# Shouldn't Electric Potential energy be $U=2kQq/r$?

Let's take 2 charges $$Q$$ and $$q$$ a meter apart. Say you want them to collide. You'll have to apply force $$F$$ on charge $$q$$ (which I know is changing with distance). I know we integrate this force and we would get something like $$\text{Work} = U_{0 \ \mathrm{m}} - U_{1 \ \mathrm{m}}$$. But it's not just $$Q$$ applying force on $$q$$. $$q$$ is also applying the same force on $$Q$$. So You'll have to get $$q$$ close to $$Q$$. But also apply the same force on $$Q$$ so that it does not run away. Right? I have had this at the back of my head for days. please help me out. In simple words, we should apply force $$F$$ on both charges to keep them in place? Thus while finding potential energy, we must integrate both forces $$F$$, making $$U = 2kQq/r$$.

What am I missing?

• Broadly, work depends on both force and displacement. Please clarify your question to include the displacement of each charge in the frame you've chosen. Commented May 28, 2023 at 6:58
• If you like this question you may also enjoy reading this and this Phys.SE posts. Commented May 28, 2023 at 7:16
• Yes we need to apply a force and the value you mentioned will be the total potential energy of the system but we are concerned with the potential energy of the single charge then its only kqQ/r. You can verify this by performing a thought experiment where you just leave the 2 charges and both of them gets repelled by kqQ/r so its the system's energy thats 2kQq/r not the individuals Commented May 28, 2023 at 8:14
• I give you an advice not an answer: print out what @Puk wrote in large print, tape it above your bed, and read it every morning to remember it when you will study the energy and work of or in the electric or magnetic fields, especially the conceptual differences between quantities such as $\mathbf {E\cdot dD}$ and $\mathbf {D\cdot dE}$ , and between $\mathbf {H\cdot dB}$ and $\mathbf {B\cdot dH}$ . Commented May 28, 2023 at 14:38

A lot of controversy so here's an example... Say you have charges Q and q that are attracted to each other. What is the force you'll have to apply to maintain the system's potential energy?

Say at distance r they apply force F on each other.

You apply force F on q to hold it in place. But Q is still moving towards q. So you have to apply another force F on Q to keep it in place too. Thus at each point, you have to apply 2F force on the system. This is what the confusion was about.

The change in potential energy would be the work done from there on. So say on a particle with charge q you apply a force F + dF, just above F so that you move it further apart.

To change a particular configuration... the net force you apply is the force to maintain the configuration in the first place + the extra dF force you apply (which is almost 0).

That brings us to 2F + dF which is just 2F for all theoretical purposes.

There you go, the work to be done is thus 2F*dx for distance dx. So the work done to change the configuration must be 2kQq/x. Bringing charges from infinity, the work IS the potential energy of the system.

Now I myself don't know if that is right, which is why I put it up as a question. WHAT AM I MISSING?

Conclusion (EDIT):

Say we apply force F on both particles (total force being 2F) which is just enough to move them apart. They both move distance dx so the total work done is 2(Fdx) and the total distance they moved apart is 2*dx (they both move dx dist. in opposite directions). That would mean moving them dx distance apart, work done is Fdx, integrating it would imply that change in Potential energy to bring them distance r apart would be kQq/r...

So to bring them at distance x from infinity, we have to move them half the distance each while applying equal force on both Q and q. Which just mathematically happens to be the same as applying force F on q and moving it all the way to x distance from Q.

Thus, U = kQq/r is correct. But we do have to apply force on both particles if we practically have to get them into a configuration. Just that we have to only make them move half the distance each. I hope it makes sense for everyone now.

– Community Bot
Commented May 28, 2023 at 9:33
• This is wrong, and so is Naveen V. Potential energy is associated with a system of charges, not each charge in the system individually. The potential energy of the system of $Q$ and $q$ separated by $r$ is the work that you must do to assemble this system, e.g. by keeping $Q$ fixed and bringing $q$ close to it from infinity, or whichever way you want to do it. If you really want to associate a potential energy with each charge separately, the only sensible way to do this is by assigning $kQq/2r$ to each.
– Puk
Commented May 28, 2023 at 9:34
• I must apologize for any misinformation that I have caused , Then what do you think the total energy of the system is. If we let loose both the charges they both tend to move with an energy and by conservation of energy that doesn't add up to the total energy of the original system @Puk ? Commented May 28, 2023 at 10:13
• @Puk I understand the assumption of keeping the charge Q fixed. But the entire point is, that would further require a force F! Anyways, I don't know if that is wrong or right. Can you please explain in terms of the example I've given in the edit? Commented May 28, 2023 at 10:19
• @NaveenV The potential energy of the system is $kQq/r$. If you release the charges so they can freely move apart under the influence of the repulsive Coulomb force, the sum of the final kinetic energies of the particles at infinity be $kQq/r$. If the charges are equal in mass, each will have a kinetic energy of $kQq/2r$. A careful calculation of the work done on each charge will tell you the same thing.
– Puk
Commented May 28, 2023 at 16:49

If you are keeping $$Q$$ fixed while bringing $$q$$ towards $$Q$$, yes, there is an external force $$F$$ on $$Q$$, but it does no work since $$Q$$ is not moving. Recall that work is the product of force and displacement.

If the separation of the charges starts at $$a$$ and finishes at $$b$$, with $$b, the change in electric potential energy of the system of two charges is $$\frac{kQq}{b}-\frac{kQq}{a}$$ and this is the total work done by the external forces acting on the system independent of how the separation from $$a$$ to $$b$$ was achieved.

One method is to make sure one of the charges does not move, thus work done by the external forces is only done by one of the external forces, the one which moves a distance $$a-b$$.

Another way is to allow both charges to move a distance $$\frac{a-b}{2}$$ so in effect each external force does the same amount of work, which is half the amount done by the moving external force when one of the charges is stationary.

How might one control the movement of the charges?

Well, in the case of one charge not moving it could have a mass much, much larger than the other charge and in the case of equal distance traveled the charges could be of equal mass.