Question

Electrostatic force $F$ on a charge $\frac{Q}{2}$ at an axial point due to a finite line charge varies as $F=\frac{18000}{x^2+4x}$, 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. $2\text{C}$
• B. $0.02\text{C}$
• C. $0.002\text{C}$
• D. $0.004\text{C}$

Answer 1

The correct option is C $0.002\text{C}$

Given:
$F=\frac{18000}{x^2+4x}=\frac{18000}{x(x+4)}---\left(1\right)$

We know that electrostatic force at an axial point of a line charge,
$F=\frac{kqQ}{L}[\frac{1}{x}-\frac{1}{(x+L)}]$
where $x$ is distance from one end.

$\Rightarrow F=\frac{kqQ}{L}\left[\frac{x+L-x}{x(x+L)}\right]$
$\Rightarrow F=\frac{kqQ}{x(x+L)}-----\left(2\right)$
Here, it is given in the question $q=\frac{Q}{2}$
From $\left(1\right)$ and $\left(2\right)$,
$kqQ=18000\Rightarrow \frac{k{Q}^2}{2}=18000$
$\Rightarrow Q^2=\frac{36000}{9\times {10}^9}\Rightarrow Q=\sqrt{4\times {10}^{-6}}=0.002\text{C}$