Now imagine the positive charge $q_O$ is initially at a location where the electric potential is 8 Volts. That means it has an electric potential energy at that location is $U_{initial}=(8 volts)(q_O)$, where $q_O$ is in SI units of Coulombs. If the positive charge $q_0$ was to move though the displacement $\Delta \overrightarrow{r}$ to the final location, it will have decreased in electric potential energy. The electric potential at the final location is 4 Volts and thus the electric potential energy is $U_{final}=(4 volts)(q_O)$. The potential energy has decreased and so, barring no other interactions, the kinetic energy must increase. This does lead to an important conclusion.

Electric Potential from Point Charges

Most charge distributions are not that of point charge but they are comprised of the electric potential from each individual particle making up the charge distribution. Think a charged plate is sheet of point charges. To understand how a collection of charges creates a field, lets first look at how each individual point particle creates a field. The electric potential of a point charge $q$ is equal to:

$V=\frac{kq}{r}$, where r is the distance from the point charge and k is a constant.

The first thing to note is that the electric potential $V$ decreases like $\frac{1}{r}$ as opposed to the electric field magnitude that falls off like $\frac{1}{r^2}$. The potential decreases more slowly than the electric field strength. This can be seen by comparing the nature of each function in the graph below where $\frac{1}{r}$ gets smaller more rapidly than $\frac{1}{r^2}$.

Another difference is the electric field is a vector field while the electric potential is a scalar field. That is very good news when it comes to calculations as it's much easier to deal with a bunch of numbers than it is to deal with a bunch of vectors - the mathematics are more simple. The below equation is more rigorous than the one above and it reads: the electric potential at point p is equal to the constant $k = \frac{1}{4\pi \epsilon{0}}$, multiplied by the charge $q$, and divided by the absolute value of the displacement between the charge $q$ and the point p.

When performing calculations of the field you get a number, positive for positive charges and negative for negative charges, for every point in space. The graph of the electric potential from a positive charge is the $\frac{1}{r}$ above while the negative sign for a negative makes the graph for a negative charge as a function of distance proportional to $-\frac{1}{r}$. Below is a graph of $-\frac{1}{r}$ to show the nature of the function.

Electric Potential for Charge Distributions

The electric potential field from a single charged particle $q$ is above. To find the field from a group of charges you can simply add up the contributions from each individual particle at the location of interest. Potentials add through the principle of superposition, which means you just add them up with no other scaling factors.

$V_{total}(p) = \Sigma V_i(p) = V_1(p) + V_2(p) + ...$

An example of a charge distribution is a set of parallel charged plates. Take two metal sheets and charge one of them positively and the other negatively. This creates an electric potential between them, after all if you put an electron or proton in between them they would both speed up, albeit in opposite direction. The 2-D figure below depicts the voltage (electric potential) in the region between the two plates. The 3-D graph uses the third (vertical) axis to describe the voltage, whereas 2-D represents it as a number.

Both figures show the equipotentials, with the 3-D projecting them onto the floor of the graph.

Key Equations and Infographics

Now, take a look at the pre-lecture reading and videos below.