Question: What is the magnitude of the electric field at the point $\left(3.00\hat{i}-2.00\hat{j}+4.00\hat{k}\right)\mathbf{\text{m}}$?

If the electric potential in the region is given by$\mathit{V}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{x}\mathit{y}{\mathit{z}}^{\mathbf{2}}$, where Vis in volts and coordinates x, y, and zare in meters?

The electric potential in the region,$V=2xy{z}^2$

Using the given data of the electric potential and the given distance in the equation of the electric field, we can get the magnitude and the directional value of the electric field at the given point.

Formulae:

The expression of the electric field due to the potential difference,

$E=-\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial Z}\hat{k}\right)$ (i)

The magnitude of a vector, $\left|\overrightarrow{E}\right|=\sqrt{{E_x}^2+{E_y}^2+{E_z}^2}$(ii)

Substituting the given value of the electric potential in equation (i), we can get the value of the electric field as follows:

$$\begin{gathered} E=-\left(\frac{\partial \left(2.00xy{z}^2\right)}{\partial x}\hat{i}+\frac{\partial \left(2.00xy{z}^2\right)}{\partial y}\hat{j}+\frac{\partial \left(2.00xy{z}^2\right)}{\partial Z}\hat{k}\right) \\ =-\left(2.00y{z}^2\right)\hat{i}-\left(2.00x{z}^2\right)\hat{j}-\left(4.00xyz\right)\hat{k} \\ =-\left(2.00\left(-2.00\text{m}\right){(4.00\text{m})}^2\right)\hat{i}-\left(2.00\left(3.00\text{m}\right){(4.00\text{m})}^2\right)\hat{j} \\ -\left(4.00\left(3.00\text{m}\right)\left(-2.00\text{m}\right)\left(4.00\text{m}\right)\right)\hat{k} \\ =\left(64\text{N/C}\right)\hat{i}-\left(96\text{N/C}\right)\hat{j}+\left(96\text{N/C}\right)\hat{k} \end{gathered}$$

Thus, the magnitude of the electric field can be calculated using equation (ii) as follows:

$$\begin{gathered} \left|E\right|=\sqrt{{\left(64\text{N/C}\right)}^2+{(-96\text{N/C})}^2+{\left(96\text{N/C}\right)}^2} \\ =150.1\text{N/C} \end{gathered}$$

Therefore, the magnitude of the electric field is 150.1 N/C.