Q: The electric potential at a point (x , y , z) is given by V = -x^{2} y – xz^{3} + 4 . The electric field at that point is

$\displaystyle (a) \vec{E} = (2 xy-z^3 )\hat{i} + xy^2 \hat{j} + 3 z^2 x \hat{k} $

$\displaystyle (b) \vec{E} = (2 xy-z^3 )\hat{i} + x^2 \hat{j} + 3x z^2\hat{k} $

$\displaystyle (c) \vec{E} = (2 xy )\hat{i} + (x^2 + y^2)\hat{j} + (3 xz – y^2 )\hat{k} $

$\displaystyle (d) \vec{E} = z^3 \hat{i} + xyz\hat{j} + 3 z^2 \hat{k}$

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^{2}y – xz

^{3}+ 4

$\displaystyle \vec{E} = -\hat{i}\frac{\partial V}{\partial x} -\hat{j}\frac{\partial V}{\partial y} -\hat{k}\frac{\partial V}{\partial z}$

$\displaystyle \vec{E} = -(-2xy-z^3)\hat{i} -(-x^2) \hat{j} -(-3z^2 x)\hat{k} $

$\displaystyle \vec{E} = (2xy+z^3)\hat{i} + (x^2)\hat{j} + (3z^2 x)\hat{k} $