Electric fields, Voltage, and Current

The first phenomenon we need to consider is how charged particles behave in electrical fields. We'll describe the classical, continuum-level (not talking about the structural details of the material) response in this section, and then later move into the influence that material structure can have.

Electric Fields

Charged particles experience a force, $F$, when in the presence electric field $\mathcal{E}$. A field is a region of space in which we can observe forces. The electric field a charged particle feels can be produced by other nearby charges or by moving magnets - but at this point we will ignore the source of the electric field and simply assume one exists.

Use NetLogo model 13.3.1 below to answer some simple questions about charges in electric fields. The simulation allows you to change the electric field direction and magnitude electric-field and the charge on the particle charge-of-particle Play around with those values, observing how the charges accelerate in the field, and answer the questions in Exercise 13.3.1.

Exercise 13.3.1: Electric Fields
Due: Fri, Oct 18, 12:30 PM

  1. Which type of particle (positively or negatively charged) moves in the direction of the electric field and which moves in the opposite direction?

  2. How does changing the electric field strength and the charge of the particle affect how quickly the particle accelerates (and its final speed upon leaving the modeled system)?

Voltage (Electric Potential Difference)

If there are forces, we can also talk about energy. When a charge is accelerated by an electric field, the field is doing work on the particle, changing its kinetic energy. Work is defined as $W=F \cdot d$, the dot product of the force over the distance that it is applied. You think of this as: each little bit of distance that the particle moves, the field pushes on it a little bit, changing the kinetic energy.

Wherever the particle is in the field, there is potential energy stored in the interaction between the particle and the field (although it might be zero). Just as there is gravitational potential energy associated with mass in a gravitational field, there is electrical potential energy associated with a charge in an electric field. Figure 13.3.1 below illustrates these two analogous types of potential energy:

  • In the case of gravity, the potential energy is equal to the mass of the particle times the force of gravity (the magnitude of the gravitational field) times the height from the ground: $U_g=mgh$. Note, that the height determines the distance of gravitational field the mass can move through. The greater the distance, the more work the field will do, since $W=F \cdot d$.
  • In the case of a charged in an electric field, the potential energy is equal to the charge times the force of the electric field times some distance the charge will move through: $U_e=q\mathcal{E}L$. The distance, $L$ is analogous to the height for gravitational potential energy.

It is useful to be able to talk electrical potential independently of the magnitude of the charge in an electric field. We do this by dividing potential energy by a unit of charge, and we call it a Voltage:

$$V=\frac{U_e}{q}=\frac{q\mathcal{E}L}{q}=\mathcal{E}L \tag{13.3.1}$$
(a) potential energy of a mass due to a gravitational field. (b) potential energy of a charge due to an electric field.

Figure 13.3.1 (a) potential energy of a mass due to a gravitational field. (b) potential energy of a charge due to an electric field.

Voltage and Current

If we set up a voltage (an electric potential difference) across a metal wire, electrons will flow through it, because they feel a force from the electric field. Current, $I$ is defined as the rate of flow of charge through the wire, and has units of amperes in charge per second: $\text{A} = \text{C}/\text{s}$.

If you change the voltage in a simple conducting metal wire and measure the current, you will find that they are proportional:

$$I \propto V$$

The standard constant used to turn this proportionality into an equation is resistance, $R$: $$I = \frac{V}{R}$$.

You'll often see the equation written $V=IR$, but it is probably more useful to think of it as $I = \frac{V}{R}$, because the voltage causes the current flow, and increasing the resistance will decrease the current.

Alternatively, we sometimes talk about the conductance, $G$, of a material which is just the inverse of resistance: $G=\frac{1}{R}$. Using conductance, the relation of voltage and current becomes $I=GV$.

The flow of current through a potential difference (a voltage) is analogous to the flow of water downhill in a pipe as shown in Figure 13.3.2. Increasing the voltage across the two ends of the wire causes charges to flow faster, thereby increasing the current.

The relationship between the flow of current voltage is analogous to the the relationship between the flow of water in a pipe and how steeply the pipe is tilted.

Figure 13.3.2 The relationship between the flow of current voltage is analogous to the the relationship between the flow of water in a pipe and how steeply the pipe is tilted.