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(D) Given: mass $m = 1 	ext{ g} = 10^{-3} 	ext{ kg}$, charge $q = 20 \mu	ext{C} = 20 	imes 10^{-6} 	ext{ C}$, length $r = 0.9 	ext{ m}$, electri

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$6$

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(D) Given: mass $m = 1 	ext{ g} = 10^{-3} 	ext{ kg}$, charge $q = 20 \mu	ext{C} = 20 	imes 10^{-6} 	ext{ C}$, length $r = 0.9 	ext{ m}$, electric field $E = 100 	ext{ NC}^{-1}$, and $g = 10 	ext{ ms}^{-2}$.The force exerted by the electric field on the charge is $F_e = qE = 20 	imes 10^{-6} 	imes 100 = 2 	imes 10^{-3} 	ext{ N}$ (upwards).The gravitational force on the ball is $F_g = mg = 10^{-3} 	imes 10 = 10 	imes 10^{-3} 	ext{ N}$ (downwards).The net effective force acting downwards is $F_{	ext{eff}} = F_g - F_e = 10 	imes 10^{-3} - 2 	imes 10^{-3} = 8 	imes 10^{-3} 	ext{ N}$.The effective acceleration due to gravity is $g_{	ext{eff}} = \frac{F_{	ext{eff}}}{m} = \frac{8 	imes 10^{-3}}{10^{-3}} = 8 	ext{ ms}^{-2}$.For a particle to complete a vertical circle, the minimum velocity at the lowest point is $v = \sqrt{5g_{	ext{eff}}r}$.Substituting the values: $v = \sqrt{5 	imes 8 	imes 0.9} = \sqrt{40 	imes 0.9} = \sqrt{36} = 6 	ext{ ms}^{-1}$.

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