A charged water drop whose radius is 0.1 μm is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be (g=10 ms−2)

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1.61 N/C

26.2 N/C

262 N/C

1610 N/C

answer is C.

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Detailed Solution

In balance condition QE=mg=43πr3ρ g⇒E=4×(3.14) (0.1×10−6)3×103×103×1.6×10−19=262 N/C

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