# Weird question in electrostatics -- Work done bringing two positive charges together

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samy4408
in a cours of electrostatic when we have a positive charge and we bring another one (also postitive)we have to do work and apply a force that equals the force of repultion over the distance which seems weird because if we do that the net force will be equal to 0 and the charge will not move can you explain to me why the charge will move dispite that the net force is 0?

weirdoguy
can you explain to me why the charge will move dispite that the net force is 0?

Do you know Newtons laws of motion? Zero net force does not mean charge is not moving. It can move uniformly.

Ibix
2022 Award
As weirdoguy says, zero net force means constant speed, not necessarily zero speed. That's Newton's first law.

If the particle is initially at rest you would need to supply a bit more force than the repulsive electrostatic force to make it start moving - the work from that extra bit of force goes into kinetic energy of the particle. But after that you can just balance the repulsion with your applied force to keep the particle moving at a constant rate.

samy4408
Gold Member
2022 Award
in a cours of electrostatic when we have a positive charge and we bring another one (also postitive)we have to do work and apply a force that equals the force of repultion over the distance which seems wierd because if we do that the net force will be equal to 0 and the charge will not move can you explain to me why the charge will move dispite that the net force is 0?
Samy, it would be nice if you would pay attention to the spell checker.

samy4408
Do you know Newtons laws of motion? Zero net force does not mean charge is not moving. It can move uniformly.
the charge is moving uniformly when the net force is 0 if it already has a kinetic energy in this case the charge is not moving and we bring it from infinity to that point

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samy4408
Samy, it would be nice if you would pay attention to the spell checker.
sory

samy4408
As weirdoguy says, zero net force means constant speed, not necessarily zero speed. That's Newton's first law.

If the particle is initially at rest you would need to supply a bit more force than the repulsive electrostatic force to make it start moving - the work from that extra bit of force goes into kinetic energy of the particle. But after that you can just balance the repulsion with your applied force to keep the particle moving at a constant rate.
thanks

berkeman
Gold Member
I think that when a charge is moving it also stores energy in its magnetic field, which acts like additional mass.

Homework Helper
Gold Member
in a cours of electrostatic when we have a positive charge and we bring another one (also postitive)we have to do work and apply a force that equals the force of repultion over the distance which seems weird because if we do that the net force will be equal to 0 and the charge will not move can you explain to me why the charge will move dispite that the net force is 0?

This is part of the usual textbook story to explain how the work you do to assemble charges is stored as electrical potential energy in the system.

By balancing the electric force due to the charges already there (and assuming no other interactions), you are able to move the new charge (as others have said) with essentially constant velocity. Usually, we imagine this as a slow quasi-equilibrium process… so that the new charge has essentially-zero (infinitesimally-small) kinetic energy…. Otherwise you will have to momentarily apply a larger force to bring the new charge to rest at its final location in the assembly.

A similar situation occurs when we slowly stretch a spring by pulling on a mass connected at one end.

samy4408
Gold Member
2022 Award
I think this is the usual argument to derive electrostatic field energy. It's not that you consider just the free motion of the charges, for which total energy and momentum are of course conserved. BTW from this you can also much more elegantly and generally derive the field energy for the general case, using Maxwell's equations and the Lorentz force between charge-current distributions.

