# Motion of a charged particle

1. Jun 25, 2007

### nicktacik

In one dimension, the motion of a charged particle (q1) will be [assume q2 is stationary]

$$\frac{d^2 x}{d t^2} =\frac{Kq_1q_2}{m x^2}$$

Is there a solution to this differential equation?

2. Jun 25, 2007

### olgranpappy

$$(At+B)^{\frac{2}{3}}$$
where
$$\frac{-2A^2}{9}=\frac{Kq_1q_2}{m}$$
and
$$B$$
is anything you want.

3. Jun 25, 2007

### nicktacik

Thanks, that seems to work. I wonder why maple couldn't give me that answer.

4. Jun 25, 2007

### olgranpappy

well... it's a pretty particular solution I gave you... it only works when the "total energy" is zero, i.e., when
$$\frac{1}{2}mv^2 + \frac{Kq1q2}{x} = 0$$.

If you rewrite A and B in terms of x(0) and v(0) you will see that the condition on A means that
1/2mv(0)^2+Kq1q2/x(0)=0... but it is also easy to show that 1/2mv^2 + Kq1q2/x is a constant in time thus it is always zero.

In general the solution is hard, but using the constants of the motion we can write
$$\int_{x(0)}^{x}dy\frac{\sqrt{m}}{\sqrt{2E-2Kq_1q_2/y}}=t$$
to find t(x) and then invert to find x(t)...