It is an interesting question:
Firstly, the Electric field "$E$" is the slope of the potential, i.e., $E=-{\frac{dV}{dx}}$. Therefore "constant electric field" means the potential is either increasing or decreasing at a constant rate (along the space).
Secondly, $E$ is a physically measurable quantity but $V$ is not. You can never know the absolute value of potential $V$, although you know the value of $E$ which is constant in your case. It is because,
$V=-\int Edx$ $=$$-Ex+C$
Note that we have an integrating constant "$C$", i.e., you can always add any constant value with your solution for the potential. It further means $V$ must be measured with respect to some reference and the measured value always depends on that reference.
An example: Let's consider a parallel plate capacitor in which Potential of one plate is $V_1$ and another plate is $V_2$ and they are located at position $x_1$ and $x_2$ respectively. Also let the distance between two plates, $dx=x_2-x_1$=2cm, and electric field (which is ideally constant in such capacitor) $E=5V/cm$.
What is the value of potential that satisfies the case? Lets check,
if $V_1=0V$ and $V_2=10V$ then $E={\frac{dV}{dx}}={\frac{V_2-V_1}{dx}}={\frac{10-0}{2}}=5V/cm$
Now lets add some constant value with the potentials. i.e.,
if $V_1=5V$ and $V_2=15V$ then $E={\frac{V_2-V_1}{dx}}={\frac{15-5}{2}}=5V/cm$
if $V_1=10V$ and $V_2=20V$ then $E={\frac{V_2-V_1}{dx}}={\frac{20-10}{2}}=5V/cm$
See!!!! All of those set of ($V_1$,$V_2$) is giving the same "Constant" electric field. Therefore you can know the potential of one plate with respect to the potential of another, but can never know the absolute value of both.
This is true not only for capacitor but also for every other cases. At this point the potential of "Earth" is usually taken to be the reference in order to measure the potential at any point. It is because earth is a very large object and it's absolute potential does not get affected by any of our action.
