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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceEasy2 minPYQ_2020

PhysicsEasysingle choice

An oil drop carrying a charge of $2$ electrons has a mass of $3.2 \times 10^{- 17} k g$ . It is falling freely in the air with terminal speed. The electric field required to make the drop move upwards with the same speed is

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Answer:

Solution:

$q = 2 e; m = 3.2 \times 10^{- 17} k g$

viscous force $6 π η r v = m g$ (initially) ......(i)

now motion upwards

$∴ v i s c o u s e f o r c e d o w n w a r d s$

$6 π η r v + m g = q E \dots \dots (i i)$

Put value from (i) in (ii)

$q E = 2 m g$

$E = \frac{2 m g}{q} = \frac{2 \times 3.2 \times 1 0^{- 17} \times 10}{2 \times 1.6 \times 1 0^{- 19}} = 2 \times 10^{3} V / m$

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

⏱ 2mℹ️ Source: PYQ_2020

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