Question

Two mutually perpendicular infinite wires along $\text{x-}$axis and $\text{y-}$axis carry charge density ${\lambda }_1$ and ${\lambda }_2$. The electric line of force at $\text{P}$ is along the line $y=\frac{x}{\sqrt{3}}$, where $\text{P}$ is also a point lying on the same line, then find $\frac{{\lambda }_1}{{\lambda }_2}$

• A. $2$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

Net electric field at point $\text{P}$ is given as,
$\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$

Electric field due to horizontal wire,
${\overrightarrow{E}}_1=\frac{{\lambda }_1}{2\pi {\epsilon }_{0}y}$ in the $\text{+y}$ direction

Electric field due to horizontal wire,
${\overrightarrow{E}}_2=\frac{{\lambda }_1}{2\pi {\epsilon }_{0}x}$ in the $\text{+x}$ direction

The angle $\left(\alpha \right)$ made by resultant electric field with $\text{x}$ direction,
$\tan \alpha =\frac{E_1}{E_2}=\frac{\frac{{\lambda }_1}{2\pi {ϵ}_{0}y}}{\frac{{\lambda }_2}{2\pi {ϵ}_{0}x}}$

$\Rightarrow \tan \alpha =\frac{{\lambda }_{1}x}{{\lambda }_{2}y}...\left(1\right)$

Now comparing the given equation, $y=\frac{1}{\sqrt{3}}x$ with the straight line equation, $y=mx$, we get, slope of line $m$ as,

$m=\tan \alpha =\frac{1}{\sqrt{3}}...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$\tan \alpha =\frac{{\lambda }_{1}x}{{\lambda }_{2}y}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{{\lambda }_{1}x}{{\lambda }_{2}y}$

$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{y}{x\sqrt{3}}$

According to problem, $y=\frac{1}{\sqrt{3}}x$
$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}$

$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{3}$

Hence, option (d) is correct answer.