In Millikan's oil drop experiment, an oil drop of radius $r$ and charge $q$ is held in equilibrium when the applied voltage is $8$ volts. If the radius of the drop is $2 r$ for the same charge the potential (in volts) required to keep it in equilibrium will be how many times the previous voltage?

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Solution

$q V = m g$
$= \frac{4}{3} π r^{3} ρ g$
or $V α r^{3}$
$V_{1} = (2 r)^{3}$ or $V_{1} = 8$ volt
Eight times the previous voltage.

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In Millikan's oil drop experiment, an oil drop of radius r and charge q is held in equilibrium when the applied voltage is 8 volts. If the radius of the drop is 2r for the same charge the potential (in volts) required to keep it in equilibrium will be how many times the previous voltage?

qV=mg=43πr3ρgor Vαr3V1=(2r)3 or V1=8voltEight times the previous voltage.

In Millikan's oil drop experiment, an oil drop of radius $r$ and charge $q$ is held in equilibrium when the applied voltage is $8$ volts. If the radius of the drop is $2 r$ for the same charge the potential (in volts) required to keep it in equilibrium will be how many times the previous voltage?

$q V = m g$
$= \frac{4}{3} π r^{3} ρ g$
or $V α r^{3}$
$V_{1} = (2 r)^{3}$ or $V_{1} = 8$ volt
Eight times the previous voltage.