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According to the definition of potential energy for a system of two point-charges: the potential energy of a configuration is equal to the work done when bringing the charges into such a fixed configuration.

In practice one has to take a point-charge $Q$ to be fixed and bring the other charge $q$ from infinity to the desired position, arriving to $$ U = \int_x^{\infty} k \frac{Qq}{x^2}\,dx = -k\frac{Qq}{x}$$

But if the charge $Q$ is kept fixed, doesn't that mean we are using additional force to keep it at that point and hence using more energy.

Why, if we assume none of the charges to be fixed, the energy of the system doesn't change?

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    $\begingroup$ Yes! It changes the energy of system. The extra force applied to keep the charge stationary changes the energy of the system. But the expression written above is just for the electrostatic energy and not the total energy of system. There would be different contributors to the total energy and electrostatic interactions are one of them $\endgroup$
    – SteelCubes
    Jan 15, 2021 at 11:31
  • $\begingroup$ The question is about understanding the definition. Not about a more realistic situation. $\endgroup$
    – ohneVal
    Jan 15, 2021 at 11:32
  • $\begingroup$ @SteelCubes if it changes the energy of the system, why then $ KQq/x$ is given as the potential energy of system, even if they're not fixed ? $\endgroup$
    – Adidas10
    Jan 15, 2021 at 12:14
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    $\begingroup$ @Adidas10- $\frac{KQq}{x}$ is taken as the potential energy of the system because the other energy is in the form of kinetic energy. Note, I said it changes the total energy of the system. It's just a game of how things are being defined. $\endgroup$
    – SteelCubes
    Jan 15, 2021 at 12:59
  • $\begingroup$ @SteelCubes What energy is provided to the fixed charge so as to stay in its position ? $\endgroup$
    – Adidas10
    Jan 15, 2021 at 14:19

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So first of all realize that you are dealing with a definition within Electrostatics, which means there is no time involved, no dynamics, everything is to be thought as frozen in time. So the question, "if we assume none of the charges to be fixed" makes no sense in this context because we are looking at one snapshot in time and calculating if you want the instantaneous potential energy.

Having said that, this idea of bringing in charges from infinity to a given point is a way to understand and compute the potential energy of a configuration, however it does not correspond to any real movement of charges.

Consider the system of two point-charges you mention. You also have to bring the first charge to the desired point from infinity, however since there is no other charge already present, you are not doing work against anything, so the first one is "free". Then to place the second one you have to some work as you computed it. And no matter how you do this since the forces here are conservative, you will obtain the same answer. Again there is no time here.

Now if we ask ourselves whether the situation is stable, the answer is obviously not. Let say both charges are positive, if you were to let the time run (unfreeze your simulation) they will fly apart. This is probably what you had in mind.

So "fixed" in electrostatics just means there are no time dependencies, charge distributions are to be thought to be frozen in time, or being examined at one instance in time. Computing the potential energy at one instant in time, can be done in many ways, but mathematically it agrees with this idea of bringing in charges from infinity.

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  • $\begingroup$ Sir/Mam let me change the question a little bit. If in running time somehow we can hold both the charges at their place then what will be the energy provided to each charge so that it remains stationary ?(in no loss of energy given) $\endgroup$
    – Adidas10
    Jan 15, 2021 at 12:17
  • $\begingroup$ The energy provided has to be equal to the energy stored in the configuration as if it were static, since at the end of the day you are considering a static situation (there is no movement if you hold the charges with some force) $\endgroup$
    – ohneVal
    Jan 15, 2021 at 13:01
  • $\begingroup$ so does that mean total energy provided is equal to $kQq/x$ hence energy given to each charge may be $kQq/2x$ ? But the potential energy of each charges is $kQq/x$ so where's the mistake ? $\endgroup$
    – Adidas10
    Jan 15, 2021 at 14:12
  • $\begingroup$ The potential energy is not associated to a charge but, to the whole configuration. To keep the configuration in place you need exert a force equal in strength but opposite to the Coulomb repulsion. This is again a statement about a fixed instant in time. If you want to take time into consideration then you need to supply that amount of force per unit of time and that is not even energy (under dimensional grounds) $\endgroup$
    – ohneVal
    Jan 15, 2021 at 14:19
  • $\begingroup$ So is energy given to each charge is $kQq/2x$ true. If it is not energy what is it? $\endgroup$
    – Adidas10
    Jan 15, 2021 at 14:37

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