Q. The electric potential at a point $(x,y,z)$ is given by $V = - x^{2}y - xz^{3} + 4$ The electric field $\overrightarrow{E}$ at that point is
A) $\overrightarrow{E} = \widehat{i}\left( 2xy + z^{3} \right) + \widehat{j}x^{2} + \widehat{k}3xz^{2}$
B) $\overrightarrow{E} = \widehat{i}\ 2xy + \widehat{j}\left( x^{2} + y^{2} \right) + \widehat{k}(3xz - y^{2})$
C) $\overrightarrow{E} = \widehat{i}\ z^{3} + \widehat{j}\ xyz + \ \widehat{k}\ z^{2}$
D) $\overrightarrow{E} = \widehat{i}\left( 2xy - z^{3} \right) + \widehat{j}\ xy^{2} + \widehat{k}3z^{2}x$
Explanation
(a) $E_{x} = - \frac{dV}{dx} = - \left\lbrack - 2xy - z^{3} \right\rbrack$
$E_{y} = - \frac{dV}{dy} = - \left\lbrack - x^{2} \right\rbrack$
$E_{z} = - \frac{dV}{dz} = - \left\lbrack - 3xz^{2} \right\rbrack$
$\therefore\overrightarrow{E} = E_{x}\widehat{i} + E_{y}\widehat{j} + E_{z}\widehat{k} = \left( 2xy + z^{3} \right)\widehat{i} + x^{2}\widehat{j} + 3xz^{2}\widehat{k}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.