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5017Fon equilibriunT \sin \theta=F_{E}andT \cos \theta=m gDividing eqn (1) & (2)\begin{aligned}\tan \theta & =\frac{F_{E}}{m g} \\F_{E} & =m g \tan \theta \\F_{E} & =\left(4.5 \times 10^{-3}\right) \times(9.81) \times \tan (5.5) \\F_{E} & =4.25 \times 10^{-3} \mathrm{~N}\end{aligned}Now,\begin{aligned}F_{E} & =\frac{K Q^{2}}{(2 L \sin \theta)^{2}}=4.25 \times 10^{-3} \\Q & =\sqrt{\frac{\left(4.25 \times 10^{-3}\right) \times(2 \times 1.3 \times \sin 5.5)^{2}}{\left(9 \times 10^{9}\right)}} \\Q & =1.71 \times 10^{-7} \mathrm{C} \\\theta & =0.171 \mathrm{CC}\end{aligned} ...