Question

The potential at a point x (measured in μm) due to some charges situated on the x-axis is given by
V(x) = 20 / $(x^2$ - 4) Volt
The electric field E at x = 4 μm is given by:
[AIEEE-2007]

• A. $\frac{5}{3}$ Volt/μm and in the positive x direction
• B. $\frac{10}{9}$ Volt/μm and in the negative x-direction
• C. $\frac{10}{9}$ Volt/μm and in the positive x-direction
• D. $\frac{5}{3}$ Volt/μm and in the negative direction

Answer 1

The correct option is C

$\frac{10}{9}$ Volt/μm and in the positive x-direction

$V\left(x\right)=\frac{20}{x^2-4}$
$E\left(x\right)=\frac{-\partial V}{\partial x}=\frac{-\partial }{\partial x}\left[20\right(x^2-4{)}^{-1}]$
$E\left(x\right)=\frac{20}{(x^2-4{)}^2}.2x$
$E(x=4\mu m)=\frac{40x}{(x^2-4{)}^2}$
$=\frac{10}{9}\frac{volt}{\mu m}$, in the +x direction.