Q. The electric potential $V$ is given as a function of distance $x$(metre) by $V = \left( 5x^{2} + 10x - 9 \right)volt.$ Value of electric field at $x = 1$ is
A) $- 20V/m$
B) $6V/m$
C) $11V/m$
D) $- 23V/m$
Explanation
(a) $E = - \frac{dV}{dx} = - \frac{d}{dx}\left( 5x^{2} + 10x - 9 \right) = - 10x - 10$
$\therefore(E)_{x = 1} = - 10 \times 1 - 10 = - 20V/m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.