# Will a point charge in space have any electric energy?

If we bring a unit positive charge from infinity to any point, let that point be A, then the potential of that point is negative of the work done in bringing it. My first question is why negative of the work done?

Secondly, let us say if we bring a point charge from infinity to a point in space (isolated), will it have any potential energy? Why?

Please help me out, I cannot understand the concept of electric potential.

Thank you for the help in advance! :)

• Potential of a point due to some specified system of charges is work done on a positive unit charge by controlled force of experimenter in bringing it from infinity to the point quasistatically, while the system of charges does not change in any way during the process. There is no negative of work in the definition. – Ján Lalinský Apr 18 '18 at 10:06
• The negative is used in definitions that use work of electric force of the system of charges instead of the work of force due to the experimenter. Those works are the same but have opposite sign. – Ján Lalinský Apr 18 '18 at 10:09

## 1 Answer

First, potential is defined for a point, not a charge! So it's non-sense to say the potential of an electron or things like this. The meaningful term is potential energy. We can say an electron has an amount of potential energy.

Say that charge is at infinity. There exists an electric field in the space. The work done on a charge $Q$ to transfer it from infinity to $\mathbf{r}$ (without the charge gaining kinetic energy) is:

$$W=\int_\mathcal{O}^\mathbf{r}\mathbf{F}_{\text{me}}\cdot d\mathbf{l}=\int_\mathcal{O}^\mathbf{r}-(Q\mathbf{E})\cdot d\mathbf{l}=Q(-\int_\mathcal{O}^\mathbf{r}\mathbf{E}\cdot d\mathbf{l})=Q V(\mathbf{r})$$

Now,

$$\require{cancel} \Delta E_\text{internal}=\cancelto{0}{\Delta K} + U_\text{f}-\cancelto{0}{U_\text{i}}=U_\text{f}=W_\text{external}=QV(\mathbf{r})$$

That is,

$$U_\text{f}=QV(\mathbf{r})=\int_\mathcal{O}^\mathbf{r}\mathbf{F}_{\text{me}}\cdot d\mathbf{l}$$

(For $V$ and $\mathbf{E}$ don't see that $Q$ charge because it can't exert force on itself.)

A single charge will not have any potential energy if you use the above formula. Because there won't be any field to exert force to the charge. You can freely move it in the space. But this doesn't mean the charge doesn't have any energy. In fact, the energy is stored in the electric field of the charge and turns out that it's infinity for a perfect point charge.

The total electrostatic energy of an arrangement is given by:

$$E_\text{total}=\frac{\varepsilon_0}{2}\int_\mathcal{V}E^2 d \tau$$

Where $d\tau$ is the volume element.

But where we are dealing with usual problems, we don't usually count the energy used to make the charges. That's why we use the above formula (charge times potential).