Electrostatics Question 353
Question: The intensity of the electric field required to keep a water drop of radius $ {{10}^{-5}}cm $ just suspended in air when charged with one electron is approximately [MP PMT 1994]
Options:
A) $ 260volt/cm $
B) $ 260newton/coulomb $
C) $ 130volt/cm $
D) $ 130newton/coulomb $
$ (g=10newton/kg,e=1.6\times {{10}^{-19}}coulomb) $
Show Answer
Answer:
Correct Answer: B
Solution:
For balance $ mg=eE $
therefore $ E=\frac{mg}{e} $ Also $ m=\frac{4}{3}\pi r^{3}d=\frac{4}{3}\times \frac{22}{7}\times {{({{10}^{-7}})}^{3}}\times 1000kg $
therefore $ E=\frac{4/3\times 22/7\times {{({{10}^{-7}})}^{3}}\times 1000\times 10}{1.6\times {{10}^{-19}}} $ = 260 N/C
