Why do voltages seem to behave differently than expected?

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Voltages behave differently than expected due to the concept of potential difference rather than absolute voltage values. When measuring voltage, it is essential to understand that the measurement reflects the difference between two points, not an absolute value approaching infinity near a power supply's positive terminal. The formula V=(kq)/r applies to point charges, which is not relevant in circuit contexts where resistors and other components influence measurements. The negative terminal is often set as a reference point, making its voltage zero, while the positive terminal's voltage indicates the work done to move charge through the circuit. Understanding these principles clarifies why voltage readings do not align with initial expectations.
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I am very confused right now about voltages. When I plug in a power supply and set it to an arbitrary voltage, I measure voltages between 0 and what I set it to. However, shouldn't it be that the closer I get to the positive terminal of the power supply, the closer the voltage gets to infinity. If V=(kq)/r then as r approaches 0, V should approach infinity; however, this is not seen. Also, why should the voltage be equal to zero at the negative pole? The negative charge is doing -work against the positive charges, so shouldn't V approach -infinity? Thanks to anyone who can clarify this for me.
 
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What do you mean by voltage? Potential or potential difference?

I think Potential difference is the significant term. Potentia difference is the measure of the energy difference of unit charge between two points. Like mgh is the difference in the gravitational potential energy of a body of mass m between two points. Otherwise the gravitational potential energy at the surface of Earth will be zero, which is not true as we know.

Now think again of your question.
 
Haftred said:
I am very confused right now about voltages. When I plug in a power supply and set it to an arbitrary voltage, I measure voltages between 0 and what I set it to. However, shouldn't it be that the closer I get to the positive terminal of the power supply, the closer the voltage gets to infinity. If V=(kq)/r then as r approaches 0, V should approach infinity; however, this is not seen. Also, why should the voltage be equal to zero at the negative pole? The negative charge is doing -work against the positive charges, so shouldn't V approach -infinity? Thanks to anyone who can clarify this for me.
I take it you're measuring voltages with two leads - a ground and the probe that gives you a voltage. What you're measuring is the voltage difference between the two leads. If the two leads are hooked up next to each other, naturally there's no difference and your measured voltage is zero. If you were to hook the ground lead up somewhere in the middle of the circuit, your probe would give you a negative voltage at the negative terminal.

The point you measure at doesn't determine the resistance. The resistors (and/or other elements) are still in the circuit regardless of where you measured from. In other words, it doesn't matter where you measure the voltage, the resistance in the circuit doesn't change. You're measuring the circuit, not modifying it (okay, technically, not quite true - it's impossible to measure a circuit without modifying it, but you're at least not modifying the values you're measuring).
 
Haftred,
It is questions like these that can send one around the bend! Let's try anyway. Firstly the formula for voltage (electric potential difference) that you refer to applies to a point charge, which is not applicable here. Secondly when we talk about potential difference one may choose your zero reference level arbitrarily (usually at the negative terminal of the circuit). Then the electric potential of the power supply would be the amount of work one needs to do in order to move 1 coulomb of positive charge from the negative terminal through the circuit to the positive terminal of the power supply. The "voltage" of the power supply is therefore the amount of work that one coulomb of positive charge would do in the circuit when it is driven by the power supply through the circuit. This "capacities" of power supplies differs due to their abilities to generate electric fields of different magnitudes in the circuit components (which pushes the charge through the components). I hope this helps to ease the pain a bit.
 
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