A) \[\vec{E}=\hat{i}2xy+\hat{j}({{x}^{2}}+{{y}^{2}})+\hat{k}(3xz-{{y}^{2}})\]
B) \[\vec{E}=\hat{i}{{z}^{3}}+\hat{j}xyz+\hat{k}{{z}^{2}}\]
C) \[\vec{E}=\hat{i}(2xy-{{z}^{3}})+\hat{j}x{{y}^{2}}+\hat{k}3{{z}^{2}}x\]
D) \[\vec{E}=\hat{i}(2xy+{{z}^{3}})+\hat{j}{{x}^{2}}+\hat{k}3x{{z}^{2}}\]
Correct Answer: D
Solution :
[d] The electric field at a point is equal to negative of potential gradient at that point. \[\vec{E}=-\frac{\partial V}{\partial r}=\left[ -\frac{\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial z}\hat{k} \right]\] \[=\left[ (2xy+{{z}^{3}})\hat{i}+\hat{j}{{x}^{2}}+\hat{k}\,\,3x{{z}^{2}} \right]\]
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