Graph for Coulomb Force vs $1/r$ My teacher told me that the graph for the coulomb force $F$ vs $1/r$ where $r$ is the distance between the 2 charges should be parabolic but I can't seem to understand why. I am aware that equations of the form $y^2=4ax$ are parabolic but why should $F$ vs $1/r$ graph be parabolic?
 A: The Coulomb force magnitude is given by
$$F=\frac{kQq}{r^2}$$
Pulling out the $(1/r)^2$ gives us
$$F=kQq\cdot\left(\frac1r\right)^2$$
So then F has a quadratic dependence on $1/r$, which is a parabola if you were to graph $F$ vs $1/r$.
If you still cannot see it, then call $1/r$ something else, like $x=1/r$. Then $F(x)=kQqx^2$
A: Assume $F$ to be $y$ and $1/r$ to be $x$
Then, according to coulomb's law
$$y = cx^2$$
where $c = \frac{q_1 q_2}{4 \pi \varepsilon_0}$
Now you can rearrange the equation as follows.
$$x^2 = c' y$$
where $c' = \frac{1}{c}$
Now you can take $c' = 4a$, and this will give you your familiar expression for a parabola with the directrix parallel to the $x$-axis
$$x^2 = 4ay$$
Note that there was no need to do all this. A quadratic expression is always a parabola.
A: You are right that functions like $y = ax^2$ have a parabolic graph of $y$ vs $x$.
The force between two charges is
$$F = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$$
So the graph of $F$ vs $r$ is not parabolic. But you are not graphing that.
You are graphing $F$ vs $1/r$. You can see how the equations can be made to be similar. $$F \to y, \space \frac{q_1q_2}{4\pi\epsilon_0}\to a, \space and \space1/r \to x$$
The confusing part might be  $1/r \to x$. Why do that? What does it mean?
The usual function $F = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$ tells you how $F$ changes as distance changes. As $r$ gets big, $F$ gets small.
But suppose you wanted a measure of "closeness" instead of "distance". $F$ gets big as the charges get close together. One way to do that is to use $1/r$ as a measure of closeness. Inventing the terminology $x = 1/r$ for closeness, we get
$$F = \frac{q_1q_2}{4\pi\epsilon_0}x^2$$
Now you can see exactly how $F$ gets big as closeness gets big. The mental shift can be confusing, but sometimes you learn things by transforming variables like that.
A: $$F = k_e\frac{q.Q}{r^2} \rightarrow (\frac{1}{r})^2 = \frac{F}{k_eq.Q} \rightarrow x^2 = 4ay$$
where $x = 1/r$ and $4a = 1/(k_e.qQ)$
Hope this helps.
