Q. The electric potential V at any point ($x,y,z$) in space is given by $V = 4x^{2}$V. the electric field at (1,0,2) m inVm$^{-1}$ is
A) 8, along negative X-axis
B) 8, along positive X-axis
C) 16,along negative X-axis
D) 16, along positive X-axis
Explanation
(a) $E = - \frac{dV}{dr} = - \frac{d}{dx}\left( 4x^{2} \right) = - 8x$
$= - 8\ (1) = - 8\ V\ m^{- 1}$
Negative sign indicates $\overrightarrow{E}$ is along negative direction of $X$-axis.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.