Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Voltages are often thought of as continuous physical quantities. I was wondering whether by zooming in a lot, they are discrete.

I feel like the answer to the above question is yes as voltages in the real world are generated by actions of electrons. Can someone give me a more formal proof or a disproof?

Whether voltages are discrete of continuous can have some impact the correctness of devices such as the analog to digital converter. For example, if voltages in the real world are continuous, then the Buridan principle[1] says that there cannot be a perfect analog to digital converter because such a device makes a discrete decision from continuous inputs.

[1] : Lamport, L. (2012). Buridan’s Principle. Found Phys 42, 1056–1066. http://link.springer.com/article/10.1007/s10701-012-9647-7

(It would be great if someone could also answer a related question http://electronics.stackexchange.com/questions/126091/is-there-an-adc-with-a-finite-bound-on-measurement-time)

share|improve this question
Since $e$ is a charge and $k$ is dimensionless (natural number), $k\cdot e$ is still a charge, not an electric potential (voltage). –  Jan Hudec Aug 20 '14 at 6:01
Thanks for pointing that embarassing mistake of mine. I removed that line. –  Abhishek Aug 21 '14 at 16:54

4 Answers 4

up vote 14 down vote accepted

For static charges, the relationship is V (voltage) = Q (charge) / C (capacitance). Capacitance is a function of the shape, size and distance between objects, which are all continuous values. (Well, I suppose you could argue that shape and size are quantized to the atomic spacing of the object's material, but you can't say the same thing for distance.) So even though the charge term is quantized, the capacitance — and therefore, the voltage — is not.

share|improve this answer
Building on Dave's response... You might think, "Aha, but for a given capacitance, voltage must be discrete because charge is discrete." Not even that is true, at least, not always. Usually we can continuously vary voltage because voltage only determines the average number of charges. Or in another sense, the number of charges varies all the time so the voltage fluctuates and its average is continuous. This sort of subtlety (about what is "voltage") appears regularly in statistical mechanics and has real effects in nanoelectronic devices which only contain a few charges. –  Nanite Aug 19 '14 at 18:58
Actually, atomic spacing/distance is quantized: en.wikipedia.org/wiki/Planck_length –  sbell Aug 20 '14 at 13:55
@sbell: Yes, I was wondering if someone would bring that up. But the Planck length doesn't imply a quantization; it simply puts an ultimate limit on our ability to measure position. –  Dave Tweed Aug 20 '14 at 14:30
@DaveTweed We just don't know at this point. From the wiki article: "it is often guessed that spacetime might have a discrete or foamy structure at a Planck length scale." –  sbell Aug 20 '14 at 14:38
@sbell: The article also says, "There is currently no proven physical significance of the Planck length". –  Dave Tweed Aug 20 '14 at 14:44

Voltage is a continuous function. If you are a certain distance from a (point) charge $q$, the potential is

$$V=\frac{q}{4\pi\epsilon_0 r}$$

By adjusting the value of $r$ to anything you want (not quantized), you can get any potential you want. And so yes, when you do any analog-to-digital conversion, you will "destroy" a certain amount of information.

The question is always "is that of practical significance"? If it is, you need to build yourself a higher resolution converter...

share|improve this answer
The problem with ADC's is more than just destruction of information. If voltages are continuous, it means that we cannot build an ADC that is guaranteed to output a logically valid digital value after any fixed (finite) amount of time (assuming the device does not trivially output the same value at all time and that the device's output is a function over the voltage (same input voltage => same output value)). –  Abhishek Aug 19 '14 at 17:40
Noise on any device ensures that for some values of the input, there will be some variation on the output. In my mind that is destruction of information (because you are adding noise to the signal, you lose information about the signal). In other words, there is no device that will give you the same (non trivial) output for a given input, for all inputs - if it has more than one state, then there is a transition region where the "same" input can lead to one of two output states. Even hysteresis won't save you. –  Floris Aug 19 '14 at 17:46

Voltage doesn't come directly from the charge of the electron. It's the energy per charge. The charge carriers may be discrete, but the energy is not.

We can easily generate a potential by moving a wire through a magnetic field. The potential is proportional to the speed of the wire, which is a continuous value. $$V = vBL\sin{\theta}$$

share|improve this answer
Is that formula just a macroscopic approximation of a discrete phenomena? I guess we will never know! –  Abhishek Aug 19 '14 at 17:35
@Abhishek QED (Quantumelectrodynamics) explains much complexer scenarios as this as well. It is the macroscopic approximation of a continous quantum phenomena. –  peterh Aug 20 '14 at 9:19

It's not a fundamental feature of electrical potential, but:

If you have a polycrystalline metal and you cut and polish a smooth surface, the differently-oriented regions will present a different lattice plane to the outside. Crystals cut along different planes may have slightly different work functions, and so the electric potential very close to such a conductor will vary randomly at the level of a few millivolts. This is sometime called the "patch effect" and it can be comparable to the Casimir force (see e.g.) and other small electrostatic effects.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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