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Question

# An electric field can just support a water droplet 1.0×10−6m in diameter carrying one electron charge. The magnitude of electric field strength is

A
3.21×104V/m
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B
2.31×104V/m
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C
1.32×104V/m
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D
6.42×104V/m
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Solution

## The correct option is A 3.21×104V/mAnswer is A.Force in electric fieldF=qEMass of water droplet=fwV=fw43πr3=fw43π(d2)3=1000×43×π×(10−62)3=1000×43×π×18×10−18=5.23×10−16kgSo, gravitational force =mg=5.23×10−16×10=5.23×10−15NNow, for equilibriumqE= gravitational forceE=5.23×10−151.6×10−19 (∵electroncharge=1.6×10−19C)=3.21×104Vm

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