Question

What are the magnitude and direction of the electric field at the dot in Figure EX26.7?

Short answer

The magnitude is $\mathbf{20}\mathbf{}\mathbf{000}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}$ and direction of the electric field isvertically downwards.

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 potential varies by$200V$over a distance of$1cm=0.01m$in our case, hence the field's magnitude is :

$E=\frac{200V}{0.01m}=\boxed{\mathbf{20}\mathbf{}\mathbf{000}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}}$

The field points downwards as the potential falls in the downward direction.

So, the Electric field points vertically downwards.