Question

If potential (in $\text{(in Volt)}$ in a region is expressed as,
$V(x,y,z)=6xy-y+2yz$
The electric field $\text{(in N/C)}$ at point $(1,1,0)$ is

• A. $-(6\hat{i}+9\hat{j}+\hat{k})$
• B. $-(3\hat{i}+5\hat{j}+3\hat{k})$
• C. $-(6\hat{i}+5\hat{j}+2\hat{k})$
• D. $-(2\hat{i}+3\hat{j}+\hat{k})$

Answer 1

The correct option is C $-(6\hat{i}+5\hat{j}+2\hat{k})$

Given:
$V(x,y,z)=6xy-y+2yz$

As we know that

$\overrightarrow{E}=-\frac{\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial z}\hat{k}$

Now,

$\frac{\partial V}{\partial x}=6y;\frac{\partial V}{\partial y}=6x-1+2z$

$\frac{\partial V}{\partial z}=2y$

So the electric field become

$\overrightarrow{E}=-6y\hat{i}-(6x-1+2z)\hat{j}-2y\hat{k}$

At $(1,1,0)\Rightarrow x=1,y=1,z=0$

$\overrightarrow{E}=-6\hat{i}-5\hat{j}-2\hat{k}$

Hence, option $\left(c\right)$ is correct.