Relationship between Potential and Potential energy

Question

I know that if we have a conservative force then: $\vec{F}(r)=-\vec{\nabla}V(r)$ where $V(r)$ is the potential. I also know that I can take potential energy from here doing: $E_{pot} (r) = \int_{S} \vec{F}\cdot d\vec{r}$, where $S$ is just a path.

Ok, but I don't understand what is the relationship between potential and potential energy.

And also: we know that electric potential is given by $V(r)=\frac{k_e Q}{r}$ so we can obtain the force using $\vec{F}(r)=-\vec{\nabla}V(r)$ it comes that $\vec{F}(r)=\frac{k_e Q}{r^2}\hat{r}$ which is wrong by Coulomb's law...what is wrong in my thinking?

Answer 1

In mechanics, we love energy: indeed, it often really simplifies solving problems. Often, we want to know how the speed of a particle varies, or, in other words, what its kinetic energy is. Luckily, for some forces, knowing the variation in kinetic energy is really easy: these forces are the conservative forces. A force $\textbf{F}$ is conservative when its work along a path $\Gamma$ only depends on the initial and final positions: then, we can introduce a potential energy, say $E_p$, so that

$$ W(\textbf{F}) = \int_\Gamma \textbf{F}\cdot\textbf{dl} = E_{pi} - E_{pf}$$

However, the kinetic energy theorem states that the total work done on a particle equals its kinetic energy variation, so

$$ \Delta KE = W(\textrm{F}) = -\Delta E_p $$

Now, let's focus on Coulomb's law. Consider a charge $Q$ in $P$, and another one, say $q$ in $M$. It appears that $Q$ creates a force $\textbf{F}_e$ on $q$, with

$$\textbf{F}_e = \frac{Qq}{4\pi\epsilon \|\textbf{PM}\|^3}\textbf{PM}$$

Since $\textbf{F}_e$ only depends on the position of $P$ and $M$, it is derived from a potential energy $E_p$, and we find that

$$ E_p = \frac{Qq}{4\pi\epsilon\|\textbf{PM}\|}$$

However, even if the notation is usefull for a few particles, it really becomes difficul when studying a complex distribution of charges. Then, we prefer to see a single charge as the source of an electric field $\textbf{E}$, which exists everywhere, with $$\textbf{E}(M) = \frac{Q}{4\pi\epsilon \|\textbf{PM}\|^3}\textbf{PM}$$

Now, multiplying $\textbf{E}$ by the charge $q$ gives the electric force created by $Q$. So, $\textbf{E}$ and $\textbf{F}_e$ are almost the same: we introduce $U$ which is the potential created by $Q$, so that $E_p = qU$.

Answer 2

$ - \nabla V$ gives you, in fact. The electric field, not the electric force. In order to obtain electric force you must do $ \vec{F} = q\vec{E} $, and that will eventually lead to coulombs law. The relation between electric potential and electric energy comes from the charge $q$. For example: Let $q$ be a charge in an electric field that will follow a trajectory $ C$. We define the work done by the electric field as $ W = qV$, where $V$ is the difference between the initial electric potential and final electric potential (a proof of this formula can be found in any college-level electromagnetism book). So, roughly, one can say that the electric potential at a point P is the amount of work it takes to move a charge $q$ from infinity to the point P (assuming $ V(\infty) = 0 $). It is something like "amount of work per unity of charge".

Answer 3

The relationship between potential $V$ and potential energy $U$ is:

$$V= \frac Uq$$

Talking electricity, $q$ is charge. Talking gravity, $q$ would instead be mass $m$. Similar for chemical, magnetic and other types of potentials.

The potential is simply potential energy per something. For example potential energy per unit of charge, or per unit of mass etc.