Potential vs. potential energy – a subtle difference

Electric potential energy is, like gravitational PE, an energy of position, but it also depends upon the charge of the particle in question, so it's not quite the potential energy we're used to. Thus, we refer to it as "potential," and later usually as "voltage."

Electric potential is defined as the potential energy of a particle divided by its charge. Therefore a +2 Coulomb (C) charged particle at one location in an electric field has half the potential as a +1 C particle at the same location.

$$V = \frac{PE}{Q}$$

We define potential that way because the magnitude of the charge also influences the potential energy.

Digression: How things move

Nothing moves in the universe unless one of two things are true:

1. An object is acted upon by a contact force, like a foot kicking a soccer ball (notice the distortion of that ball — that's cool) or


2. An object is subject to a gradient in an invisible force field, like a gravitational field or an electric field.

In the upper photo, the ball will experience a momentary force, but after it loses contact with the foot, it can receive no more force from it, and it continues on only under the forces of gravity and air friction.

Other forces work at a distance, like gravity or electrostatic attraction or repulsion. Think about the attraction or repulsion between two magnets. They don't have to touch to exert forces on each other, but we can say that there's a force field between them. However, there can be no movement unless the forces acting on an object are unbalanced – greater or smaller in one direction than another.

In the case of the skydiver, he falls because the gravitational force is weaker above him than below. That is, he follows the slope or gradient of gravitational strength as he falls. If the skydiver were the same distance from two planets with equal mass (called the Lagrangian point), there would be no net force on him (the pulling forces would be balanced) and he wouldn't move.

Movement in an electric field

Consider a simple electric field formed by charging two parallel metal plates, one positively and the other negatively, as shown below:

Now we take a "test charge," typically a particle of charge +1, and move it around in the field. The test charge is a thinking device – a handy theoretical way to think about motion and forces in fields.

Now if we move the test charge toward the positive side of the field, it will feel repulsion from the positive plate and attraction to the negative plate, so this will be an "uphill" trip, requiring work to be done.

The work will be equal to the potential energy gained, but recall that the force of moving any charge will be multiplied by the charge of any particle.

Ground

Potential is a relative term – potential compared to what? When we talk about the electric potential we're always talking about the potential difference between two points, separated by some distance.

Often, the point of comparison is "ground." We view Earth as an infinite source or sink (willing acceptor) of electrons, and its potential is stable. We refer all electrical activity to Earth or "ground" as being of zero potential, or V = 0.

The British call ground "Earth" and speak of whether an electric circuit is "Earthed" rather than "grounded."

We compare most potentials to ground. In an American home, power outlets operate at a potential of about 110 Volts above the ground potential. Homes generally have two sources of potential relative to ground, ±110 V, and we can also tap the potential across these two, for a total potential difference of 220 V.

Home wiring and grounding

A typical U.S. home electrical service box

Fields

Here is another look at electric fields and how they create potential differences. Consider the two illustrations below, a positive charge on the left and a negative charge on the right. The set of circles surrounding each are a kind of topographic map describing the force that another charge would feel if placed a certain distance from the central charge. The closer the lines, the greater the force, just like how the steepness of a mountain or valley slope is shown on a topographic map.

Here is a 3-D view of the same two scenarios. Now the strength of the attractive or repulsive force is modeled by the height of the circles. Positive charges are drawn into a potential well created by the negative charge, and they fall down a "hill". The closer a test charge is to the negative charge, the faster it accelerates toward it because the Coulomb attraction scales as the inverse square of the distance between them (Coulomb's law).

For charges all of this behavior is modeled by Coulomb's law, an inverse-square law analogous to the universal law of gravitation. Here's a comparison of the two laws, two of the most important relationships in physics:

It's worth comparing these two force laws:

• The constants G and k, the universal gravitation constant (G = 6.674 x 10^-11 N·m²Kg^-2) and Coulomb's constant (k = 8.98 x 109 N·m²C^-2) respectively, are proportionality constants, and you can really think of them as just being there to get the units right.


