The electric potential at a point $(x, y, z)$ is given by $V=-x^2 y-x z^3+4$ The electric field $\vec{E}$ at…
The electric potential at a point $(x, y, z)$ is given by
$V=-x^2 y-x z^3+4$
The electric field $\vec{E}$ at that point is
$\vec{E}=\hat{i}\left(2 x y+z^3\right)+\hat{j} x^2+\hat{k} 3 x z^2$
$\vec{E}=\hat{i} 2 x y+\hat{j}\left(x^2+y^2\right)+\hat{k}\left(3 x z-y^2\right)$
$\vec{E}=\hat{i} z^3+\hat{j} x y z+\hat{k} z^2$
$\vec{E}=\hat{i}\left(2 x y-z^3\right)+\hat{j} x y^2+\hat{k} z^2 x$
Solution
Key Idea Electric field at a point is equal to the negative gradient of the electrostatic potential at that point.
Potential gradient relates with electric field according to the following relation $E=\frac{-d V}{d r}$
$\begin{aligned}
\vec{E} & =-\frac{\partial V}{\partial r}=\left[-\frac{\partial V}{\partial x} \hat{i}-\frac{\partial V}{\partial y} \hat{j}-\frac{\partial V}{\partial x} \hat{k}\right] \\
& =\left[\hat{i}\left(2 x y+z^3\right)+\hat{j} x^2+\hat{k} 3 x z^2\right]
\end{aligned}$