Question

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

• A. $\frac{5}{2}V/\mu m$ and in the - ve $x$-direction
• B. $\frac{5}{2}V/\mu m$ and in the + ve $x$-direction
• C. $\frac{10}{9}V/\mu m$ and in the - ve $x$-direction
• D. $\frac{10}{9}V/um$ and in the $+$ ve $x$-direction

Answer 1

Answer: (D)

$E=-\frac{\partial V^{\wedge }}{\partial x}i-\frac{\partial V^{\wedge }}{\partial y}-\frac{\partial V}{\partial z}\hat{k}$
$\Rightarrow E_x=-\frac{\partial V}{\partial x}=-\frac{d}{dx}\left[\frac{20}{x^2-4}\right]=\frac{40x}{{\left(x^2-4\right)}^2}$
$\Rightarrow E_x\text{ at }x=4\mu m=\frac{10}{9}V/\mu m$
$\text{ and is along + ve }x\text{-direction. }$