91Ó°ÊÓ

1. The potential does not depend on z.
2. The scale of the vertical axis,$V_s=500\text{V}$
3. The coordinates of the point,$(x,y)=(0.4\text{m,0.5m)}$

Using the concept of the electric potential, we can get the unit-vector notation of the electric field by using the given data. Now, using the same electric field, we can get the electrostatic force value through an electric field and force relation using the charge value.

Formulae:

The equation for the electric field as x-and y-coordinates related to electric potential is given by, $\stackrel{→}{E}=-\frac{∂V}{∂x}\hat{i}-\frac{∂V}{∂y}\hat{j}$ (i)

In this case, the potential does not depend on z.

The electrostatic force of the particle, $\stackrel{→}{F}=q\stackrel{→}{E}$ (ii)

Substitute the values of potential and distance using the above graphs in the above equation (i), we get the electric field in unit vector notation as follows:

$$\begin{gathered} \stackrel{→}{E}=-\left(\frac{-2V_s}{0.4\text{m}}\right)\hat{i}-\left(\frac{V_s}{0.5\text{m}}\right)\hat{j} \\ =-\left(\frac{-2\left(500\text{V}\right)}{0.4\text{m}}\right)\hat{i}-\left(\frac{500\text{V}}{0.5\text{m}}\right)\hat{j} \\ =\left(2500\hat{i}-1000\hat{j}\right)\text{V/m} \end{gathered}$$

The equation for the electric force on the electron due to the electric field is givenusing equation (ii) as follows:

$$\begin{gathered} \stackrel{→}{F}=\left(-1.6×{10}^{-19}C\right)\left[\left(2500\hat{i}-1000\hat{j}\right)\left(\text{V/m}\right)\right] \\ =-\left(4×{10}^{-16}\text{N}\right)\hat{i}+\left(1.6×{10}^{-16}\text{N}\right)\hat{j} \end{gathered}$$

Therefore, the electric force on the electron is $-\left(4×{10}^{-16}\text{N}\right)\hat{i}+\left(1.6×{10}^{-16}\text{N}\right)\hat{j}$.