A) \[\vec{E}=\hat{i}\,(2xy+{{z}^{3}})+\hat{j}\,{{x}^{2}}+\hat{k}\,3x{{z}^{2}}\]
B) \[\vec{E}=\hat{i}\,2xy+\hat{j}({{x}^{2}}+{{y}^{2}})+\hat{k}\,(3xz-{{y}^{2}})\]
C) \[\vec{E}=\hat{i}\,{{z}^{3}}+\hat{j}\,xyz+\hat{k}\,{{z}^{2}}\]
D) \[\vec{E}\,=\,\hat{i}\,(2xy-{{z}^{3}})+\hat{j}\,x{{y}^{2}}+\hat{k}\,3{{z}^{2}}x\]
Correct Answer: A
Solution :
| [a] Key Idea 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 \[E=\frac{-dV}{dr}\] |
| \[\vec{E}=-\frac{\partial \,V}{\partial \,r}=\left[ -\frac{\partial \,V}{\partial \,x}\hat{i}-\frac{\partial \,V}{\partial \,y}\hat{j}-\frac{\partial \,V}{\partial x}\hat{k} \right]\] |
| \[=[\hat{i}\,(2xy+{{z}^{3}})+\hat{j}\,{{x}^{2}}+\hat{k}\,3x{{z}^{2}}]\] |
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