Can work be 0 but the potential energy still increase?

Question

Let's say that we have two point charges of same magnitude charge, $+Q$ and $-Q$, separated by distance $d$, and charge +Q is fixed in place. If I move -Q away from the +Q charge by giving it an initial velocity directly away from the +Q and then always match the electric force so that the net force that the +Q experiences is 0, the -Q charge will move at a constant velocity equal to the initial velocity given.

Let's say I take -Q to a distance 2d from +Q and then let go of the charge. At that moment, the charge still has the initial velocity given to it, so no $ \Delta $ KE but an increase in potential energy (since the product of +Q and -Q is negative):

$PE_f = -kQ^2/2d$ and $PE_i = -kQ^2/d$.

But since the integral of work from the force I exert and the electric force cancel out, there is no work. I feel like somehow I'm leaving out the work that my hand is doing, but even when my hand disappears there is still a change in PE but no net work (since the net force is 0). What is going on here?

Answer 1

Can work be 0 but the potential energy still increase?

The net work on a charge can be zero while the potential energy of the system of charges increases.

But since the integral of work from the force I exert and the electric force cancel out, there is no work.

There is no net work done.

Per the work energy theorem, the net work done on an object equals its change in kinetic energy. Since there was no change in kinetic energy, the net work done is zero. You did positive work since your force was in the direction of the displacement, while the electric field did an equal amount of negative work, since its force was opposite to the direction of the displacement, for a net work of zero. The end result is the electric field took the energy you gave the charge and stored it as electrical potential energy in the two charge system.

The gravity analogy is lifting an object at constant velocity from point A to point B where the vertical distance between A and B is $h$ and the force of gravity is constant. The change in KE is zero but there is an increase in gravitational potential energy of $mgh$ of the Earth-object system.

Keep in mind that potential energy is a system property. Neither the single negative charge nor the single object being lifted owns the potential energy. In the case of electrical potential energy, it belongs to the configuration of charges. In the case of gravitational potential energy, it belongs to the configuration of gravitating bodies.

Hope this helps.

Answer 2

There are just too many different definitions of work for you to definitively say which is what, unless you clarify things enough to pin the result down.

For example, if you ask what is the work done by a specific force under a specific displacement, then by the definition of $W=\int\vec F\cdot\vec{\mathrm dr}$, this work will be non-zero in your situation, corresponding to the gain in PE.

The naïve definition used in the Work Energy Theorem (WET) is only about changes in KE. Then, yes, your problem will arise, that the work is zero yet PE raised. Needless to say, this naïvety is so cumbersome that almost nobody ever means to use this.

Instead, what is really happening is that, for every kind of PE that a specific form can be defined, we would select their contributions out from the WET and choose to consider them separately. Then, we would consider changes in the total energy as the new definition of work.

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I know that you have concurrently asked a question about thermodynamic work. This is related to that.

I have to point out to you that, especially when magnetism is considered, the concept of work itself is ill-defined. The sooner you can move away from stupid word games related to work, the sooner you can be much less confused. Just keep track of energies as they arise in the discussion, being clear to state them in as precise as you can state them to be, and the confusions will dissipate.