Question

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

• A. $\frac{5}{3}V{\mu m}^{-1}$ and in positive x direction
• B. $\frac{10}{9}{\mu m}^{-1}$ and in negative x direction
• C. $\frac{10}{9}{\mu m}^{-1}$ and in positive x direction
• D. $\frac{5}{3}{\mu m}^{-1}$ and in negative x direction

Answer 1

The correct option is C $\frac{10}{9}{\mu m}^{-1}$ and in positive x direction

Here, $V\left(x\right)=\frac{20}{x^2-4}=20{(x^2-4)}^{-1}$
$E=-\frac{dV}{dx}=-20(-1){(x^2-4)}^{-2}\left(2x\right)=\frac{40x}{{(x^2-4)}^2}$
At $x=4\mu m,\mu E=\frac{40\times 4}{{(4^2-4)}^2}=\frac{160}{144}=\frac{10}{9}V\mu m^{-1}$
$\overrightarrow{E}$ is along positive x-direction