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If potential (in volts) in a region is expressed as $V (x,y,z) =6xy-y+2yz $ the electric field (in $N/C$) at point $(1, 1, 0)$ is
$-(6$$\hat i + 9\hat j + \hat k)$
$-(3$$\hat i + 5\hat j + 3\hat k)$
$-(6$$\hat i + 5\hat j + 2\hat k)$
$-(2$$\hat i + 3\hat j + \hat k)$
Solution
Given, $V=6 x y-y+2 y z$
$\vec{E}=\left[\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}+\frac{\partial V}{\partial z} \hat{k}\right]$
$\vec{E}=-\left\{\frac{\partial}{\partial x}[6 x y-y+2 y z] \hat{i}+\frac{\partial}{\partial y}[6 x y-y+2 y z] \hat{j}\right.$
$\left.+\frac{\partial}{d z}[6 x y-y+2 y z) \hat{k}\right\}$
$=-\{(6 y) \hat{i}+(6 x-1+2 z) \hat{j}+(2 y) \hat{k}\}$
$\vec{E}_{1,1,0}=-\{(6 x 1) \hat{i}+(6 \times 1-1+2 \times 0) \hat{j}+(2 \times 1) \hat{k}]$
$=-(6 \hat{i}+5 \hat{j}+2 \hat{k})$
