An oil drop having a mass 4.8 times 10^{-13} | vedclass

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(D) Initially,the oil drop is in equilibrium between the plates. The downward gravitational force $mg$ is balanced by the upward electric force $qE$.

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) Initially,the oil drop is in equilibrium between the plates. The downward gravitational force $mg$ is balanced by the upward electric force $qE$. Thus,$qE = mg$.When the polarity of the plates is reversed,the electric force $qE$ now acts downwards,in the same direction as gravity.The net force on the drop becomes $F_{net} = mg + qE$.Since $qE = mg$,we have $F_{net} = mg + mg = 2mg$.The instantaneous acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{2mg}{m} = 2g$.Given $g = 10 ~ms^{-2}$,the acceleration $a = 2 	imes 10 = 20 ~ms^{-2}$.

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