Question

The potential at a point $x$ (measured in pm) due to some charges situated on the $x$ -axis is given by
$V(x)=\frac{20}{x^2-4}$ volt
The electric field $E$ at $x=4\mu m$ is given by

• A. $\left(\frac{10}{9}\right)volt/\mu m$ in the $+vex-$ direction
• B. $\left(\frac{5}{3}\right)volt/\mu m$ in the $-vex-$ direction
• C. $\left(\frac{5}{3}\right)volt/\mu m$ in the $+vex-$ direction
• D. $\left(\frac{20}{9}\right)volt/\mu m$ in the $-vex-$ direction

Answer 1

Answer: (A)

Given, $V(x)=\frac{20}{(X^2-4)}$ volt
We know that
$E=-\frac{dV}{dx}$
$=-\frac{d}{dx}\left[\frac{20}{\left(X^2-4\right)}\right]$
$=\frac{20\times (2x)}{(X^2-4{)}^2}$ volt
At $x=4\mu m$
$v=\frac{20\times (2\times 4)}{[(4{)}^2-4]}=\frac{10}{9}V\mu m$
Since, $V$ decreases as $x$ increases, it follows from
$E=-\frac{dV}{dx}$ that $E$ is along the $+$ ve $x$-direction.