What are the magnitude and direction of the electric field at the dot in Figure EX26.8?

Given figure :

Theory used :

The electric field is directed towards the decreasing potential and has a magnitude equal to the rate of change of the potential.

The electric field will be equal to the potential difference divided by the distance between these potentials. That is to say,

$E=\frac{∆V}{d}$

The electric field strength will have a Magnitude of $\left|\overrightarrow{E}\right|=\boxed{\mathbf{20}\mathbf{}\mathit{k}\mathit{V}\mathbf{/}\mathit{m}}$because the potential difference reduces by $200V$over a distance of $1centimeter.$

The field will be oriented in such a way that the potential will drop in that direction.

It is evident from the diagram that this occurs in the direction perpendicular to the dotted lines, which is to the left and below. This direction is:

$\frac{\overrightarrow{E}}{\left|\overrightarrow{E}\right|}\propto -\hat{i}-\hat{j}$

Normalizing when expressed vectorially,

$\frac{\overrightarrow{E}}{\left|\overrightarrow{E}\right|}=\boxed{\mathbf{-}\frac{\sqrt{\mathbf{2}}}{\mathbf{2}}\mathbf{(}\hat{\mathbf{i}}\mathbf{+}\hat{\mathbf{j}}\mathbf{)}}$

So, the direction is $-135°$with the $x$-axis.