Question

Electric field at an axial point due to a finite line charge varies as $E=\frac{90}{x^2+4x}$, where $x$ is in metres and is the distance along axial line from one of the ends. The total charge on the line is

• A. $10\text{nC}$
• B. $11\text{nC}$
• C. $12\text{nC}$
• D. $13\text{nC}$

Answer 1

The correct option is A $10\text{nC}$

Given:
$E=\frac{90}{x^2+4x}=\frac{90}{x(x+4)}-----\left(1\right)$

We know that electric field at an axial point of a line charge,
$E=\frac{kQ}{L}[\frac{1}{x}-\frac{1}{x+L)}]$
where $x$ is distance from one end.

$\Rightarrow E=\frac{kQ}{L}\left[\frac{x+L-x}{x(x+L)}\right]$
$\Rightarrow E=\frac{kQ}{x(x+L)}-----\left(2\right)$

From (1) and (2),
$kQ=90$
$\Rightarrow Q=\frac{90}{9\times {10}^9}={10}^{-8}=10\text{nC}$