Question

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

• A. $\vec{E}=\hat{i}\left(2 x y+z^{3}\right)+\hat{j} x^{2}+\hat{k} 3 x z^{2}$
• B. $\vec{E}=\hat{i} 2 x y+\hat{j}\left(x^{2}+y^{2}\right)+\hat{k}\left(3 x z-y^{2}\right)$
• C. $\vec{E}=\hat{i} z^{3}+\hat{j} x y z+\hat{k} z^{2}$
• D. $\vec{E}=\hat{i}\left(2 x y-z^{3}\right)+\hat{j} x y^{2}+\hat{k} 3 z^{2} x$

Answer 1

Answer: (A)

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:
$E=\frac{-d V}{d r}$
$\vec{E}=-\frac{\partial V}{\partial r}$
$=\left[-\frac{\partial V}{\partial x} \hat{i}-\frac{\partial V}{\partial y} \hat{j}-\frac{\partial V}{\partial x} \hat{k}\right]$
$=\hat{i}\left(2 x y+z^{3}\right)+\hat{j} x^{2}+\hat{k} 3 x z^{2}$