A) \[\vec{E}=\hat{i}\left( 2xy+{{z}^{3}} \right)+\hat{j}{{x}^{2}}+\hat{k}3x{{z}^{2}}\]
B) \[\vec{E}=i\,2xy+\hat{j}\left( {{x}^{2}}+{{y}^{2}} \right)+\hat{k}\left( 3xz-{{y}^{2}} \right)\]
C) \[\vec{E}=\hat{i}{{z}^{3}}+\hat{j}\,xyz+\hat{k}{{z}^{2}}\]
D) \[\vec{E}=\hat{i}\left( 2xy-{{z}^{3}} \right)+\hat{j}x{{y}^{2}}+\hat{k}3{{z}^{2}}x\]
Correct Answer: A
Solution :
Electric field at a point is equal to the negative gradient of the electrostatic potential at that point. Potential gradient relates with electric field according to the following relation \[E=\frac{-dV}{dr}\] \[\vec{E}=-\frac{\partial V}{\partial x.}\left[ -\frac{\partial V}{\partial x.}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial x}\hat{k} \right]\] \[=\left[ \hat{i}\left( 2xy+{{z}^{3}} \right)+\hat{j}{{x}^{2}}+\hat{k}3x{{z}^{2}} \right]\]
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