There is a negatively charged wall($q$), which has equal charge density. There is a positively charged particle($Q$) with a mass of $m$, the distance between them is $d$.
How much time would it take for the particle to get to this wall?
EDIT: I messed up on viewing the wall as a stationary charged particle, so I would rather ask the question regarding such one.
I have started solving this exercise by the following equation, which is true for the system in its initial position: $F=k*\frac{q*Q}{r^2}$
We also know that $F=ma$, thus $a=\frac{k}{m}*\frac{q*Q}{r^2}$
The problem is, that now we have the value of $a$ against $r$, but we are interested in the time $t$.
So what I did was drawing the plot for $a$ against $r$.
I am not sure at all in the following step of mine: I concluded that the area under the graph equals $\frac1t$, as the area under the curve equals the integral of the acceleration, which is the speed $v$ divided by the distance $r$.
$\frac{v}{r}=\frac1t$
As it is the area under the curve, I looked at the integral of $a$ in the respect of $r$, but the definite integral from $0$ to $d$ has given me strange results (infinity), thus I for some strange reason concluded that the indefinite integral might be useful as well (I have not learnt calculus, only watched some videos about it, so please bear with me) . This resulted in in the following: $$\int a dr=-\frac{k*q*Q}{m*r}$$
From the previous paragraph I "know" that it might be equal to $\frac1t$, thus I have answered the question as $t=-\frac{m*d}{k*q*Q}$. I am kind of sure that this is not a correct way of doing things, so please correct me, but even though this result at least seems reasonable, as if I were to increase the mass of the particle the time would increase as well, the same is true for the distance, and if I were to increase the charges on any of the two charged objects, the time would decrease.
