0
$\begingroup$

Note: For the below question: I am ignoring exotic circumstances such as superconductors.

My Problem: I came across this question regrading the ability to have a voltage without having current. Current without Voltage and Voltage without Current? The accepted answer in the above question stated the following:

“for instance if you have a single charge, that charge induces a voltage in space, even if it's empty”.

This statement implies that there is always voltage if there is a non-zero charge (Coulombs). In all, the answer above essentially stated you can have voltage but no current within a system. In addition I have read many post online also stating that you can absolutely have voltage without current. Below I list two reasons why I think you must have current if you have voltage.

Reason 1: The derived SI units of Voltage are Joules/Coulmbs. We also know that a Joule is work done by a force to create displacement on an object. The work done (Joules) can be calculated by $$Joules = F * Cos(θ)* s$$ where F = force, s = displacement, θ = angle between force and direction of displacement.

So basically, from the SI units, we can see that voltage is simply the work done per Coulomb (per charge). The above statement said the voltage would be induced in empty space but how could work be done in empty space if there is nothing to move(nothing to displace) in the empty space?

My point is how can voltage be present without current flow when the very definition of voltage requires displacement of electrons (work done per coulomb). To prove my point, lets calculate the voltage of an ideal isolated insulator in free space where it has a net negative charge. I am using the example of an ideal insulator so the electrons cannot move from one valance shell to another in the system, thus no possible electron flow. We know there would be an electrostatic force produced from the net negative charge but since there is no where for the electrons to move in the system, there would be no displacement. Since there is no displacement the work(Joules) done would be zero (via the equation to calculate joules above). So if we assume the system has x Coulombs of net charge, then we could calculate its voltage by the following: $$Voltage = \frac{F*Cos(θ) *0}{X Coulombs}= 0$$ But the moment there is even the smallest amount of displacement of electrons, we will than have both voltage and current since we would have both rate of coulomb flow (Current) and work done per coulomb (Voltage).

Reason 2: (This would only apply to Ohmic materials): If we have a circuit with a constant resistance and we have zero current flow, by Ohms law how could you have a non-zero voltage? $$V=IR$$ $$V=0R$$ $$V=0$$

Summary: I understand how one would think you can have voltage without current when you think of voltage as water pressure. My issue is I don’t think this analogy works when we look at the units of Voltage. Rather in the water analogy for electricity, it seems that water pressure would better correspond to Coulombs, since a charge will have a electrostatic force that will propagate through free space which is independent of the flow of electrons. Where voltage by its units is dependent upon displacement which would mean voltage could only be non-zero when there is electron flow. Thus an isolated net charged object in free space with no electron flow (no displacement) can not have a non-zero voltage.

My Question: Is correct to say from the above argument that you cannot have voltage without current? (Again, excluding superconductors)

$\endgroup$
6
  • 14
    $\begingroup$ The depth of a moat is the distance a person would fall if he jumped into the moat. Does it follow that if there happen to be no people around, the moat has no depth? $\endgroup$
    – WillO
    Commented Sep 8, 2021 at 4:52
  • $\begingroup$ @WillO So this illustration is very similar the water pressure one. The answer to your question is of course no. My point is not those types of illustration don't make sense but rather those illustrations (like water pressure) demonstrate the concept of Coulombs not Volts. Since Volts is dependent upon displacement which in you illustration corresponds to people moving into the moat. Thanks for the comment! $\endgroup$ Commented Sep 8, 2021 at 13:14
  • $\begingroup$ @Cabbage Champion About reason 2: What if the resistance of the material is very large, $R \to \infty$? There is no way to evaluate $0 \cdot \infty$. (It's not necessarily $0$) $\endgroup$ Commented Sep 8, 2021 at 15:26
  • $\begingroup$ Voltage between A and B is the work to be done (or that would be done) per unit charge to move a positive charge from A to B. Since such work depends on the amount of charge that would be transferred, the dependence on the amount of charge cancels. Notice voltage is work to be done, not work being done; so, even if the charge(s) isn’t transferred from A to B (no current), the voltage between A and B is still defined. $\endgroup$
    – alejnavab
    Commented Mar 16, 2022 at 5:07
  • $\begingroup$ Also, forget about Ohm’s law. Ohm’s law applies to conductors, so of course if there’s no voltage then there’s no current, and vice versa. But voltage is defined independently of current and resistance, just like how current is defined independently of voltage and resistance. Suppose I put a single proton (or electron) in empty space. Now imagine two points anywhere near that proton. Certain work has to be done to move a hypothetical proton between those points. By dividing such work by the charge of the hypothetical charge, you get the voltage between those points, and no current is involved. $\endgroup$
    – alejnavab
    Commented Mar 16, 2022 at 5:12

2 Answers 2

3
$\begingroup$

Yes it does make sense. Voltage is a measure of potential. When you say it has units of Joules per coulomb, that does not mean that you cannot have a potential without a charge. It simply means that were a charge to be present at that potential, then it would posses a given PE. Consider the density of gold, say, which is defined in kilograms per cubic metre. Can you say that the density of gold cannot exist as a value where there is no gold?

Likewise, Ohms law defines the voltage drop that occurs when a given current flows through a given resistor, or, conversely, the current that flows when a given resistance is used to connect two regions of differing potential where the potential difference is V. If you don't connect the two regions, you get no current, but the Voltage difference remains.

When you plug a zero current in to Ohms law to say V=0R so V=0 you are positing either a scenario in which there is no voltage (eg a flat battery), or where there is some voltage V but no current because R is infinite.

Electrical potential is analogous to gravitational potential. There is a gravitational potential difference, gh, between the top and bottom of a cliff of height h. Its value can be expressed in Joules/kilogram. That simply means that if you have Z kilograms at the top of the cliff it will have a PE of Zgh. If the mass is zero, the PE is zero but the potential remains gh.

$\endgroup$
2
$\begingroup$

This becomes more clear if you consider the units involved.

Voltage is an effort variable, which can exist in the absence of a flow variable. This is the open circuit condition, as for example when you measure the voltage of a battery with a high-impedance voltmeter (where the current- the flow variable- is zero, or nearly so).

$\endgroup$
2
  • $\begingroup$ You say voltage is a effort variable, but if the displacement is zero isn't the work done zero, thus the voltage would be zero? (Joules/Coulombs) Do I have a misunderstanding of the units? $\endgroup$ Commented Sep 8, 2021 at 13:08
  • $\begingroup$ @CabbageChampion You do. Key point in the definition: you put a "test charge" in that field and it acts upon it. Voltage is a measure of electrostatic potential, "potential" indicating that it can do work but has not necessarily done so. Otherwise, it is akin asking of a falling tree sound when there's no one around to hear it. Voltage says what happens once you put a charge in and the field starts acting upon it. What you are asking gets more convoluted at the QED level ("how do electrons know where to emit photons for them to be absorbed"?), but that's a whole other can of worms. $\endgroup$
    – Lodinn
    Commented Sep 8, 2021 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.