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. $\left(10/9\right)volt/\mu m$ and in the $+ve$ x direction
• B. $\left(5/3\right)volt/\mu m$ and in the $-ve$ x direction
• C. $\left(5/3\right)volt/\mu m$ and in the $+ve$ x direction
• D. $\left(10/9\right)volt/\mu m$ and in the $-ve$ x direction

Answer 1

The correct option is A $\left(10/9\right)volt/\mu m$ and in the $+ve$ x direction

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

Electric field , $E=-\frac{dV}{dx}=-\frac{-20}{(x^2-4{)}^2}\left(2x\right)=\frac{40x}{(x^2-4{)}^2}$

At $x=4\mu m,E=\frac{40\left(4\right)}{(4^2-4{)}^2}=\frac{160}{144}=\left(10/9\right)volt/\mu m$

Positive sign indicates that $\overrightarrow{E}$ is in the +ve x-direction.