Relationship between charge and electric potential
At any point in an electric field the electric potential is the amount of electric potential energy divided by the amount of charge at that point.


Break it down, and consider the units for each quantity:


the electric potential (volt, V)
is the amount of electric potential energy (joule, J)
divided by the amount of charge at that point (coulomb, C)


Mathematically:

$$ \text{the electric potential} = \frac{\text{the amount of electric potential energy}}{\text{the amount of charge at that point}} \\
$$

or:

$$ \mathrm V = \frac{\mathrm J}{\mathrm C} $$

That is, the the definition of the volt.

It might make more sense to think of a rearrangement of this relationship:

At any point in an electric field, the amount of electric potential energy is the electric potential, multiplied by the amount of charge at that point.

$$ \text{electric potential energy} = \text{electric potential} \cdot \text{charge at that point} 
$$

$$ \mathrm J = \mathrm V \mathrm C $$

That is, voltage gives you an idea of how much potential energy you could have at some point, independent of how much charge you put there. Of course, if you are moving around a significant amount of charge, you are going to change the field, and thus the potential at each point.

Analogously, gravitational potential is measured in J/kg. Height times acceleration due to gravity yields gravitational potential, so if we assume a constant acceleration to gravity (such as approximately true for Earth's surface), then we can think about gravitational potential as height. Lifting a mass to some potential (height) takes some work, lifting twice that mass to the same potential takes twice as much work.

For another explanation, see this previous answer of mine.