The thought experiment is as follows: You consider point charges first. In the thought experiment all the point charges are kept at infinity. Then you put charge ##Q_1## to its place ##\vec{x}_1##. For this you don't need any work, because there's no force acting on it. Then you put the 2nd charge at its place ##\vec{x}_2##, keeping the 1st charge fixed at its position (!). On this 2nd charge the Coulomb force is acting, i.e.,
$$\vec{F}_2=-Q_2 \nabla \Phi_1, \quad \Phi_1=-\frac{Q_1 Q_2}{4 \pi \epsilon_0 |\vec{x}-\vec{x}_1|}.$$
The total work is independent of the path, because the electrostatic force is conservative. Thus the work done finally is
$$E_{\text{mech}}=-\frac{Q_1 Q_2}{4 \pi \epsilon_0 |\vec{x}_2-\vec{x}_1|}.$$
So if ##Q_1 Q_2>0## you have to use energy (because of the repulsive force) and if ##Q_1 Q_2## you get energy back (because of the attractive force). That means the change in total field energy is
$$E_{\text{field}}^{(2)}=\frac{Q_1 Q_2}{4 \pi \epsilon_0 |\vec{x}_1-\vec{x}_2|}.$$
Now you keep the two charges fixed and move the 3rd one at its position ##\vec{x}_3##. Now you have the superposition of the Coulomb forces from particle 1 and 2 and thus for three particles you get
$$E_{\text{field}}^{(3)}= \frac{Q_1 Q_2}{4 \pi \epsilon_0 |\vec{x}_1-\vec{x}_2|} + \frac{Q_1 Q_3}{4 \pi \epsilon_0 |\vec{x}_1-\vec{x}_3|}+ \frac{Q_2 Q_3}{4 \pi \epsilon_0 |\vec{x}_2-\vec{x}_3|},$$
where the first term is from bringing the 2nd charge at its position in presence of the charge 1 as discussed before and the other 2 terms from bringing the 3rd charge at its position in the presence of charges 1 and 2. This argument you can repeat. So the total change of field energy is
$$E_{\text{field}}^{(N)}=\sum_{j=1}^N \sum_{k<j} \frac{Q_j Q_k}{4 \pi \epsilon_0 |\vec{x}_j-\vec{x}_k|} = \frac{1}{2} \sum_{j \neq k} \frac{Q_j Q_k}{4 \pi \epsilon_0 |\vec{x}_j-\vec{x}_k|}.$$
It is important to note that we did not count the "self-energy" of each point charge, i.e., not the energy contained in the electric field of charge 1 in the first step of the gedanken experiment nor the self-energies of the other charges. The reason is that these self-energies are infinite and also unimportant for the energy balance between the field an charges, since only energy differences between different configurations are physically observable but not absolute energies.

That's also the reason that for continuous charge distributions ##\rho## you can simply write
$$E_{\text{field}}=\frac{1}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 r \int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\rho(\vec{r}) \rho(\vec{r}')}{4 \pi \epsilon_0 |\vec{r}-\vec{r}'|},$$
where now the "self-energy" is contained in this expression, but it fulfills the purpose for the energy balance as well and is not divergent, because the singularities at ##|\vec{r}-\vec{r}'|=0## are integrable.

This expresses the field energy from the point of view of the charge distribution making up this field, which is of course due to our "mechanistic" approach defining it. As usual in electrodynamics you can express such quantities also using the fields (in macroscopic electrodynamics you can to a certain degree even arbitrarily shuffle parts of the quantities between "matter" and "field" descriptions, but that's another story). To see this we use the fact that the electrostatic potential is given by
$$\Phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\rho(\vec{r}')}{4 \pi \epsilon_0 |\vec{r}-\vec{r}'|}.$$
Plugging this in our expression for the field energy we get
$$E_{\text{field}}=\frac{1}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 r \rho(\vec{r}) \Phi(\vec{r}).$$
On the other hand from Gauss's Law we have
$$\rho(\vec{r})=\epsilon_0 \vec{\nabla} \cdot \vec{E}(\vec{r}).$$
Plugging this into the above integral and integrating by parts (assuming that the charge distribution vanishes outside a finite region), we get
$$E_{\text{field}}=\frac{\epsilon_0}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 r \vec{E}^2(\vec{r}).$$
This suggests that
$$u_{\text{field}}(\vec{x})=\frac{\epsilon_0}{2} \vec{E}^2(\vec{x})$$
is the energy density of an electrostatic field, and this can indeed also justified by using the local form of energy conservation to derive the general expressions for energy density and energy-current density from Maxwell's equations and the Lorentz-force Law, but that's another story.