# Electric Potential

Electric potential energy U is potential energy (similar to gravitational potential energy) that is assigned to a system of charged particles when conservative electrostatic force acts between two or more particles within that system. Electric potential energy, as other types of energy, has units of Joules, J. If one of the particles moves from an initial position i to a final position f, a potential energy difference results:
$\Delta U=U_f-U_i=-W$ (1)

Recall that
$W=F\cdot d$

For convenience, it is assumed that the particle is initially brought in from infinity resulting in the initial potential energy of zero. This reduces the previous equation to:
$U=-W_\infty$ (2)

Electric potential V is a measure of electric potential energy per unit charge, and has the units of Volts (1V = 1J / 1C). It is defined as:

$V=\frac{U}{q} (3)$
It is important to keep in mind that electric potential is a scalar and not a vector quantity.

The electric potential difference is defined by:
$\Delta V=V_f-V_i=\frac{\Delta U}{q}$ (4)

Since elementary charges are very small, they have very small energy levels associated with them. We assign a special unit called the electron-volt (eV), to measure the work required to move a single elementary charge e through a potential difference of 1 V: 1 eV = 1.60 * 10^-19 J.

When observing a map of the electric field of a certain arrangement of particles, adjacent points with the same electric potential form an equipotential surface.

 Red dashed lines represent equipotential surfaces; black solid lines represent electric field lines.
 Blue lines represent equipotential surfaces; red lines represent electric field lines.

Electric potential difference can be calculated from the electric field and the path that a particle follows in the field. For a definite path between an initial point and a final point:
$V=-\int_i^f \overrightarrow{E}\cdot \overrightarrow{ds}$ (5)

where the initial point is taken to be infinitely far away. Potential due to a point charge is defined by:
$V=\frac{1}{4\pi \epsilon_0}\frac{q}{r}$ (6)

Note that the potential depends only on the charge that creates the electric field and the distance from that charge. For a group of point charges, the resulting potential is the summation of all the particles in question, where the sign of the charge directly corresponds to the sign of the resulting potential.

An electric dipole is a configuration of a positive charge and a negative charge a distance d apart:
$V=\frac{1}{4\pi \epsilon_0}\frac{p\cos \theta}{r^2},\ p=qd$ (7)

where p is the magnitude of the electric dipole moment, a vector pointing from the negative charge to the positive charge.

 (a) An electric dipole. The electric field vectors E(+) and E(-) at point P on the dipole axis result from the dipole's two charges. Point P is at distances r(+) and r(-) from the individual charges that make up the dipole. (b) The dipole moment p of the dipole points from the negative charge to the positive charge.

For discrete charge distributions one could use the method of summation, but in the continuous case the integration saves the day. For a continuous line of charge we consider a differential length element dx on the line with a differential charge element dq associated with it and equal to the linear charge density times the differential length element (dq = lambda*dx). The general equation is then
$V=\frac{1}{4\pi \epsilon_0}\int \frac{dq}{r}$ (8)

For a disk with continuous charge distribution we derive a similar result. This time, however, we deal with surface charge density instead of linear charge density (dq=s(2pr)(dr)). The general equation is then
$V=\frac{\sigma}{2\epsilon_0}\int_0^R \frac{rdr}{\sqrt{z^2+r^2}}=\frac{\sigma}{2\epsilon_0}(\sqrt{z^2+R^2}-z)$ (9)

 A nonconducting disk of radius R, charged on its top surface to a uniform charge density sigma. We wish to find the potential V at point P on the central axis of the disk at distance z.

Previously we found how to calculate the potential from the electric field, so now we can explore finding the electric field from the potential. We start with equating the two expressions for work:
$-q_0dV=q_0E(\cos \theta)ds,\ E\cos \theta=-\frac{dV}{ds},\ E_s=-\frac{\partial V}{\partial s}$

This states that the component of E in any direction is the negative of the rate at which the electric potential changes with distance in that direction. Therefore, if the potential is given as a function of x, y and z V (x,y,z) - we can take the partial derivative with respect to x, y and z to find the electric field at a specified point.

Finding the total electric potential energy of a system of point charges is similar to the process of finding the potential of a group of charges. Using a superposition principle, consider the work needed to bring each individual charge to its position as if it were the only one in space. For instance, take four point charges. Pick one charge to be the reference charge. This first one is "free" of work because there is no other charge for it to interact with. Now take the second charge and compute the work needed to bring this charge in contact with the first one. Repeat this for each successive charge and sum to find the total electric potential energy of the system.
$U=W=q_2V=\frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r}$ (10)

Consider a charged isolated conductor. The charge distributes itself along the outer surface of the conductor. Since the electric field is zero at all points inside this conductor, all possible points inside have the same potential:
$V_f=V_i$

Example 1:

Example 2:
(a) What is the electric potential energy of two electrons separated by 2.00 nm?
(b) If the separation increases, does potential energy increase or decrease?

Example 3:
What is the excess charge on a conducting sphere of radius r = 0.15m if the potential of the sphere is 1500V and V=0 at infinity?

And now that you've read the chapter, here are a couple of tasty treats for you: