A) \[\sqrt{\frac{16\pi {{\varepsilon }_{0}}{{d}^{3}}mg}{\sqrt{{{R}^{2}}-{{d}^{2}}}}}\]
B) \[\sqrt{\frac{8\pi {{\varepsilon }_{0}}{{d}^{3}}mg}{\sqrt{{{R}^{2}}-{{d}^{2}}}}}\]
C) \[\sqrt{\frac{8\pi {{\varepsilon }_{0}}{{d}^{3}}mg}{\sqrt{{{R}^{2}}+{{d}^{2}}}}}\]
D) \[\sqrt{\frac{16\pi {{\varepsilon }_{0}}{{d}^{3}}mg}{\sqrt{{{R}^{2}}+{{d}^{2}}}}}\]
Correct Answer: A
Solution :
[a] \[{{F}_{e}}=N\,\,\sin \theta \] \[mg=N\cos \theta \] \[{{F}_{e}}=mg\,\tan \theta \] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{4{{d}^{2}}}=mg\frac{d}{\sqrt{{{R}^{2}}-{{d}^{2}}}}\] \[q=\sqrt{\frac{16\pi {{\varepsilon }_{0}}{{d}^{3}}mg}{\sqrt{{{R}^{2}}-{{d}^{2}}}}}\]You need to login to perform this action.
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