Question

If potential (in volts) in a region is expressed as $V(x,y,z)$ = $6xy-y+2yz$, the electric field $(inN/C)$ 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

Answer: (D)

The electric field, $\overrightarrow{E}$ and potential V in a region are related as
$\overrightarrow{E}=-[\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}]$
Here, $V(x,y,z)=6xy-y+2yz$
$\therefore \overrightarrow{E}=-[\frac{\partial }{\partial x}(6xy-y+2yz)\hat{i}+\frac{\partial }{\partial y}(6xy-y+2yz)\hat{j}$
$+\frac{\partial }{\partial z}(6xy-y+2yz)\hat{k}]$
$=-[(6(y))\hat{i}+(6x-1+2z)\hat{k}+(2(y))\hat{k}]$
At point (1, 1,0),
$\overrightarrow{E}=-[(6(1))\hat{i}+(6(1)-1+2(0))\hat{j}+(2(1))\hat{k}]$
$=-(6\hat{i}+5\hat{j}+2\hat{k})$