Assume that a negative charge in the space near of it a small positive charge, Then for sure that the negative charge will attract the positive charge naturally because it has an electric field around it ,So the point which the positive charge at will have an electric potential that equal to $V = \frac {kq}{r}$.
But, If there is no any electric field in a point then the force must be zero because $E = \frac {F}{q}$, Since $E = 0$ then the force must be zero, So there is no any potential energy provided to any point around.
When I begin to study electric potential inside a charged hollow or solid sphere I really surprised, How is this? There is electric potential without electric field ! Can any one explain this for me?
Best Answer
Okay, I'll ellaborate my comment with the purpose of making it clearer. It's true it was very chaotic, but that's because I wanted to address many thing in few words.
I think there are many issues here. I'll try to clear them up.
1.- Potential energies can start wherever you want.
The other answers have already pointed this out with a nice example: a ball on a platform. This is because we are more used to gravitational potential energy: $E_{pg}=mgh$.
We are assuming that we are on usual situations on Earth, and we consider that $g$ doesn't vary.
So, imagine we have a ball on the ground. We would say that its potential energy is $0$ because it is at $0m$ height.
However, now someone excavates a big hole $10m$ down in the ground. What if the ball falls into the hole?
You have two options:
a) The new potential energy will be negative: $m\cdot g\cdot (-10m)$. We maintain that the ball was originally at $0m$.
b) We redefine our origin of coordinates. Now the hole's bottom is height=0. Consequently, the ball was at $h=+10m$ before falling.
Well, the thing is that both options are absolutely valid. You can choose any of those options, and the energy will keep being conserved. All physical laws and processes will behave exactly the same way.
As a consequence: we can choose where we start counting the potential energy. We can decide where we have $E_p=0$. This does not happen to kinetic energy, only potential energy.
2.- The origin of optential energies MUST in fact be DECIDED BY YOU
There is no "special point" where the potential energy (PE) has to be 0. Nothing can force you to choose any special point. It's you who must decide where to set $E_p=0$.
This means you have freedom to choose that. This can be good or bad, but that's how it is.
3.- Potentials are relative quantities, not absolute
The electric potential energy of a point charge is not
$$V=K \frac{q}{r}$$
That would be quite absolute. You cannot actually get an absolute potential. What you can obtain is potential differences. The real formula you can obtain is:
$$ V=\left( K \frac{q}{r} - K \frac{q}{r_0} \right) =Kq \left(\frac{1}{r}-\frac{1}{r_0}\right)$$
Where $r_0$ is the point you chose as reference. This means that you cannot calculate any "absolute potential", rather, you calculate "the difference of potentials between your point and the one you chose as reference". You can only caculate differences between 2 points.
And what's that $r_0$? The one you chose! Just like you choose what height is $E_{pg}=0$.
4.- Usual gauge
What we normally do is: we set $r_0$ to be at infinity. That means stablishing that $V=0$ at infinity. It also implies that $V\neq0$ unless the particle is infinitely far away from us.
If we chose this one, then the secnod term vanishes, and so
$$ V==Kq \left(\frac{1}{r}-\frac{1}{r_0}\right) \rightarrow K\frac{q}{r}$$
This is the usal formula. But you must keep in muind that "points do not have an absolute potential". The potential depends on where you chose to put $V=0$. Only if you put $V=0$ at infinity you have this formula.
You cannot use this formula if therea re charges at infinity.
By the way, check that, if you choose another $r_0$, you will have
$$ V=Kq \left(\frac{1}{r}-const\right) = Kq/r + constant $$
5.- Differences of potential are what matters.
The electric field is
$E= dV/dx$
and $V=Kq/r + constant$.
when you derive, the constant goes out. The electric field does not depend on where you chose the origin, as it must be.
In the same way, if you let a ball fall 10m high, you will have $\Delta E_p= mg\Delta h$, and that $\Delta E_p$ will become KE. Only differences of energy matter. It doesn't matter if the ball falls from 90m to 80m, or from 0m to minus 10m. Only differences of potential matter.
Same thing here. The force cannot depend on where you chose the origin, and it does not depend. Derivatives cancel the constant.
In conclusion:
I'll end with a wise sentence from a wise teacher: Energy is free, what you pay for is DIFFERENCS OF ENERGY. Do you want to have 1000 joules right now? Okay, now you have. I just moved the origin. That's free. If you wanted to CHANGE the value with a fixed origin, then that's adding a potential difference, and that's work done, that's expensive. Work costs money, energy does not, as it depends on the origin. If energy weren't free, then you can be sure that the electric company would have a term on the bill called "PE origin", and of course, it would be more and more expensive every month.