Q: 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 :

(a) $\displaystyle (-6 \hat{i} + 9 \hat{j} + \hat{k} ) $

(b) $\displaystyle (-3 \hat{i} + 5 \hat{j} + 3\hat{k} ) $

(c) $\displaystyle -(6 \hat{i} + 5 \hat{j} + 2\hat{k} ) $

(d) $\displaystyle (-2 \hat{i} + 3 \hat{j} + \hat{k} ) $

**Click to See Answer : **

$\displaystyle E_x = – \frac{\partial}{\partial x}(6xy-y + 2yz) = -6 y $

$\displaystyle E_y = – \frac{\partial V}{\partial y} $

$\displaystyle E_y = – \frac{\partial}{\partial y}(6xy-y + 2yz) = -(6 x – 1 + 2 z) $

$\displaystyle E_z = – \frac{\partial V}{\partial z} $

$\displaystyle E_z = – \frac{\partial}{\partial z}(6xy-y + 2yz) = -2 y $

$\displaystyle \vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k} $

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

$\displaystyle \vec{E}_{(1,1,0)} = -6 \hat{i} – 5 \hat{j} – 2 \hat{k} $

$\displaystyle \vec{E}_{(1,1,0)} = -( 6 \hat{i} + 5 \hat{j} + 2 \hat{k}) $