Electrostatic Potential and Capacitance - Result Question 10
11. The electric potential at a point $(x, y, z)$ is given by $V=-x^{2} y-x z^{3}+4$. The electric field $\vec{E}$ at that point is
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(a) $\overrightarrow{{}E}=\hat{i} 2 x y+\hat{j}(x^{2}+y^{2})+\hat{k}(3 xz-y^{2})$
(b) $\overrightarrow{{}E}=\hat{i} z^{3}+\hat{j} x y z+\hat{k} z^{2}$
(c) $\overrightarrow{{}E}=\hat{i}(2 x y-z^{3})+\hat{j} x y^{2}+\hat{k} 3 z^{2} x$
(d) $\overrightarrow{{}E}=\hat{i}(2 x y+z^{3})+\hat{j} x^{2}+\hat{k} 3 x z^{2}$
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Answer:
Correct Answer: 11. (d)
Solution:
- (d) The electric field at a point is equal to negative of potential gradient at that point.
$ \begin{aligned} \overrightarrow{{}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}] \\ & =[(2 x y+z^{3}) \hat{i}+\hat{j} x^{2}+\hat{k} 3 x z^{2}] \end{aligned} $
