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Engineering
Electrical Engineering
Irrelevance of electrical potential
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[QUOTE="SlowThinker, post: 5709209, member: 572662"] I don't know how much math you have done, and I'm probably not the best person to explain things, but... For example, the force exerted by gravity on an item of mass ##m## is ##F=-m\frac{\Delta\Phi}{\Delta h}## (generally only for small ##\Delta h##). For example at height 20 meters the gravitational potential can be ##9.8{\times}20 \text{ J/kg}##. At height 23 meters the potential is then ##9.8{\times}23\text{ J/kg}##. Thus the force on an item of mass 2 kg is $$F=-2\times\frac{9.8\times23-9.8\times20}{23-20}\text{ J/m}=-2\times9.8\text{ N}$$ Now if the potential at 20 meters was ##9.8\times20+12345\text{ J/kg}## and at 23 meters it was ##9.8\times23+12345\text{ J/kg}##, the force would stay the same. This is how all potentials work. You can add a constant to it, and the numbers change, but the measurable results (forces, acceleration, energy used) stay the same. You can pick any constant you like. Sometimes it's such that objects at sea level have zero potential, other times objects at your table have zero potential, other times it's objects at infinity that have zero potential. Once you pick a reference point, all other values are determined. Again, this is a feature of every potential, not just electric or gravitational potential. Electric field between 2 points is given by the difference of potential between the 2 points (if the path between them is clear). Same as before: the field is the same, whether one point has ##\text{2 J/C}## and the other has ##\text{1 J/C}##, or one has ##\text{2+34567 J/C}## and the other has ##\text{1+34567 J/C}##. The movement of electrons is the same in both cases. [/QUOTE]
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Irrelevance of electrical potential
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