The electric potential at a point $(x, y, z)$ is given by $\mathrm{V}=-x^2 y-x z^3+4$ The electric field…

The electric potential at a point $(x, y, z)$ is given by $\mathrm{V}=-x^2 y-x z^3+4$ The electric field $\overrightarrow{\mathrm{E}}$ at that point is :
  1. $\vec{E}=\hat{i}\left(2 x y-z^3\right)+\hat{j} x y^2+\hat{k} 3 z^2 x$
  2. $\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}}\left(2 x y+\mathrm{z}^3\right)+\hat{\mathrm{j}} \mathrm{x}^2+\hat{\mathrm{k}} 3 \mathrm{x} \mathrm{z}^2$
  3. $\vec{E}=\hat{i} 2 x y+\hat{j}\left(x^2+y^2\right)+\hat{k}\left(3 x z-y^2\right)$
  4. $\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} z+\hat{\mathrm{j}} x y z+\hat{\mathrm{k}} z^2$

Solution

$\begin{aligned} V & =-x^2 y-x z^3+4 \\ \vec{E} & =-V=-\left(\hat{i} \frac{\delta}{\delta x}+\hat{j} \frac{\delta}{\delta y}+\hat{k} \frac{\delta}{\delta z}\right) \\ & \left.=\left(2 x y+z^3\right) \hat{i}+x^2 \hat{j}+3 x z^2+4\right) \end{aligned}$

Asked in: NEET 2009 (Mains)

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