In Millikan's oil drop experiment, an oil drop of mass $16 \times 10^{-6} \mathrm{~kg}$ is balanced by an electric field of $10^6 \mathrm{~V} / \mathrm{m}$. The charge in coulomb on the drop, assuming $g=10 \mathrm{~m} / \mathrm{s}^2$ is
- A$6.2 \times 10^{-11}$
- B$16 \times 10^{-9}$
- ✓$16 \times 10^{-11}$
- D$16 \times 10^{-13}$
Answer: C.
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