Answer
The magnitude of the electric field strength is $20,000~V/m$ and it points at an angle of $45^{\circ}$ below the -x axis.
Work Step by Step
The electric field points from positions of higher potential to positions of lower potential. Therefore, the electric field points at an angle of $45^{\circ}$ below the -x axis.
We can find the electric field strength:
$E = -\frac{\Delta V}{\Delta x}$
$E = -\frac{400~V}{0.020~m}$
$E = -20,000~V/m$
The magnitude of the electric field strength is $20,000~V/m$ and it points at an angle of $45^{\circ}$ below the -x axis.
