If potential (in volts) in a region is expressed as $V(x, y, z)=6 x y-y+2 y z$, the electric field (

Question

If potential (in volts) in a region is expressed as $V(x, y, z)=6 x y-y+2 y z$, the electric field (in N/C) at point $(1,1,0)$ is

Professor PC Thomas Classes · Answer

The electric field $\vec{E}$ and potential $V$ in a region are related as $\vec{E}=-\left[\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}+\frac{\partial V}{\partial z} \hat{k}ight]$
Here, $V(x, y, z)=6 x y-y+2 y z$
$$
\begin{aligned}
	herefore \vec{E}=-\left[\frac{\partial}{\partial x}(6 x y-y+2 y z) \hat{i}+\frac{\partial}{\partial y}(6 x yight. & -y+2 y z) \hat{j} \
& \left.+\frac{\partial}{\partial z}(6 x y-y+2 y z) \hat{k}ight]
\end{aligned}
$$
$$
=-[(6 y) \hat{i}+(6 x-1+2 z) \hat{j}+(2 y) \hat{k}]
$$

At point ( $1,1,0$ ),
$\vec{E}=-[(6(1)) \hat{i}+(6(1)-1+2(0)) \hat{j}+(2(1)) \hat{k}]=-(6 \hat{i}+5 \hat{j}+2 \hat{k})$