Question

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 meters. The electric force experienced by a charge of $2$
coulomb situated at point $(1,1,1)$ is

• A. $6\sqrt{5}\text{N}$
• B. $30\text{N}$
• C. $24\text{N}$
• D. $4\sqrt{35}\text{N}$

Answer 1

The correct option is D $4\sqrt{35}\text{N}$

Relation between Electric field and potential gradient is
$\overrightarrow{E}=-\frac{\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial z}\hat{k}$

$\overrightarrow{E_x}=-\frac{\partial V}{\partial x}\hat{i}=-(6-8y)\hat{i}$
$\overrightarrow{E_y}=-\frac{\partial V}{\partial y}\hat{j}=-(-8x-8+6z)\hat{j}$
$\overrightarrow{E_z}=-\frac{\partial V}{\partial z}\hat{k}=-\left(6y\right)\hat{k}$
$\overrightarrow{E}=\overrightarrow{E_x}+\overrightarrow{E_y}+\overrightarrow{E_z}$
$\overrightarrow{E}=-\left[\right(6-8y)\hat{i}+(-8x-8+6z)\hat{j}+(6y\left)\hat{k}\right]$

At $(1,1,1),\overrightarrow{E}=2\hat{i}+10\hat{j}-6\hat{k}$

Magnitude of Electric Field is
$\left|\overrightarrow{E}\right|=\sqrt{2^2+{10}^2+6^2}=\sqrt{140}=2\sqrt{35}$

Force$=qE=2\times 2\sqrt{35}=4\sqrt{35}\text{N}$