In a Millikan's oil drop experiment the charge on an oil drop is calculated to be 6.35 × 1 0 − 19 C . The number of excess electrons on the drop is

As we know, Q = ne Number of electron n = e Q 1.6 × 1 0 − 19 6.35 × 1 0 − 19 = 3.9 = 4

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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceEasy2 minPYQ_2011

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In a Millikan's oil drop experiment the charge on an oil drop is calculated to be $6.35 \times 1 0^{- 19} C$ . The number of excess electrons on the drop is

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Solution:

As we know, $Q =$ ne Number of electron $n = \frac{Q}{e}$ $\frac{6.35 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 3.9 = 4$

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

⏱ 2mℹ️ Source: PYQ_2011

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