Electrostatics Question 388

Question: A charged water drop whose radius is 0.1μm 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=10ms1) [RPET 1997]

Options:

A) 1.61N/C

B) 26.2N/C

C) 262N/C

D) 1610N/C

Show Answer

Answer:

Correct Answer: C

Solution:

In balance condition QE=mg=(43πr3ρ)g

therefore E=4×(3.14)(0.1×106)3×103×103×1.6×1019

=262N/C



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