Question

Two mutually perpendicular wire carry charge densities ${\lambda }_1$ and ${\lambda }_2$. The electric line of force makes angle $\alpha$ with second wire, then the ratio $\frac{{\lambda }_1}{{\lambda }_2}$ is equal to (in terms of angle $\alpha )$

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

Answer 1

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


$E_1=\frac{{\lambda }_1}{2\pi {\epsilon }_{0}x}$

$E_2=\frac{{\lambda }_2}{2\pi {\epsilon }_{0}y}$

$\tan \alpha =\frac{E_2}{E_1}=\frac{{\lambda }_2}{{\lambda }_1}\left(\frac{x}{y}\right)----\left(1\right)$

From the figure,

From eq. $\left(1\right)$ $\tan \alpha =\frac{y}{x}$

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{\tan \alpha }\frac{1}{\left(\frac{y}{x}\right)}=\frac{1}{{\tan }^2\alpha }$

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