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PhysicsElectrostaticsPotential & potential energyMedium2 minQB

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1000 identical drops of mercury are charged each to a potential of 1 V each. They join to form a single drop. The potential of this drop will be

Answer:

Solution:

$\frac{Kq}{r} = 1 V \Rightarrow 1000 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} V = \frac{K ( 1000 ) q}{10 r} \Rightarrow R = 10 r = 100 V$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Potential & potential energy

⏱ 2mℹ️ Source: QB

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