• In the ULG, we multiply masses; in Coulomb's law it's charges. Notice that mass is always a positive quantity, while charge can be negative or positive. That means that the force between charges can be negative (by convention that's attractive) or positive (repulsive). Recall that like charges repel and opposite charges attract. On the other hand, no one ever suddenly gets ejected from the surface of a planet because gravity becomes repulsive. It's only attractive.

Electric fields – dipoles

We just need to do a little more thinking about electrostatic forces and electric fields before we can really understand electric potential.

Here is a rendering of the electric field produced by a positive and negative charge moved close enough to affect one another. The arrangement is called a dipole — two "poles," positive and negative, like the north and south poles on a bar magnet.

The blue lines are called electric field lines, or just field lines. They indicate the direction of the electrostatic force that would be experienced by a single hypothetical positive charge called a test charge.

In the lower figure, three locations of our test charge are shown along with vectors representing the forces applied to them. The red vectors represent the repulsive force of the positive pole. Note that the red vectors are longer (representing a stronger force) because the test charge is closer to the positive pole, and they always point away from it. Likewise, the black vectors are the attractive forces due to the negative pole. The blue vector is the sum of these for each location, the net force on the test charge. These net force vectors form the field lines an give us their direction.

We use the test charge method to map out all electric fields, and we draw in the resulting field lines so that they're close together where the force is high, farther apart where it is low.

Gradients

Finally, let's look at a 3D view of that dipole we discussed above. The bottom axis represents the position of the test charge along the axis connecting the two charges. The right axis represents position at a 90˚ angle to that, and on the vertical axis we plot the force; up is repulsive, and down is attractive. This plot makes clear the force felt by our test charge and how it would move if we placed it somewhere and let go.

Force field gradients accelerate particles. If there is no gradient, there is no net force, so there can be no acceleration. The steeper the gradient, the greater the acceleration.

"Voltage" ... or how lousy language persists

This is more of a peeve of mine than anything else: I don't like how we append the suffix "age" onto words to make them mean what can be better expressed by another perfectly fine word. When we say "voltage" we really mean electric potential. When we say "square footage" we really mean "area", and when say "percentage" we really mean “fraction” (which can be expressed as percent).

Oh, and never, ever say "amperage" when you mean current or "ohmage" when you mean resistance. That's strictly bush-league.

Fields and calculus

The force on a particle (particle is a generic word for "object" in physics) in a force field is the slope of the field or potential energy function in some direction. When potential energy functions get more complicated, like the mock-2D potential below, we generalize that concept of slope to the gradient, slope that has to be specified by more than one direction.

For a function f(x) of one variable, the slope is given by

$$\frac{df}{dx}$$

For functions of two variables, we use the gradient operator, given the symbol ∇ ("nabla"), often read "grad".

$$\nabla f(x, y) = \frac{df}{dx} \hat{i} + \frac{df}{dx} \hat{j}$$

Ohm's law and the flow of current

In most applications of electricity, what we're really interested in is regulating the flow of electric current to do useful things. We'd like to make the flow of current predictable and to be able to manipulate it.

It turns out the the flow of current through a material is directly proportional to the potential, and it's inversely proportional to the resistance of the material to the flow of current. That's known as Ohm's law, and it's written like this:

Ohm's law is one of the most important relationships in all of the field of electricity and magnetism.

$$I = \frac{V}{R}$$

Example 1

How much work is required to move a -2.0 μC charge from ground (0.0 V) to a position where the potential is +60V ?

Solution: We know that potential is the amount of work done (in Joules) divided by the amount of charge (in Coulombs) being moved. We can rearrange that to calculate the work:

$$V = \frac{w}{Q} \; \longrightarrow \; w = VQ$$

Example 2

The work to move a charge of 1.0 μC from point A to point B is 4.5 × 10^-4 J. If the particle started from rest and has kinetic energy, KE = 1.2 × 10^-4 J at point B, what must the potential difference be between point A and point B ?

Solution: First, we'll just calculate the potential from the relationship w = QV:

$$ \begin{align} V &= \frac{w}{Q} = \frac{4.5 \times 10^{-4} \; J}{1 \times 10^{-6} \; J} \\[5pt] &= 450 \; V \end{align}$$

Now we actually did more work than necessary, because we accelerated the particle so that it has excess kinetic energy, KE = 1.2 x 10^-4 J.