Electrostatics Question 786
Question: The electric potential at a point (x, y, z) is given by $ V=-x^{2}y-xz^{3}+4. $ The electric field E at that point is
Options:
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}xy^{2}+\hat{k}3z^{2}x $
D) $ \vec{E}=\hat{i}(2xy+z^{3})+\hat{j}x^{2}+\hat{k}3xz^{2} $
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Answer:
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}=[ -\frac{\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial z}\hat{k} ] $
$ =[ (2xy+z^{3})\hat{i}+\hat{j}x^{2}+\hat{k}3xz^{2} ] $
