# Graph for Coulomb force vs $1/r^2$

We know that when there is a relation like x inversely proportional to y^2, the graph would be like this:

But in this question"Plot a graph showing variation of Coulomb force(F) versus $1/r^2$, where $r$ is distance between two charges of each pair of charges($1\,\mu C$, $2\,\mu C$) and ($1\,\mu C$, $-3\,\mu C$). Interpret the graphs obtained.".. The answer in my book is given in a strange way like this:

I don't think this is a wrong answer as I have checked it in many books. So, why did they do it like that (that graph in answer looks like x directly proportional to y)?

• x directly proportional to y but x = 1/r² , see the plot.
– user46925
Commented Feb 23, 2016 at 18:24
• And you might ask, why on earth would you do this? Well, plot $y = 1/x$, $y=1/x^{2.5}$, $y=7/x^{8}$, etc. It's hard to eyeball which one is which, right? But straight lines are easy to eyeball. Commented Feb 23, 2016 at 18:30

Coulomb force is inversely proportional to the distance squared:

$$F = \frac{k}{r^2}$$

So if you plot $F$ versus $r$, as $r \rightarrow \infty$, $F \rightarrow 0$ (like the first image).

In the second image, $F$ is drawn versus $\frac{1}{r^2}$. As $r \rightarrow \infty$, $r^{-2} \rightarrow 0$ and $F \rightarrow 0$. "But why there are linear", you ask.

Because if you plot $y$ versus $x$ according to the equation $y = mx$, you'll have a line with slope $m$. So if you plot $F$ versus $r^{-2}$, you'll have a line with slope $k$.

\begin{align} y &= mx \\ &\updownarrow \\ F &= k r^{-2} \end{align}

• Right...so basically the graph can be plotted in either way(like 1st image as well as 2nd image)..But just bcoz we are comparing the slope of two graphs we draw the graph like the 2nd image...Am i right,sir? Commented Feb 23, 2016 at 15:42
• @hungry_dude: Yes, working with linear graphs is easier and it gives an immediate impression of the result. If you plot them in the first way, it is really hard to tell the difference especially at large $r$. Commented Feb 23, 2016 at 17:56

A graph between $x$ and $y$ is different from that of $x$ and $(1/y)$.

Simply let $x$ and $y$ be related as follows:

$$y=k/x$$ where $k$ is a constant. You can see that $x$ is indirectly proportional to $y$. Now cross multiply and turn the equation into $(1/y)=(x/k)$ and consider $1/y$ as one variable $z$ (say). As a result the equation becomes $z=x/k$ where you can clearly see that $z$ is directly (not indirectly) proportional to $x$. So it is a straight line because $z$ is nothing but $1/y$.