Question

Two mutually perpendicular infinite wires carry positive charge densities ${\lambda }_1$ and ${\lambda }_2$. The electric lines of force makes angle $\alpha$ with second wire then $\frac{{\lambda }_1}{{\lambda }_2}$ is

• A. ${\tan }^2\alpha$
• B. ${\cot }^2\alpha$
• C. ${\sin }^2\alpha$
• D. ${\cos }^2\alpha$

Answer 1

The correct option is B ${\cot }^2\alpha$

The direction of net electric field due to both infinite charged wires will be along the electric lines of force.

The directions of electric fields due to both wires at any point $\text{P}$ at distance from the point of intersection of both wires is shown in the given figure.
Here,

$E_1=$ electric field due to infinite charged wire of charge density ${\lambda }_1;$

$E_2=$ electric field due to infinite charged wire of charge density ${\lambda }_2;$

So, from the figure,

$\tan \alpha =\frac{E_2}{E_1}$

$\Rightarrow \tan \alpha =\frac{2k{\lambda }_2/r_2}{2k{\lambda }_1/r_1}[\because E=\frac{2k\lambda }{r}]$

$\Rightarrow \tan \alpha =\frac{{\lambda }_2\times r_1}{{\lambda }_1\times r_2}$

$\Rightarrow \tan \alpha =\frac{{\lambda }_2}{{\lambda }_1}\times \frac{r\cos \alpha }{r\sin \alpha }=\frac{{\lambda }_2}{{\lambda }_1}\times \cot \alpha$

$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}={\cot }^2\alpha$

Hence, option $\left(b\right)$ is the correct answer.