The electric potential at a point (x, y, z) is given by V=-x^2 y-x z^3+4 The electric field E at that point is :

Correct answer: B. E = i (2 x y+z^3)+ j x^2+ k 3 xz z^2 aligned V & =-x² y-x z³+4 E & =-V=-( i ( )/( x)+ j ( )/( y)+ k ( )/( z)) & .=(2 x y+z³) i +x² j +3 x z²+4) aligned

NEET 2009

Exam: NEET
Year: 2009
Subject: Physics
Chapter: Electrostatics

NEET 2009 · Physics · Electrostatics

The electric potential at a point (x, y, z) is given by V=-x^2 y-x...

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 :

Options

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

Correct answer

B. $\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}}\left(2 x y+\mathrm{z}^3\right)+\hat{\mathrm{j}} \mathrm{x}^2+\hat{\mathrm{k}} 3 \mathrm{xz} \mathrm{z}^2$

Step-by-step solution

\begin{aligned} V & =-x² y-x z³+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+z³\right) \hat{i}+x² \hat{j}+3 x z²+4\right) \end{aligned}

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