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NCERT Class XII Chapter
Electric Charges and Fields
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Question : 25 of 34
Marks: +1, -0
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10410^4 N C−1C^{-1} in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm–3. Estimate the radius of the drop.
(g = 9.81 m s−2s^{-2}; e = 1.6 × 10−1910^{-19} C)
Solution:  
In equilibrium, force due to electric field on the drop balances the weight mg of drop
i.e., qE = mg or qE = Vρg
or qE = 43πr3ρg\frac{4}{3\pi r^{3}\rho g} or r3r^3 = 3qE4πρg\frac{3qE}{4\pi\rho g} = 3×12e×E4πρg\frac{3 \times 12e \times E}{4\pi\rho g}
or r3r^3 = 36×1.6×10−19×2.55×1044×3.14×1.26×103×9.8\frac{36 \times 1.6 \times 10^{-19} \times 2.55 \times 10^{4}}{4 \times 3.14 \times 1.26 \times 10^{3} \times 9.8} = 0.947 × 10−1810^{-18}
or r = 0.981 × 10−610^{-6} m = 9.81 × 10−710^{-7} m = 9.81 × 10−410^{-4} mm
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