Why is Coulomb's force law a $1/r^2$ dependence, while the Coulomb potential has a $1/r$ potential? My source of confusion comes from reading the following except from my general chemistry textbook, 
"The energy of interaction between a pair of ions can be calculated using Coulomb's law"
$$V = \frac{k Q_1 Q_2}{r}$$
Now, as I learned coulumbs law in general physics, the quantity is (scalar form),
$$F= \frac{kqq}{r^2}$$
How can I rationalize the difference here? Why is potential energy $1/r$ dependence, but Coulomb's law a $1/r^2$ dependence?
 A: The force (vector) is given by the gradient of the potential energy:
$$
\vec F = -\vec\nabla V.
$$
Even without calculating the gradient explicitly, dimensional analysis (that is, consistency of the units) implies
$$
|\vec F|\propto \frac{V}{r}
$$
with an $r$-indepednent coefficient, because the gradient $\vec \nabla$ has the same units as $1/r$. Therefore, $V\propto 1/r$ implies $|\vec F|\propto 1/r^2$. This explains the difference. If desired, the details can be filled in using the identity
$$
\vec\nabla r = \frac{\vec x}{r}
$$
with $r\equiv |\vec x|$.
A: The voltage is the work done per unit charge so
$$
V=\frac{1}{q}\int \vec F \cdot  d\vec r
$$
so if $F\sim ~1/r^2$ then $V\sim \int \frac{dr}{r^2}\sim 1/r$.
Note that, in the cylindrical geometry - say an infinite line of charge -  $F\sim 1/\rho$ so $V\sim \int \frac{d\rho}{\rho}\sim \log(\rho)$, which brings up interesting when one integrates to $0$ or from $\infty$.
A: It is simply because they are different concepts. One is a force and the other is potential energy. Take, for example, a spring where $F=-kx$ but $U=-\frac{1}{2}kx^2$. They are related, in the one dimensional case by  $-\frac{d U}{d x}=F$. This becomes the gradient in three dimensions. If this doesn't seem to make sense to you, consider learning mechanics and not looking deeper into electromagnetism.
