The electric potential at a point (x, y, z) is given by V=-x² y-x...

The electric potential at a point

(x, y, z)

is given by V=-x² y-x z³+4 The electric field

\vec{E}

at that point is

Options

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 $\vec{E}=\hat{i}\left(2 x y-z^3\right)+\hat{j} x y^2+\hat{k} z^2 x$

Correct answer

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

Step-by-step 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+z³\right)+\hat{j} x²+\hat{k} 3 x z²\right] \end{aligned}

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