Question

The electric potential at a point $\left(x , y , z\right)$ is given by: $V=-x^{2}y-xz^{3}+4$
The electric field $\overset{ \rightarrow }{E}$ at that point is:-

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

Answer 1

Answer: (B)

$V=-x^{2}y-xz^{3}+4$
$E_{x}=\frac{- \partial V}{\partial x}=-\frac{\partial \left(- x^{2} y - x z^{3} + 4\right)}{\partial x}=+2xy+z^{3}$
$E_{y}=\frac{- \partial V}{\partial y}=-\frac{\partial \left(- x^{2} y - x z^{3} + 4\right)}{\partial y}=+x^{2}$
$E_{z}=\frac{- \partial V}{\partial z}=-\frac{\partial \left(- x^{2} y - x z^{3} + 4\right)}{\partial z}=+3xz^{2}$
$so=\overset{ \rightarrow }{E}=E_{x}\hat{i}+E_{y}\hat{j}+E_{z}\hat{k}$