Electric Field: Practice Problem & Solution
In Millikan oil drop experiment, a charged drop falls with a terminal velocity V. If an electric field E is applied vertically upwards it moves with terminal velocity 2V in upward direction. If electric field reduces to E/2 then its terminal velocity will be: (1990)
Solution Explained:
To solve this problem, we apply the core principles of Electric Field. Understanding the underlying formula is key to arriving at the correct answer below:
Downward falling: $mg = 6\pi\eta r V$. Moving upwards: $qE - mg = 6\pi\eta r (2V) = 2(mg) \implies qE = 3mg$. If field is E/2, upward force is $qE/2 = 1.5mg$. Net upward force = $1.5mg - mg = 0.5mg = 6\pi\eta r V'$. Thus $V' = V/2$.
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