An oil drop carries six electronic charges, h | vedclass

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Question

For the first case, there is only gravitational force to balance drag froce $F=m g$

For the second case, drop is moving upward so drag force will b

Vedclass · Accepted Answer

$32.7$

Vedclass · Answer

For the first case, there is only gravitational force to balance drag froce $F=m g$

For the second case, drop is moving upward so drag force will be in downward direction $F+m g=6 e E$ from the first case we have $F=m g$

$2 m g=6 e E$

$E=\frac{m g}{3 e}=\frac{1.6 	imes 10^{-15} 	imes 10}{3 	imes 1.6 	imes 10^{-19}}=3.27 	imes 10^{4} \mathrm{NC}^{-1}$

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

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