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(A) When the drop is balanced, the electric force equals the gravitational force: $qE = mg$ $(1)$.When the field is switched off, the drop falls with

Vedclass · Accepted Answer

$8 	imes 10^{-19} \,C$

Vedclass · Answer

(A) When the drop is balanced, the electric force equals the gravitational force: $qE = mg$ $(1)$.When the field is switched off, the drop falls with terminal velocity $v$. According to Stokes' Law, the drag force equals the gravitational force: $mg = 6 \pi \eta r v$ $(2)$.From $(1)$ and $(2)$, we have $qE = 6 \pi \eta r v$, so $r = \frac{qE}{6 \pi \eta v}$.The mass of the spherical drop is $m = \frac{4}{3} \pi r^3 ho$.Substituting $m$ into $(1)$: $qE = \frac{4}{3} \pi \left( \frac{qE}{6 \pi \eta v} ight)^3 ho g$.Solving for $q$: $q^2 = \frac{3 	imes (6 \pi \eta v)^3}{4 \pi E^2 ho g} = \frac{3 	imes 216 \pi^3 \eta^3 v^3}{4 \pi E^2 ho g} = \frac{162 \pi^2 \eta^3 v^3}{E^2 ho g}$.Substituting the values: $q^2 = \frac{162 	imes \pi^2 	imes (1.8 	imes 10^{-5})^3 	imes (2 	imes 10^{-3})^3}{(\frac{81}{7} \pi 	imes 10^5)^2 	imes 900 	imes 9.8}$.Calculating this yields $q^2 = 64 	imes 10^{-38} \,C^2$, so $q = 8 	imes 10^{-19} \,C$.

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