Question

Two mutually perpendicular infinite wires along $x$-axis and $y$-axis carry charge densities ${\lambda }_1\text{and}{\lambda }_2$. The electric line on force at $P$ is along the line $y=\frac{1}{\sqrt{3}}x$, where $P$ is also a point lying on the same line, then find $\frac{{\lambda }_2}{{\lambda }_1}$.

Answer 1

Net electric field at point $P$ is $\overrightarrow{E}$

$\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$

Field due to the horizontal and vertical wire is $\overrightarrow{E_1}$ and $\overrightarrow{E_2}$ respectively,

$\overrightarrow{E_1}=\frac{{\lambda }_1}{2\pi {\epsilon }_{0}y}$ (in + y direction)

$\overrightarrow{E_2}=\frac{{\lambda }_2}{2\pi {\epsilon }_{0}x}$ ( in x direction)

Let $\alpha$ be angle of electric field with $x$-axis.

$\tan \alpha =\frac{E_1}{E_2}=\frac{\frac{{\lambda }_1}{2\pi {\epsilon }_{0}y}}{\frac{{\lambda }_2}{2\pi {\epsilon }_{0}x}}$

Given, $y=mx=\frac{1}{\sqrt{3}}x$

$\therefore \tan \alpha =m=\frac{1}{\sqrt{3}}$

$\frac{1}{\sqrt{3}}=\frac{{\lambda }_{1}x}{{\lambda }_{2}y}\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{y}{\sqrt{3}x}$

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