Electrostatics Question 372
Question: In Millikan’s oil drop experiment an oil drop carrying a charge Q is held stationary by a potential difference $ 2400V $ between the plates. To keep a drop of half the radius stationary the potential difference had to be made $ 600V $ . What is the charge on the second drop [MP PET 1997]
Options:
A) $ \frac{Q}{4} $
B) $ \frac{Q}{2} $
C) $ Q $
D) $ \frac{3Q}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
In balance condition therefore $ QE=mg $
therefore $ Q\frac{V}{d}=( \frac{4}{3}\pi r^{3}\rho )g $
therefore $ Q\propto \frac{r^{3}}{V} $
therefore $ \frac{Q _{1}}{Q _{2}}={{( \frac{r _{1}}{r _{2}} )}^{3}}\times \frac{V _{2}}{V _{1}} $
therefore $ \frac{Q}{Q _{2}}={{( \frac{r}{r/2} )}^{3}}\times \frac{600}{2400}=2 $
therefore Q2 = Q / 2
BETA
