# General relationships of variables in expression

I am asked to determine how certain modifications to the variables in Coulomb's equation will effect the resultant force:

$$F=k\frac{Q_1Q_2}{r^2}$$ The question asks me what will happen with $Q_1$ doubles, and I determine $F$ is doubled. Then I am asked what happens when $Q_1$ and $Q_2$ are doubled, and I get $F$ is quadrupled. But then I am asked what happens when $r^2$ is tripled and I come across some confusion. I calculate $F$ is one third, using $1m$ for $r$ and then solving the equation once for $r=1^2$ and then $r=(1^2)3$, and comparing the resulting values in magnitude. The answer states that $F$ is a ninth.

Additionally, I am asked what will happen to $F$ when $Q_1$ and $Q_2$ are doubled and $r^2$ is halved. I am hoping that someone cant hint me in the right direction.

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First, you shouldn't write things like $F=2F$, when what you really mean is something like "new $F$ equals twice old $F$," because $F=2F$ implies $F=0$. Anyway, you are right about what happens when $r^2$ is tripled; the 1/9 answer is what happens when $r$ is tripled. –  Gerry Myerson Jul 18 '12 at 6:50
Thanks, I have edited the post. –  Kurt Jul 18 '12 at 6:53

Let $F(Q_1,Q_2,r) = k \frac{Q_1 Q_2}{r^2}$.

I'm not sure if you intend for $r$ or $r^2$ to be halved.

If $r$ is halved, then $F(2Q_1, 2Q_2, \frac{1}{2} r) = k \frac{2Q_1 2Q_2}{(\frac{1}{2} r)^2} = k \frac{4Q_1 Q_2}{\frac{1}{4} r^2} = k \frac{16Q_1 Q_2}{r^2} = 16\, F(Q_1,Q_2,r)$.

If $r^2$ is halved, then $F(2Q_1, 2Q_2, \frac{1}{\sqrt{2}} r) = k \frac{2Q_1 2Q_2}{(\frac{1}{\sqrt{2}} r)^2} = k \frac{4Q_1 Q_2}{\frac{1}{2} r^2} = k \frac{8Q_1 Q_2}{r^2} = 8\, F(Q_1,Q_2,r)$.

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Doubling $Q_1$ and $Q_2$ quadruples $F$, as you know. Then halving $r^2$ doubles the force again. The net result is to multiply the force by 8.