We'll put a little subscript e so that we know we're talking about electrical potential energy and not gravitational Like PE would've made sense, too, because that's the first two letters of the words potential energy. So why u for potential energy? I don't know. Physicists typically choose to represent potential energies is a u. It's coming from theĮlectrical potential energy. But the total energy in this system, this two-charge system, Potential energy decreases, the kinetic energy increases. So originally in this system, there was electrical potential energy, and then there was lessĮlectrical potential energy, but more kinetic energy. We would say thatĮlectrical potential energy is turning into kinetic energy. So where is this energy coming from? What is the source of this kinetic energy? Well, the source is theĮlectrical potential energy. Gaining kinetic energy, where is that energy coming from? I mean, if you believe inĬonservation of energy, this energy had to come from somewhere. The Q2's gonna get pushed to the right, and the Q1's gonna get pushed to the left. They're gonna fly apart because they repel each other. Sitting next to each other, and you let go of them, We'll call this one Q1Īnd I'll call this one Q2. If you had two charges, and we'll keep these straightīy giving them a name. The actual formula he uses for U requires calculus to derive, because the electric field (and thus the force) varies as the distance between the charges varies. Or, to turn it around, the U in this case represents the work you would have to do in order to move the two charges from an infinite separation to their position in the problem. So the U in this case decreases as the charges fly apart, and if we let them fly apart forever, then the U would approach 0, and all the potential energy would become kinetic energy. Also, it means that if you tried to push the charges so close together that r = 0, then U would become infinite! This should "make sense", since the formula for the FORCE is F = k*q1*q2/r^2, which means the force of repulsion becomes infinite as the distance between two positive charges approaches 0. Very often, we choose U = 0 at the beginning, or maybe at the end, but in this type of question, we usually use the formula for PE that he gives us, U = k*q1*q2/r, which means that U only equals 0 when the two charges are infinitely far apart (r = infinity). It's important to always keep in mind that we only ever really deal with CHANGES in PE - in every problem, we can choose the point where U = 0. Great question! You are exactly correct, with the small clarification that the work done moving a charge against an electric field is technically equal to the CHANGE in PE. Thank you for taking the time to review my question, Or is there a possibility that on one side there is a high potential pushing and on the other side of the load there is actually a pull? One answer I found was " there is always 1 millivolt left over after the load to allow the current be pushed back to the power source.”Īnother stated, “It returns because of momentum.”ĭoes electricity flow back to its source because of high voltage on one side of the load and no or maybe low voltage on the other as commonly stated. Electricity flows because of a path available between a high potential and one that is lower seems too obvious. The question was "If voltage pushes current how does current continue to flow after the source voltage dropped across the load or circuit device". I had a DC electrical question from a student that I was unsure on how to answer. I am not a science or physics teacher, I teach automotive.
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