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. $\frac{10}{9} V / u m$ 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}{d x}\left[\frac{20}{x^2-4}\right]=\frac{40 x}{\left(x^2-4\right)^2}$
$ \Rightarrow E_x \text { at } x=4 \mu m=\frac{10}{9} V / \mu m $
$\text { and is along + ve } x \text {-direction. }$