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



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