# How can the electric potential be zero at a point where the electric field isn't, if that field can give a test charge kinetic energy?

If the charge comes from a point of $V_3=0$ and gains kinetic energy, then it is because the value of potential energy at the new point is **lower**. Then potential energy can be converted into kinetic energy. Don't worry about the actual potential energy values - only the **differences** in value matter.

Think of having a box on the floor. You might say that there is zero (gravitational) potential energy associated with it. But that is only because you chose to consider the floor as the reference.

- Lift the box to a shelf, and there is positive potential energy stored.
- Put it in a hole in the floor, and there is negative potential energy stored.

The values don't matter. What matters is only that some values are smaller than others. Because the box will always want to fall towards lower values. It will fall from the high shelf to the floor at zero potential energy. And it will fall from the floor at zero to the hole at negative potential energy. **It always wants to move towards lower values - the actual value doesn't matter.**

You are free to pick whichever point you want as the zero-value reference. It doesn't matter, only the difference between points matters.

The same is the case for electrical potential energies. You could place a positive charge at the shown equipotential line and say that zero (electrical) potential energy is stored. Then surely, the charge will want to move towards the neighbour locations where the potential energy stored is *less than zero*. That somebody chose the potential energy values at this particular equipotential line to be zero, doesn't matter. It could have been anything else.

This tendency to move towards lower values of potential energy is what the field lines show. At all points on the equipotential line, there are field lines showing the direction that the charge wants to move along.

In general, you should forget about the actual values of potential energies and only care about the *differences* in the value between points. This is why **voltage** is the main parameter in these cases; voltage is the difference in electrical potential between two points. Just pick whichever reference that makes it easier to work with in your specific scenarios.

If I place a positive charge in between the 2 charges, it has electric potential energy because it is feels a force towards the negative charge and repelled by the positive and gains kinetic energy.

It's true that a positive test charge feels a force towards the negative charge since a positive charge 'rolls downhill' in the (electric) potential, and the $V = 0$ line is not the lowest point in the potential. Take a look at this surface plot of an electric dipole potential:

Image credit

How can $V_3=0V$ if a charge placed in between the two charges will gain kinetic energy?

The potential is *negative* in the 'depression' - the charge has minimum potential energy at the bottom of the depression.

Now, you might wonder why the zero in potential isn't assigned to the bottom of the depression (since a global change in the potential leaves the physical electrical field unchanged). Such a global change would require the potential 'at infinity' to be non-zero. However, it seems more logical to set the potential energy of a test charge, 'infinitely' far away from the dipole, to be zero.

There seems to be a basic misunderstanding between the electric potential energy of a system of charges and the electric potential at a point due to a system of charges.

To make things easier let's assume that the zero of electric potential is at infinity and that the electric potential energy of a system of charges is zero when all the charges are infinitely far away from one another.

The diagram is a good one in that shows an important feature in that the electric field lines (yellow) are at right angles to the equipotential lines (dashed).

This means that when a test charge is moved along an equipotential line the direction of the force (the direction of the tangent to the electric field line) on that test charge is at right angles to the direction of motion to the charge and so no work needs to be done moving the test charge.

This means that moving a test charge from infinity (at zero potential) along the dashed line labelled $V_3$ requires no work to be done.

So dashed line labelled $V_3$ is at zero potential.

Now does that test charge contribute to the potential energy of the system of charges?

In this special case it does not but also in general a test charge does not contribute to the potential energy of a system of charges because the definition of an electric field at a point is $\vec E = \lim\limits _{q\to 0} \frac{\vec F}{q}$ where $\vec F$ is the force on test charge $q$ as it tends to zero.

Going back and now considering the potential energy of the system of charges.

The work done to assemble the $+q$ and $-$ charges from infinity to a separation $r$ is $- \frac{kq^2}{r^2}$ and this is the potential energy of that system of two charges.

If one brings in another charge $Q$ along line $V_3$ and places it exactly between the othe rtwo charges the potential energy of the system of three charges is still $- \frac{kq^2}{r^2}$.

This might seem rather strange but perhaps better understood if one assembled the tree charges in a different way.

First just have charge $+q$ present and bring up charge $Q$ from infinity to be $\frac r2$ away from it.

The work done to do this is $\frac{2kQq}{r}$.

Now bring up charge $-q$ from infinity to be $r$ away from charge $+q$ with charge $Q$ midway between them.

The work done to do this is $-\frac{kq^2}{r} - \frac{2kQq}{r}$.

So the total work done to assemble the three charges is $\frac{2kQq}{r}-\frac{kq^2}{r} - \frac{2kQq}{r}= - \frac{kq^2}{r}$ which is the potential energy of the system of three charges is the same as before.

Note that the equipotential line $V_3$ is a special case and bringing a charge $Q$ to a position on equipotential line $V_2$ and the potential energy of the system of three charges will now be $QV_2 - \frac{kq^2}{r^2}$.