# 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. – dmckee Jul 1 '18 at 20:23
• Yes, I agree, but I doubt if this is what was worrying our questioner! – Philip Wood Jul 1 '18 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. – Alfred Centauri Jul 2 '18 at 2:42

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.