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In electrostatics it depends on the distance from the charges, should it also be in the circuit? But in practice, the voltage depends on the resistance, for example on a resistor. And the distance itself does not play any role. Why so?

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    $\begingroup$ The distance does play a role. The distance from one charge to another, that is. In a flowing current in a circuit, there are many many charges all exerting electric force on each other. $\endgroup$ – Steeven Nov 19 '18 at 19:59
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In electrostatics, we are usually dealing with a small number of discrete charges (such as putting specific charges on the corners of a cube), or some charge in a very constrained location (charged sphere). These constraints allow us to calculate potentials and other results directly from the charges and their geometry.

Circuits and other dynamical systems tend to be much messier. Instead of a small number of fixed charges, there are conductors that have an enormous amount of free charges, all of which can migrate in response to changes in the system. Direct calculation of potentials is not possible because we don't know where all the charges are any longer. Instead the general behavior of certain materials is modeled (like conductors, capacitors, etc.)

It is this ease of motion along a particular path (conductor) that makes the distance almost irrelevant when we study simple dynamical systems. We don't normally care which faucet in our house is closest to the water main, because it flows so easily through the pipe that differences aren't noticeable.

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  • $\begingroup$ This means that we cannot calculate the potential on the wire, as there are so many charges and they kind of distort it and they also move, but we can calculate what energy is needed to overcome the resistance, that is, the voltage? $\endgroup$ – Vitalii Khanas Nov 20 '18 at 15:24
  • $\begingroup$ Because the charges move, they will migrate so that all parts of the wire have the same potential. By attaching it to something of fixed potential (a voltage source or the ground), we can extend that potential to a new location. $\endgroup$ – BowlOfRed Nov 20 '18 at 16:58
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What may be confusing you is the fact that we teach people about ideal circuits most of the time, in which "ideal wires," with no voltage drop across them, connect electrical components together. Ideal wires are not actually physical objects. They are only there to show how the other (real) components are connected together. You could solder the components of a circuit directly together, without using any wires at all, and in basically every real situation, the predictions of the circuit diagram would be no worse, despite the fact that you "left out all the wires."

So let's move away from circuits as they appear in ideal circuit diagrams for a moment, and take a look at realistic circuits. In a realistic circuit, every component has a nonzero electrical resistance. This resistance details how much voltage drop is experienced by charges that move through that component in a given electrical current. The resistance is linked to a more fundamental property, the resistivity ($\rho$). This is defined in terms of the resistance $R$ and the geometry of the object (which has length $L$ and cross-sectional area $A$) by the following relationship:

$$R=\frac{\rho L}{A}$$

Unlike the resistance, the resistivity is independent of the length or cross-sectional area of the material; it tells you how much "voltage drop density" one can expect at a given current. It's not a coincidence that this sounds a lot like a description of the electric field (which can be described as the "voltage drop per unit distance"); in fact, this relationship is codified in the vector form of Ohm's Law:

$$\vec{J}=\frac{\vec{E}}{\rho}$$

for current density $\vec{J}$, electric field $\vec{E}$, and resistivity $\rho$. So the resistance is in fact directly linked to the electric field within the component. The higher the resistivity, the more electric field exists within a component for a given current. A higher electric field leads to a higher voltage drop across the component, which is exactly what is expected from a component with higher resistance.

So, to answer your question, in any real circuit the voltage drop does depend on, say, the length of the wire used to connect components. The wire has nonzero resistivity, and so it has a resistance that depends on its length; therefore, the voltage drop across the wire also depends on its length. Equivalently, the wire has nonzero resistivity, and so a current induces an electric field inside it; therefore, the voltage drop across the wire depends on the distance a charge moves in this electric field, which corresponds with the length of the wire. Usually, these resistivities, electric fields and voltage drops are quite small, and so they are either ignored or lumped together as a single resistive element in the case of ideal circuits, without much cost in accuracy of the model's predictions. However, for full conceptual understanding, you must return to the more realistic description of an electric circuit.

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But in practice, the voltage depends on the resistance, for example on a resistor. And the distance itself does not play any role. Why so?

Because in practice, often voltage in circuits is determined by external agent, such as battery or a power grid. Distance between elements or their length does influence voltages a little, but in small devices this is usually extremely small and it is OK to neglect it.

For very long DC power lines carrying current, voltage between plus and minus wire will depend on how far from the source it is measured: the farther, the lower voltage. This is because there voltage loss due to resistance of the wires.

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I'm not quite sure, what do you actually mean. Voltage is voltage, it does not "know" where it comes from. Also in case of induced electricity. Ohm's law (a relationship between voltage, resistance and current) remains in force in both cases too.

The main difference is that although a static voltage can be quite high, it typically involves relatively small charges because capacities in normal circumstances are very low. So if it has an opportunity to discharge - through a spark or by touching a grounded element - the voltage decreases very rapidly, even to zero, and the current consequently decreases very rapidly as well. On the other hand, in case of circuits an element is being constantly recharged from the other parts of the circuit and the voltage remains high. That's why although a woolen sweater can charge to thousands volts without doing any real harm to you (integrated circuits are a different story... they are very sensitive), while a mere 240V from the mains can kill you.

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  • $\begingroup$ If you're not sure what the OP is asking for, you should leave a comment, rather than trying to answer. I don't think this actually answers the question. $\endgroup$ – FGSUZ Nov 19 '18 at 21:35

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