Answer
At point 1, the magnitude of the electric field is 2500 V/m and the direction is straight down.
At point 2, the magnitude of the electric field is 5000 V/m and the direction is straight up.
Work Step by Step
Note that the electric field points from places of higher potential to places of lower potential.
Near point 1, we can see that the potential changes by 25 V when the position changes vertically by 1 cm. We can find $E$ at point 1:
$E = -\frac{\Delta V}{\Delta x}$
$E = -\frac{25~ V}{0.01~m}$
$E = -2500~V/m$
At point 1, the magnitude of the electric field is 2500 V/m and the direction is straight down.
Near point 2, we can see that the potential changes by 50 V when the position changes vertically by 1 cm. We can find $E$ at point 1:
$E = -\frac{\Delta V}{\Delta x}$
$E = -\frac{50~ V}{0.01~m}$
$E = -5000~V/m$
At point 2, the magnitude of the electric field is 5000 V/m and the direction is straight up.
