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) \[\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{3\pi a{{\in }_{0}}}\]
Correct Answer: C
Solution :
[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 \[\therefore {{F}_{net}}={{F}_{x}}\] |
\[{{F}_{x}}=\int\limits_{-\infty }^{+\infty }{\frac{{{\lambda }_{1}}{{\lambda }_{2}}a}{2\pi {{\in }_{0}}({{a}^{2}}+{{y}^{2}})}dy}\]\[\Rightarrow \frac{{{\lambda }_{1}}{{\lambda }_{2}}}{2\pi {{\in }_{0}}a}\] |
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