Question

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

• A. $\frac{5}{3}V/\mu m$ and in the negative x direction
• B. $\frac{5}{3}V/\mu m$ and in the positive x direction
• C. $\frac{10}{9}V/\mu m$ and in the negative x direction
• D. $\frac{10}{9}V/\mu m$ and in the positive x direction

Answer 1

Answer: (D)

$\frac{10}{9}V/\mu m$ and in the positive x direction

Given $V\left(x\right)=\frac{20}{x^2-4}$

We know, $E=-\frac{dV}{dx}$

$\Rightarrow E=-\frac{d}{dx}\left(\frac{20}{x^2-4}\right)$

$\Rightarrow E=-20\frac{-1\left(2x\right)}{(x^2-4{)}^2}$

$\Rightarrow E=\frac{40x}{(x^2-4{)}^2}$

$\Rightarrow E=\frac{40\times 4\mu m}{12\times 12}$

$\Rightarrow E=\frac{10}{9}V/\mu m$