0
$\begingroup$

I was doing my homework when I came across this question:

Three equal point charges, each with charge $1.40 \, \rm\mu C$ , are placed at the vertices of an equilateral triangle whose sides are of length $0.250 \,\rm m$. What is the electric potential energy $U$ of the system? (Take the potential energy of the three charges as zero when they are infinitely far apart)

I had been thinking about voltage and electric potential in terms of gravitational potential energy, so I decided to consider the system as a single point charge. I used $$k\frac{q\cdot q}r$$ to get the potential from each charge at the center of the system. This was incorrect. The correct method was to get the potential between each of the charges and add them up.

Why was my approach incorrect? Does calculating gravitational potential energy of a system work the same way? improved formattingWhy?

Improve this question
$\endgroup$
5
  • $\begingroup$ Since you were asked to determine the potential energy of the system, I don't see why each contribution of a point charge (without taking into account that they have an effect on one another) summed up at the center should be equivalent? en.wikipedia.org/wiki/… $\endgroup$
    – Maxim
    Commented Apr 21, 2019 at 20:03
  • $\begingroup$ @Maxim For the same reason that you can represent planets as a point mass at its center of gravity. I was thinking the same logic would apply here, but it doesn't. Why doesn't that logic work? $\endgroup$
    – Byte11
    Commented Apr 21, 2019 at 20:44
  • $\begingroup$ What do you mean when you say you tried to "get the potential from each charge at the center of the system"? If you are considering the system as a single point charge then there is no other particle from which you can calculate the energy. $\endgroup$
    – user195162
    Commented Apr 21, 2019 at 23:45
  • $\begingroup$ @Quantumness I imagined that there was some sort of point charge in the center. Then I added the potential energy between that charge and the other three charges on the outside. $\endgroup$
    – Byte11
    Commented Apr 22, 2019 at 0:19
  • $\begingroup$ You can represent planets as a point mass because of their special symmetry, you can't do that always, like when your planet is not a sphere (you have to do an integration in that case) same applies here, 3 point charges can't be summed to one place, because there is no enough symmetry $\endgroup$
    – Paradoxy
    Commented Apr 22, 2019 at 3:34

2 Answers 2

Reset to default
1
$\begingroup$

Why was my approach incorrect?

Your calculation gives you the potential energy for the imagined charge in the center, not the total potential energy, which would be the sum of the PE for each particle, as stated.

In the analogy of representing an extended object as a point at the center of gravity, the question is to find the particle's self energy, equivalent to finding the energy that one part of an object has on another part of the same object.

Does calculating gravitational potential energy of a system work in the same way? Why?

In the Newtonian approximation, yes: The mathematics of gravitation is the same as electrostatics, apart from the proportionality constant.

Improve this answer
$\endgroup$
0
$\begingroup$

I think I was getting confused by what the potential of the system was asking.

Conventionally, when we think of potential, we think of how much energy does this particle have in this state versus what it had before. With gravity, that would be how far a particle was lifted in the air.

In this problem, if we brought another test charge close to the system (or put it in the center), we would see a distribution of potential energies based on where the particle is. When you think of potential like this, the potential of a system makes no sense. Is it the potential at the center? Is it the sum of all potential radiating outwards?

The flaw in my thinking was thinking about the position of a test particle compared to the rest of the system, just like raising a ball above the earth. In reality, The potential energy of a system of particles is determining how much kinetic energy the particles would have if they were let go and traveled to infinity. Each of the particles are influenced by each other's electric field and are pushed away from each other, but at infinity, all of this force from the charges is converted to kinetic energy and it's that energy we are measuring.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.