A) \[\frac{7\sqrt{7}}{2\sqrt{3}}\]
B) \[\frac{5\sqrt{7}}{\sqrt{3}}\]
C) \[\frac{14\sqrt{7}}{\sqrt{3}}\]
D) \[\frac{7\sqrt{7}}{5\sqrt{3}}\]
Correct Answer: B
Solution :
\[\therefore \] \[F=\frac{1}{r\pi {{\varepsilon }_{0}}}\frac{Q(Q-q)}{{{R}^{2}}}\] \[\frac{dF}{dq}=0\] Represents a perpendicular bisector of the line segment joining \[\frac{d}{dq}\,\left[ \frac{Kq(Q-q)}{{{R}^{2}}} \right]=0\] and \[\Rightarrow \] is \[[Q-2q]=0\] Equation of perpendicular bisector of AB is \[q=\frac{Q}{2}\] \[U=\frac{{{Q}^{2}}}{2C}\] \[C'=KC\] \[U'=\frac{{{Q}^{2}}}{2C'}=\frac{1}{2}\,\frac{{{Q}^{2}}}{KC}\] \[\therefore \] \[\frac{U}{U'}=\frac{{{Q}^{2}}/2C}{{{Q}^{2}}/2KC}=K\] \[{{R}_{\min }}=\frac{r}{n}\] \[{{R}_{\max }}=nr\] Minimum value of \[\therefore \] = Perpendicular distance of centre of circle to the line - Radius of circle \[\frac{{{R}_{\min }}}{{{R}_{\max }}}=\frac{r}{n}\times \frac{1}{nr}=\frac{1}{{{n}^{2}}}\] \[=0.1\times 0.05=5\times {{10}^{-3}}\,{{m}^{2}},\] \[\Delta \phi =-0.05\times {{10}^{-3}}\]You need to login to perform this action.
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