Electrostatics – Why Is Coulomb’s Force Law a 1/r^2 Dependence While Coulomb Potential Is 1/r?

coulombs-lawelectrostaticsforcespotentialpotential energy

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?

Best Answer

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 $r\equiv |\vec x|$.