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

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(D) Initially,the oil drop is in equilibrium,so the electric force balances the gravitational force: $qE = mg$. Here,$m = 4.8 	imes 10^{-10} \,g = 4.

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) Initially,the oil drop is in equilibrium,so the electric force balances the gravitational force: $qE = mg$. Here,$m = 4.8 	imes 10^{-10} \,g = 4.8 	imes 10^{-13} \,kg$,$q = 2.4 	imes 10^{-18} \,C$,and $g = 10 \,m/s^2$. When the polarity of the plates is changed,the direction of the electric force $qE$ reverses and acts downwards along with the gravitational force $mg$. The new net force on the drop is $F_{net} = qE + mg$. Since $qE = mg$,we have $F_{net} = mg + mg = 2mg$. The instantaneous acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{2mg}{m} = 2g$. Substituting $g = 10 \,m/s^2$,we get $a = 2 	imes 10 = 20 \,m/s^2$.

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