Answer
$E_1 = -10~V/m$
$E_2 = -10~V/m$
Work Step by Step
We can see that the electric potential increases by 10 V for each 1 meter increase in position. This is a graph we would see with a uniform electric field.
We can find the electric field at all points on the graph:
$E = -\frac{\Delta V}{\Delta x}$
$E = -\frac{40~V}{4~m}$
$E = -10~V/m$
Therefore, $E_1 = -10~V/m$ and $E_2 = -10~V/m$
That is, the uniform electric field has a magnitude of $10~V/m$ and it is directed in the -x direction.
