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Question

An electric field can just support a water droplet 1.0×106m 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/m
Answer is A.
Force in electric field
F=qE
Mass of water droplet
=fwV
=fw43πr3
=fw43π(d2)3
=1000×43×π×(1062)3
=1000×43×π×18×1018
=5.23×1016kg
So, gravitational force =mg
=5.23×1016×10
=5.23×1015N
Now, for equilibrium
qE= gravitational force
E=5.23×10151.6×1019 (electroncharge=1.6×1019C)
=3.21×104Vm

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