Question

Determine the magnitude and direction of the electric field at points 1 and 2 in Figure EX26.10.

Short answer

The magnitude and direction of the electric field

a) At point 1 : $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathit{k}\mathit{V}\mathbf{/}\mathit{m}$, Downwards

b) At point 2 :$\mathbf{5}\mathbf{}\mathit{k}\mathit{V}\mathbf{/}\mathit{m}$, Upwards

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 will be diminishing in direction, thus point 1 will be down and point 2 will be up.

The magnitude will be the difference in potential between the two adjacent surfaces divided by their distance, assuming the field is constant between the dotted equipotential surfaces. That is,

$\left|E_1\right|=\frac{75-25}{0.02}=\boxed{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathit{k}\mathit{V}\mathbf{/}\mathit{m}}$for point 1, and

$\left|E_2\right|=\frac{75-25}{0.01}=\boxed{\mathbf{5}\mathbf{}\mathit{k}\mathit{V}\mathbf{/}\mathit{m}}$for point 2.