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NCERT Class XII Chapter
Electric Charges and Fields
Questions With Solutions

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Question : 25
Total: 34
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C–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–2; e = 1.6 × 10–19 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 =
4
3
πr3ρg or r3 =
3qE
4πρg
 =
3×12e×E
4πρg
or r3 =
36×1.6×10−19×2.55×104
4×3.14×1.26×103×9.8
 = 0.947 × 10−18
or r = 0.981 × 10−6 m = 9.81 × 10−7 m = 9.81 × 10−4 mm
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