# How is voltage proportional to current?

The question came up in another forum - "What causes current to flow". Many knowledgeable people explained that what causes the flow of charge are electric field forces or Coulomb forces. If there is a distribution of charge such that there is a deficit of neg. charge in one place and an excess in another place, the neg. charge will toward the deficit, due to the repulsive force acting on the charge at the place of excess.

From what I interpreted from the answers to the question, potential difference was a way of measuring this force that causes charge to flow, since if there is a force acting on that charge, it has the capability to do work, and potential difference is a measure of how much work a charge does as it flows between two points.

However, it seems to me that potential difference would not be a good way to measure this electric force, since work is also proportional to distance.

For example, I could separate some amount of charge and create some voltage. Now if I separate the same amount of charge, but separate it a longer distance, I will create a larger voltage since more work was needed to do this. Now, the instance where I separated the charge a smaller distance will have a lower voltage, but a higher force acting on that charge according to coulomb's law, and the Instance where I separated the charge a longer distance will have a bigger voltage, but a smaller force acting on it.

How is potential difference an accurate measure of the force that causes charge to flow, if we can manipulate it by simply changing the distance the charge is separated?

• The introduction of non-zero curl in the electric field in electrodynamics really throws a spanner in any attempt to understand potential as fundamental. See physics.stackexchange.com/questions/75349/… and links therein. Jul 1, 2018 at 20:23
• Yes, I agree, but I doubt if this is what was worrying our questioner! Jul 1, 2018 at 22:23
• "How is voltage proportional to current?" - unless the current is through an ideal resistor, it just isn't the case that voltage is proportional to current. For example, the voltage across an ideal inductor is proportional to the time rate of change of the current through. Jul 2, 2018 at 2:42

To understand how a current is created we need to understand the dynamics of the charges. To do this, it is easier to calculate how much work the system can do on a charge rather than just the force in the static case. The potential different can be seen as a measure of the system's capacity to do work on a charge. The current is then directly proportional to the systems ability to di work over the resistance that work must overcome.

The work done by the electric field on a charge, q, moving a distance $\Delta x$ in the x direction is $$\text{Work}=F_x\ \Delta x.$$ Dividing both sides of this equation by q we have $$\frac{\text{Work}}{q}=\frac{F_x\ \Delta x}{q}.$$ The left hand side is, by definition the potential drop ($-\Delta V$) over $\Delta x$, so we can write $$\Delta V=-\frac{F_x\ \Delta x}{q}$$Dividing both sides by $\Delta x$,

$$\frac{\Delta V}{\Delta x}=-\frac{F_x}{q}.$$ So it's not the potential difference, $\Delta V$, that gives the force (per unit charge) on the charge, but $\frac{\Delta V}{\Delta x}$, which is called the potential gradient.

The last equation is usually written as$$\frac{\Delta V}{\Delta x}=-E_x\ \ \ \ \ \text{or better still}\ \ \ \ \frac{\partial V}{\partial x}=-E_x$$in which $E_x$ is the x component of the electric field strength, defined by $E_x=\frac{F_x}{q}$, the x component of the force per unit charge on a charge placed in the field.

• Doesn't your first step assume that the force applied to the unit charge as it flows through the electric field is constant, when in fact it isn't, according to coulomb's law? Also, if I'm correct in interpreting your answer, you're saying that potential difference isn't a measure of the force acting on the charge, but rather "Change in V" / "Distance" is. So how then is potential difference proportional to current in an ideal resistor? Jul 2, 2018 at 4:41
• (1) "Doesn't your first step assume that the force applied to the unit charge as it flows through the electric field is constant, when in fact it isn't, according to coulomb's law?" We're ignoring the complicated fields that exist on an atomic scale in the conductor. On a macroscopic scale the field in a uniform conductor with a pd between its ends is uniform. (2) "pd isn't a measure of the force acting on the charge, but rather "Change in V" / "Distance" is. So how then is pd proportional to current in an ideal resistor? " Because the distance from end to end is constant! Jul 2, 2018 at 7:02

The question is very curious to my older generation who were brought up as kids imagining voltage in a circuit like pressure. However the modern syllabus strongly emphasizes voltage as energy or work per coulomb, which can seem less intuitive for some kind of viewpoints. Also voltage fields in empty space from a set of coulomb charges versus voltages around a circuit are extreme cases with their own special kind of simplifications with which it seems we need to be careful to avoid confusion. Eg. In a circuit loop the conduction electrons move like an incompressible fluid and so the total voltage does count overall even though local voltage drops do relate to local currents too. In the Coulomb apparent “paradox” identified by the questioner, an in-falling electron at the higher voltage receives less force BUT when it falls in it will eventually experience the higher force at the lower voltage. If this somehow be part of a circuit, the incompressible electric fluid would get all contributions working together simultaneously giving overall effect dependent on overall voltage. Perhaps even simpler argument slightly modifying a previous answer is to observe that the voltage drop is related to the spatially averaged force multiplied by the distance, so bigger voltage gives more current, rather heuristically for the same distance or length of conductor.

Voltage is proportional to current only in resistive circuits, in a different situation such as a circuit with a capacitor there will be no current at all between its plates if the is a source of continuous voltage.

For a resistive circuit with a source of continuous voltage, the same current must run through every section of the circuit, this means that ΔV at each section of the circuit will be proportional to its resistivity so that the potential lost in the resistance equals the potential provided by the source, V_o: $$\int \triangledown V dl = -V_o$$

Where V_o is the source voltage.

As pointed out by Blanci, the current in the circuit behaves like a fluid where the potential is the pressure and therefore the current is proportional to the potential and inversely proportional to the resistance. The more potential, the bigger will be the pressure ΔV and the more current will move.

Because the work done on the current, as it crosses the resistance, is also proportional to ΔV·J, the power done on the circuit will be:

$$\int \textbf{J}·\triangledown V dl = V_o*I =V_o^2/R$$