Question

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

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

Answer 1

The correct option is D $\frac{10}{9}\text{V/}\mu \text{m}$ and in the positive $\text{x}-$direction.

Given,

Electric potential , $V=\frac{20}{(x^2-4)}$

The relation between electric field intensity $\left(E\right)$ and electric potential $\left(V\right)$ at any point is given by,
$E=-\frac{\partial V}{\partial x}$

Substituting the values we get,
$E=-\frac{d}{dx}\left[20\right(x^2-4{)}^{-1}]$

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

when $x=4\mu \text{m}$,
$\Rightarrow E=\frac{40\times 4}{(4^2-4{)}^2}$

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

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

It will act in positive $x-$direction.

Hence, option (d) is correct answer.