0
$\begingroup$

When we calculate electric potential energy of a two particle system, say first I bring $+q_1$ and then I bring $+q_2$ against $q_1$'s electric field. Say I get that $q_2$ charge to a point $r$ distance away from $q_1$. Now potential $V_1$ at that point (where $q_2$ resides) due $q_1$ is $kq_1/r^2$ and hence the potential energy of $q_2$ is $q_2V_1= kq_1q_2/r$.

Similarly, at the point where $q_1$ resides, which is again $r$ distance away from $q_2$, the potential $V_2$ at that point due to $q_2$ is $kq_2/r^2$. Hence the potential energy of $q_1$ becomes $q_1V_2= kq_1q_2/r$.

So, the potential energy of both the charges are same and hence when we talk about the potential energy of the system, it should be $2(kq_1q_2/r)$. But in books it's written that the potential energy of a two particle system is $kq_1q_2/r$. Why is it not twice that?

I agree that the time when only charge $q_1$ arrived (with no other charge present) the potential energy of charge $q_1$ at that time will be zero but after the arrival of charge $q_2$ in the presence of $q_2$, then of course the potential of $q_2$ exists but at the same time potential of the charge $q_1$ will also exist as $q_2$ has its potential created at the point where $q_1$ resides, and hence $q_1$ would also have potential energy. And hence potential energy of the system should be the potential energy of $q_1$ and $q_2$ and it is $2(kq_1q_2/r)$, but why do books claim it is $kq_1q_2/r$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

In short, while $q_2$ does exert a force against $q_1$, this force does not perform any work because $q_1$ does not move. The work performed is the product of the force exerted times the distance the particle moves against (or with) it; since $q_1$ does not move there is no work performed on it. The potential energy is defined to be the work required to get to the physical configuration under consideration from some "stock" conditions, and since $q_1$ does not accrue any work it is not counted towards the total potential energy.

In general, the force $kqq'/r^2$ is intrinsically tied to the pair of particles - it is not counted as two independent forces on the two particles. Similarly, the pair of particles has a shared potential energy of $kqq'/r$, but this is an interaction energy and it is not a property of an individual particle - it is a property of the system as a whole. This is the case with all interaction energies.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Sir, you said that force is intrinsically tied to the particles and is no counted as two independent forces, then actually today I was solving a question and in that question we needed to draw a free body diagrams o a charged sphere which was hung by a spring and another charge sphere was beneath this system. In this case while drawing FBD the solution showed electrostatic force independently on the charged sphere which was hanging. Accurately with the the spring force, mg, it also showed the electrostatic force independently on that hanging sphere due to the other charged sphere beneath it. $\endgroup$
    – Krishna behera
    Commented Jun 14, 2015 at 6:07
  • $\begingroup$ I cannot take responsibility for the writings of other authors. Nevertheless in that case there is a corresponding force on the other charge which cannot be ignored in a full treatment of the system such as the one you attempted in the question. $\endgroup$
    – Emilio Pisanty
    Commented Jun 14, 2015 at 15:42
  • $\begingroup$ 'q1 does not move'. Does some force holds it in rest in inertial frame or do we work in q1's frame? $\endgroup$
    – Tony Stark
    Commented May 10, 2021 at 3:34
  • $\begingroup$ @Tonysta As described here, an external force needs to hold that charge in place. $\endgroup$
    – Emilio Pisanty
    Commented May 10, 2021 at 6:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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