Q. The electric potential $V$ at any point $O(x,y,z$ all in metres) in space is given by $V = 4x^{2}\ volt.$ The electric field at the point $(1m,\ 0,\ 2m)$ in $volt/metre$ is
A) 8 along negative $C - axis$
B) 8 along positive $X - axis$
C) 16 along negative $X - axis$
D) 16 along positive $Z - axis$
Explanation
(a) The electric potential $V(x,\ y,\ z) = 4x^{2}volt$
Now $\overrightarrow{E} = - \left( \widehat{i}\frac{\partial V}{\partial x} + \widehat{j}\frac{\partial V}{\partial y} + \widehat{k}\frac{\partial V}{\partial z} \right)$
Now $\frac{\partial V}{\partial x} = 8x,\ \frac{\partial V}{\partial y} = 0$ and $\frac{\partial V}{\partial z} = 0$
Hence $\overrightarrow{E} = - 8x\widehat{i}$, so at point $(1m,\ 0,\ 2m)$
$E = - 8\widehat{i}\ volt/metre$ or 8 along negative $X$-axis
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.