A) \[\frac{\lambda q}{2{{\pi }^{2}}{{\varepsilon }_{0}}R}\]
B) \[\frac{\lambda q}{{{\pi }^{2}}{{\varepsilon }_{0}}R}\]
C) \[\frac{\lambda q}{4{{\pi }^{2}}{{\varepsilon }_{0}}R}\]
D) \[\frac{\lambda q}{4\pi {{\varepsilon }_{0}}R}\]
Correct Answer: B
Solution :
[b] \[{{F}_{net}}=\int_{{}}^{{}}{dq\,\,E\cos \theta }\] \[=\int\limits_{-\pi /2}^{\pi /2}{\left( \frac{q}{\pi R} \right)Rd\theta \frac{\lambda }{2\pi \varepsilon R}\cos \theta }\] \[=\frac{\lambda q}{2{{\pi }^{2}}{{\varepsilon }_{0}}R}\,\int\limits_{-\pi /2}^{\pi /2}{\cos \theta d\theta =\frac{\lambda q}{2{{\pi }^{2}}{{\varepsilon }_{0}}R}\left[ \sin \theta \right]_{-\pi /2}^{\pi /2}}\] \[=\frac{\lambda q}{2{{\pi }^{2}}{{\varepsilon }_{0}}R}[1-(-1)]=\frac{\lambda q}{{{\pi }^{2}}{{\varepsilon }_{0}}R}\]You need to login to perform this action.
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