Think of a parallel plate capacitor with potential difference of 12V. If you place a test charge in between the electric field will accelerate it to the (-) end. Work is done here by the Electric field on the charge.
You know W = F x d
and E = F / q
so W = Eq x d
this is potential energy and the units will be in joules
By definition electric potential (voltage) is PE / q [Volts]
Voltage is potential difference of a test charge. It will be high near the (+) end of the plate and low near the (-)
It is difficult to really explain this without a diagram so look at a capacitor and see what I have written above. Hope that helps!