An oil drop carries six electronic charges,ha | vedclass

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Question

(D) In the first case,the drop falls with terminal velocity. The gravitational force $mg$ is balanced by the upward drag force $F$. Thus,$F = mg$.In t

Vedclass · Accepted Answer

None of the above

Vedclass · Answer

(D) In the first case,the drop falls with terminal velocity. The gravitational force $mg$ is balanced by the upward drag force $F$. Thus,$F = mg$.In the second case,the drop moves upward with the same terminal velocity. The drag force $F$ now acts downwards. The electric force $qE$ acts upwards,where $q = 6e$.The equation of motion is $qE = F + mg$.Substituting $F = mg$,we get $6eE = mg + mg = 2mg$.Therefore,$E = \frac{2mg}{6e} = \frac{mg}{3e}$.Given $m = 1.6 	imes 10^{-15} 	ext{ kg}$,$g = 10 	ext{ m/s}^2$,and $e = 1.6 	imes 10^{-19} 	ext{ C}$.$E = \frac{1.6 	imes 10^{-15} 	imes 10}{3 	imes 1.6 	imes 10^{-19}} = \frac{10^{-14}}{3 	imes 10^{-19}} = 0.333 	imes 10^5 	ext{ N/C} = 33.3 	imes 10^3 	ext{ N/C} = 33.3 	ext{ kN/C}$.Since $33.3 	ext{ kN/C}$ is not among the options,the correct choice is $D$.

An oil drop carries six electronic charges,has a mass of $1.6 \times 10^{-12} \text{ g}$ and falls with a terminal velocity in air. The magnitude of the vertical electric field required to make the drop move upward with the same speed as it was formerly moving is ........$kN/C$.

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