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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceEasy2 minPYQ_2020

PhysicsEasysingle choice

A charged oil drop falls with terminal velocity $v_{0}$ in the absence of electric field. An electric field $E$ keeps it stationary. The drop acquires charge $3 q$ , it starts moving upwards with velocity $v_{0}$ . The initial charge on the drop is

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Answer:

Solution:

When drop is stationary, then

$q_{1} E = 6 π η r v_{0}$ or $q_{1} = 6 π η r v_{0} / E$

When drop moves upwards, then

$3 q = \frac{6 π η r (v_{0} + v_{0})}{E} = 2 \times (\frac{6 π η r v_{0}}{E}) = 2 q_{1}$

$∴ q_{1} = \frac{3}{2} q$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Electric Field & Force

⏱ 2mℹ️ Source: PYQ_2020

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