To start, remember that for a constant electric field the change in electric potential energy would be:
WARNING: that is only for a constant electric field. I know you will be tempted later to use this for a different electric field, but DON'T DO IT. But if not that, then how do find the change in electric potential for a point charge? Let me start with a conceptual question. Suppose there were two point charges, both positive but one is held in place. If I hold the other point charge a distance r away from the other charge and let go, what will happen?
This might seem like a simple problem. Find the force on the charge, find the acceleration (using the mass) and then use the kinematic equations. This seems simple, but it would not work. Here is the problem. As the charge moves away, the electric force gets smaller in magnitude. This means that the acceleration is not constant and the kinematic equations do not work.
Really, the only way to deal with this problem is to consider the work-energy principle and the work done by the electric field. Oh, I know. It is still not easy to find the work done by the electric force because it changes. However, if you use a little calculus you would find that as the charge moves from r to a very far distance away (infinity), the work done by the electric force would be:
This is the work done by the electric force going from r to infinity. It should be positive because the force is in the same direction as the displacement. With this work, I could calculate the change in kinetic energy assuming the free charge started from rest.
But, if I like, I can instead of having work done by the electric force I can have an electric potential energy. For this case, the change in potential would be:
Of course, if I were going from an infinite distance away to the location a distance r away, I would have the negative of that change.
Now, what if I instead took a negative charge of the same magnitude (and mass) and started it from infinity and let it be pulled towards the other stationary charge? If you do that problem, it will have the same change in kinetic energy (but it is going the opposite way). It is useful to think of the change in potential energy per unit charge - the electric potential.
To make things a little bit simpler, physicists refer to the potential energy with respect to infinity - or how much energy would it have if you moved it from infinity to that location. Using this, we drop the Δ notation on the potential. The potential (with respect to infinity) of a point charge is then:
If you have more than one point charge, the total potential at some location is just the sum of the electric potentials due to the individual charges. Be careful. Remember that electric potential is a scalar quantity and not a vector. You do not have to consider the direction (because there is no direction for potential).
While this is of course correct, you should probably point out one of the major pitfalls of converting between work and electric potential - Double counting. While V is indeed the potential energy/ work needed to move a test charge in to location r from infinity _if all other charges are fixed_, the total potential energy of a system of point charges is a sum over all _pairs_, which is half the sum of the charge times potential of all of the points, WE = 1/2 Sum(q V), since k q^2/r is the work needed to bring a pair of point particles together to a distance r from infinity.
I learned a lot from your post..Physics is quite hard and I want to study more about it..Thanks for the share!
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