• # question_answer Two mutually perpendicular infinitely long lines of charge having charge per unit length as ${{\lambda }_{1}}$ and ${{\lambda }_{2}}$ are located in air at a distance "a" from each other. The force of interaction between them is - A) $\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{4{{\in }_{0}}\pi a}$                       B) $\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{\pi a{{\in }_{0}}}$ C) $\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{2{{\in }_{0}}\pi a}$           D) $\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{3\pi a{{\in }_{0}}}$

[C]  Consider an element of thickness dy at distance y Charge on element $=dq={{\lambda }_{1}}dy$ E at the position of element = E Force on element $=dF=dqE$ $E=\frac{{{\lambda }_{2}}}{2\pi {{\in }_{0}}r}=\frac{{{\lambda }_{2}}}{2\pi {{\in }_{0}}\sqrt{{{a}^{2}}+{{y}^{2}}}}$           $dF=\frac{{{\lambda }_{2}}\,\,\times \,\,{{\lambda }_{1}}}{2\pi {{\in }_{0}}\sqrt{{{a}^{2}}+{{y}^{2}}}}.dy$
 Taking component of dF in x and y direction By symmetry y component get cancel out $d{{F}_{x}}=dF\cos \theta$ Where $\cos \theta =\frac{a}{\sqrt{{{a}^{2}}+{{y}^{2}}}}$ and ${{F}_{x}}=\int\limits_{-\infty }^{+\infty }{d{{F}_{x}}}$ ${{F}_{net}}$ is in direction of a-axis $\therefore {{F}_{net}}={{F}_{x}}$ ${{F}_{x}}=\int\limits_{-\infty }^{+\infty }{\frac{{{\lambda }_{1}}{{\lambda }_{2}}a}{2\pi {{\in }_{0}}({{a}^{2}}+{{y}^{2}})}dy}$$\Rightarrow \frac{{{\lambda }_{1}}{{\lambda }_{2}}}{2\pi {{\in }_{0}}a}$