Question

In a region, the potential is represented by $V(x,y,z)=6x-8xy-8y+6yz$, where $V$ is in $\text{Volts}$ and $x,y,z$ are in $\text{meters}$. The electric force experienced by a charge of $2\text{coulombs}$ situated at the point $(1,1,1)$ is:-

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

Answer 1

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

Given:
$V(x,y,z)=6x-8xy-8y+6yz$

$\therefore \frac{dV}{dx}=6-8y;\frac{dV}{dy}=-8x-8+6z$

$\frac{dV}{dz}=6y$

Now we know that,

$\overrightarrow{E}=-\frac{dV}{dx}\hat{i}-\frac{dV}{dy}\hat{j}-\frac{dV}{dz}\hat{k}$

After substituting the value, we get

$\overrightarrow{E}=-(6-8y)\hat{i}-(-8x-8+6z)\hat{j}-6y\hat{k}$

So at $(1,1,1)\Rightarrow x=1,y=1,z=1$

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

$\Rightarrow \left|\overrightarrow{E}\right|=\sqrt{2^2+{10}^2+6^2}=\sqrt{140}=2\sqrt{35}$

So force experienced by charge is

$F=Q\left|\overrightarrow{E}\right|=2\times 2\sqrt{35}=4\sqrt{35}\text{Newton}$

Hence, option $\left(d\right)$ is the correct answer.