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Question: A charged water drop whose radius is \(0.1\mu m\) is in equilibrium in an electric field. If charge ...

A charged water drop whose radius is 0.1μm0.1\mu m 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=10ms−1)(g = 10ms^{- 1})

A

1.61N/C1.61N/C

B

26.2N/C26.2N/C

C

262N/C262N/C

D

1610N/C1610N/C

Answer

262N/C262N/C

Explanation

Solution

In balance conditionQE=mg=(43πr3ρ)gQE = mg = \left( \frac{4}{3}\pi r^{3}\rho \right)g

⇒ E=4×(3.14)(0.1×10−6)3×103×103×1.6×10−19=262N/CE = \frac{4 \times (3.14)(0.1 \times 10^{- 6})^{3} \times 10^{3} \times 10}{3 \times 1.6 \times 10^{- 19}} = 262N/C