Question

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

Short answer

The magnitude is $\left|\overrightarrow{E}\right|=\boxed{\mathbf{20}\mathbf{}\mathit{k}\mathit{V}\mathbf{/}\mathit{m}}$and direction of the electric field is $-135°$with the $x$-axis.

Step-by-step solution

Given information and formula used  

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,

$E=\frac{∆V}{d}$

Determining the magnitude and direction of the electric field at the dot in the figure 

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.