Q. The potential at a point $x$ (measured in $\mu m$) due to some charges situated on the $x$-axis is given by $V(x) = 20/(x^{2} - 4)\ Volts.$ The electric field $E$ at $x = 4\mu m$ is given by
A) $5/3\ Volt/\mu m$ and in the $- ve\ x$ direction
B) $5/3\ Volt/\mu m$ and in the $+ ve\ x$ direction
C) $10/9\ Volt/\mu m$ and in the $- ve\ x$ direction
D) $10/9\ Volt/\mu m$ and in the $+ ve\ x$ direction
Explanation
(d) $V_{x} = \frac{20}{x^{2} - 4}$
$E = \frac{- dV}{dx} = \frac{20}{\left( x^{2} - 4 \right)^{2}}(2x - 0) = \frac{160}{144} = \frac{10}{9}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.