it's a case of seperating the variables to solve the equation.”
It is this part that I don’t know how to do, could you solve that, please.

4. Oct 24, 2004

### HallsofIvy

"Separating the variables" means exactly what it says:
$$\frac{dr}{dt} = \sqrt{\frac{-2C}{mr} + A}$$
Now get the variable r on one side, the variable t on the other:

$$dr= \sqrt{\frac{-2C}{mr}+A}dt$$
$$\frac{dt}{\sqrt{\frac{-2C}{mr}+A}}= dt$$

and integrate both sides.

5. Oct 25, 2004

### Homo sapiens sapiens

Could you explain to me how you passed from $$dr= \sqrt{\frac{-2C}{mr}+A}dt$$
to $$\frac{dt}{\sqrt{\frac{-2C}{mr}+A}}= dt$$
, when you talk abut integrating the last equation e have no problem integrating the right side, but on the left side there is a dt so the independent variable is t, and I am to solve the integral between two instants t, but on the function that is being integrated there is no t, this doesn’t makes sense to me.
I’m sorry if I am doing trivial questions, but I know few about calculus, thanks for your help and patience.

6. Oct 25, 2004

### arildno

It's quite simple; HallsofIvy made a typo.
On the left-hand side, it should be dr not dt