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When I've been asked to find the relationship between two variables in a formula (assuming the other variables are constant), it's generally been things like find the relationship between $F$ and $R$ in $F=\frac{kQq}{R^2}$, where the relationship is easily identifiable as $F\propto \frac{1}{R^2}$.

However, what if I'm asked to find the relationship between two variables on the same side of the equals sign? For example $Q$ and $R$. Would I do something like re-writing the formula to be in terms of $Q$ or $R$? For example I could re-write it to be $Q=\frac{FR^2}{kq}$, which then would tell me that $Q\propto R^2$.

Is this the right way to do it, or is there some other method?

I'm sorry if the tag is incorrect, I didn't know which to use.

Harry Peter
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Threethumb
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2 Answers2

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Be aware that the symbol $\propto$ implies that all other variables are being held constant. This is crucial. For example, if you use $F=\frac{kQq}{R^2}$ and then write $F \propto \frac{1}{R^2}$, you mean that force is inversely proportional to the distance squared if the charges $Q$ and $q$ are constant. The process that corresponds to this process is actually useful: you're using the same two point charges $q$ and $Q$.

If you want to find a proportionality between $Q$ and $R^2$, you write $Q=\frac{FR^2}{kq}$ and then $Q\propto R^2$, just as you did. Again, by writing this, it is implied that all other variables are being help constant. For this case, it means you are holding the force between the two charges constant and one of the charges ($q$) constant. Now what process would correspond to? Looks like you'd want the same force exerted on your charge $q$ as you vary the distance $R$, and you're seeing how the required charge $Q$ is affected by changing the distance $R$.

Why is it so crucial to keep track of what's constant? Well, we could write $F\propto Q$ from the force relation. Afterward, we could then write $Q\propto R^2$ after rearranging. (Try it.) Then it seems like we might write $F\propto Q \propto R^2$ which might simplify to $F\propto R^2$. This is wrong, as you can see from comparing your original proportionality between $F$ and $R^2$.

BMS
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  • Ah, yes, I realize the other variables have to be constant. One follow-up question, though. Would the graph for the relationship $F\propto \frac{1}{R^2}$ be an upside down exponential one, if $F\propto R^2$ is this: http://puu.sh/6Ozcx.jpg. Or rather, would there be any difference between the graph of $F\propto \frac{1}{R^2}$ and this one: http://puu.sh/6OzgI.jpg ? – Threethumb Feb 08 '14 at 18:04
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This is a fine approach. Another way to think about it (of course, with the same result) is that if you are looking for the relationship between $Q$ and $R$, everything else is a constant. This gives $\frac Q{R^2} \propto $ constant, which also tells you $Q \propto R^{-2}$

Ross Millikan
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