In a region, the potential is represented by | vedclass

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Question

Here, $V(x, y, z)=6 x-8 x y-8 y+6 y z$

The $x, y$ and $z$ components of electric field are

$E_{x} =-\frac{\partial V}{\partial x}=-\frac{\partia

Vedclass · Accepted Answer

$4$$\sqrt {35} $ $N$

Vedclass · Answer

Here, $V(x, y, z)=6 x-8 x y-8 y+6 y z$

The $x, y$ and $z$ components of electric field are

$E_{x} =-\frac{\partial V}{\partial x}=-\frac{\partial}{\partial x}(6 x-8 x y-8 y+6 y z)$

$=-(6-8 y)=-6+8 y$

$E_{y} =-\frac{\partial V}{\partial y}=-\frac{\partial}{\partial y}(6 x-8 x y-8 y+6 y z)$

$=-(-8 x-8+6 z)=8 x+8-6 z$

$E_{z} =-\frac{\partial V}{\partial z}=-\frac{\partial}{\partial z}(6 x-8 x y-8 y+6 y z)=-6 y $

$\vec{E} =E_{x} \hat{i}+E_{y} \hat{j}+E_{z} \hat{k}$

$=(-6+8 y) \hat{i}+(8 x+8-6 z) \hat{j}-6 y \hat{k}$

At point $(1,1,1)$

$\bar{E}=(-6+8) \hat{i}+(8+8-6) \hat{j}-6 \hat{k}=2 \hat{i}+10 \hat{j}-6 \hat{k}$

The magnitude of electric field $\vec{E}$ is

$\vec{E}=\sqrt{E_{x}^{2}+E_{y}^{2}+E_{z}^{2}}=\sqrt{(2)^{2}+(10)^{2}+(-6)^{2}}$

$=\sqrt{140}=2 \sqrt{35} \,{N} {C}^{-1}$

Electric force experienced by the charge $F=q E=2 \mathrm{C} 	imes 2 \sqrt{35} \mathrm{\,NC}^{-1}=4 \sqrt{35}\mathrm{\,N}$

In a region, the potential is represented by $V(x, y, z) = 6x - 8xy - 8y + 6yz$, where $V$ is in volts and $x, y, z$ are in metres. The electric force experienced by a charge of $2$ coulomb situated at point $( 1, 1, 1)$ is

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