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1. Electric Charges and Fields
hard
In Millikan's oil drop experiment, a charged drop falls with terminal velocity $V$. If an electric field $E$ is applied in vertically upward direction then it starts moving in upward direction with terminal velocity $2V$.If magnitude of electric field is decreased to $\frac{E}{2}$, then terminal velocity will become
A
$\frac{V}{2}$
B
$V$
C
$\frac{{3V}}{2}$
D
$2V$
Solution
In the absence of electric field $(i.e. E = 0)$
$mg = 6\pi \eta rv$ …$(i)$
In the presence of Electric field
$mg + QE = 6\pi \eta r(2v)$ …$(ii)$
When Electric field to reduced to $\frac{{E}}{{2}}$
$mg + Q\,\left( {E/2} \right) = 6\pi \eta r(v')$ $(iii)$
After solving $(i)$ ,$(ii)$ and $(iii)$
We get $v' = \frac{3}{2}v$
Standard 12
Physics
