Question

In Millikan's oil drop experiment ; a drop of charge $Q$ and radius r is kept constant between two plates of potential difference of $800V$. Then charge on other drop of radius $2r$ which is kept constant with a potential difference of $3200$ is

• A. Q/2
• B. 2Q
• C. 4Q
• D. Q/4.

Answer 1

Answer: (B)

Using $QE=\frac{4\pi }{3}{r}^3\rho g$ i.e . Q$\frac{V}{t}=\frac{4\pi }{3}{r}^3\rho g$ ...(i)
also Q $\times \frac{V^{\prime }}{l}=\frac{4\pi }{3}(2r{)}^3\rho g$ ....(ii)
Dividing equation $(i)$ by equation $(ii)$ $\frac{QV}{Q^{\prime }{V}^{\prime }}=\frac{1}{8}$
or $Q^{\prime }=\frac{QV}{V^{\prime }}\times 8=\frac{Q\times 800\times 8}{3200}$ = 2Q