Electrostatics Question 388
Question: A charged water drop whose radius is $ 0.1\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=10m{{s}^{-1}}) $ [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=( \frac{4}{3}\pi r^{3}\rho )g $
therefore $ E=\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 $
