Electrostatics Question 397
Question: An oil drop having charge $ 2e $ is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is$ 900kg/m^{3} $ , the radius of the drop will be [AMU 1999]
Options:
A) $ 2.0\times {{10}^{-6}}m $
B) $ 1.7\times {{10}^{-6}}m $
C) $ 1.4\times {{10}^{-6}}m $
D) $ 1.1\times {{10}^{-6}}m $
Show Answer
Answer:
Correct Answer: B
Solution:
In equilibrium QE = mg therefore $ Q.\frac{V}{d}=mg=( \frac{4}{3}\pi r^{3}\rho )g $
therefore $ 2\times 1.6\times {{10}^{-19}}\times \frac{12000}{2\times {{10}^{-2}}}=\frac{4}{3}\pi r^{3}\times 900\times 10 $
therefore $ ~r=1.7\times 10^{6}m $
BETA
