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## Physics library

### Course: Physics library>Unit 11

Lesson 3: Electric potential energy, electric potential, and voltage

# Electric potential energy of charges

In this video David explains how to find the electric potential energy for a system of charges and solves an example problem to find the speed of moving charges. To see the calculus derivation of the formula watch this video. Created by David SantoPietro.

## Want to join the conversation?

• Only if the masses of the two particles are equal will the speed of the particles be equal, right? If the particles have different masses, they wouldn't have the same speeds... • Near the end of the video David mentions that electrical potential energy can be negative. I don't understand that. Can someone describe the significance of that and relate it to gravitational potential energy maybe? •  Really old comment, but if anyone else is wondering about the same question I find it helps to remember that all potential energy is relative. The significance of the negative in this particular scenario is that the forces are attracting as opposed to repelling.
For a relation to gravitational potential, imagine I have a ball suspended in the air. It has positive potential energy in reference to the surface of the Earth. If I think about a ball in a hole, it would have negative potential energy in reference to surface of the Earth. If I wanted to return it to the surface of the Earth, I would have to invest kinetic energy. Imagine I then suspend the ball at that point and dig the hole beneath it even deeper. If I release the ball, it will fall further down the hole to the new bottom. The gravitational potential energy has moved from a negative state to a more negative state. In doing so, it has converted the potential energy into kinetic energy.
• what if the two charges will have different masses?
how will we then calculate the velocity of the charge? • Not sure if I agree with this. The two particles will experience an equal (but opposite) force, but not necessarily equal kinetic energy. Here's why:
F = m*a means the smaller mass will accelerate more than the larger mass, and thus have a higher velocity. Because KE = 1/2 m*v^2, it will rapidly gain more kinetic energy because velocity is squared.
At the start the small mass will have less kinetic energy. There will only be an instant when KE for both are the same. After that, small mass has more kinetic energy.
Because Force is a vector, and Force will change as the particles move, it is not as straightforward to calculate as David's energy conservation method. Calculus is required.
• If the two charges have different masses, will their speed be different when released? • I am not a science or physics teacher, I teach automotive. I had a DC electrical question from a student that I was unsure on how to answer. The question was "If voltage pushes current how does current continue to flow after the source voltage dropped across the load or circuit device". Electricity flows because of a path available between a high potential and one that is lower seems too obvious.
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.”
Another stated, “It returns because of momentum.”
My question is:
Does 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.
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?
Hope this makes sense.
Thank you for taking the time to review my question,
Randy Nussler
Automotive Instructor
New Market Skills Center
Tumwater, WA • Potential energy is basically, I suppose, the work done in moving a charge thro' a distance r against an electric field. So, am I moving one of the charges so close to the other? (zero distance b/w) and calculating their U there? Or is it something else? It's confusing me? I don't get it. • 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. 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. 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). 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.

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. 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. 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.
• If the charges are opposite, shouldn't the potential energy increase since they are closer together? • At , he talks about how since the masses are the same, they have the same velocity and masses and writes (mv^2)/2 for both particles. I understand intuitively that this is true by symmetry, but what if the masses were different? Would the kinetic energy of both charges still be the same; if so, then I think I know how to calculate the velocity (equate the KE's sub in masses, solve for one velocity in terms of the other, equate the result to the U_i, solve for both velocities)? If the kinetic energy of both charges aren't equal, then I am very confused. Thanks! • There may be tons of other interesting ways to find the velocities of the different charges having different masses, but I like to do this. At first you find out the v for the total of the mass(I mean msub1+msub2). Then distribute the velocity between the charges depending on their mass ratios.

For example one of the charges have a mass of 2 kg and another 3 kg. So you find out the v for 2+3=5 kg of mass. Let's say you have got 10 m/s. Now the charge having mass 2 kg will have a velocity of (3/5)*10=6 m/s.
I think you got it. Thank you.
• I mean, why exactly do we need calculus to derive this formula for U? We can explain it like this:
``U = F * d   = K|Q1 * Q2| / d^2 * d   = K|Q1 * Q2| / d``  