Question

If potential (in volts) in a region is expressed as $V(x,y,z)=6xy-y+2yz$, the electric field $\left(\frac{N}{C}\right)$ at point $(1,1,0)$ is

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

Answer 1

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

Given that $V=6xy-y+2yz$
Partially differentiating this with respect $x,y$ and $z$ separately
we get
${\overrightarrow{E}}_x=\frac{-\partial V}{\partial x}=-6y\hat{i}$
${\overrightarrow{E}}_y=\frac{-\partial V}{\partial y}=(-6x+1-2z)\hat{j}$
${\overrightarrow{E}}_z=\frac{-\partial V}{\partial z}=-2y\hat{k}$
$\overrightarrow{E}=-6y\hat{i}+(-6x+1-2z)\hat{j}+(-2y)\hat{k}........\left(i\right)$
Given point $(1,1,0)$ represents $(x,y,z)$
Putting these values in eq$\left(i\right)$
we get
$\overrightarrow{E}=-6\hat{i}-5\hat{j}-2\hat{k}$