A) \[\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]
B) \[\frac{q}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]
C) \[\frac{q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\]
D) \[\frac{q}{2\pi {{\varepsilon }_{0}}{{R}^{2}}}\]
Correct Answer: A
Solution :
From figure\[dl=Rd\theta \] Charge on\[dl=\lambda Rd\theta \left\{ \lambda =\frac{q}{\pi R} \right\}\] Electric field at centre due to\[dl\]is \[dE=k\cdot \frac{\lambda Rd\theta }{{{R}^{2}}}\] We need to consider only the component\[dE\,\,\cos \,\theta \], as the component \[dE\,\sin \,\theta \] will cancel out because of the field at \[c\] due to the symmetrical element\[dl\], Total field at centre \[=2\int_{0}^{\pi /2}{dE\,\,\cos \theta }\] \[=\frac{2k\lambda }{R}\int_{0}^{\pi /2}{\cos \theta \,\,d\theta }\] \[=\frac{2k\lambda }{R}=\frac{q}{2\pi {{r}^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]You need to login to perform this action.
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