Exams › JEE Main › Physics
n identical small charged drops, each at potential V, merge to form one large spherical drop. The potential of the resulting big drop is:
- V/n
- Vn
- Vn^(1/3)
- Vn^(2/3)
Correct answer: Vn^(2/3)
Solution
Charge scales as n and radius as n^(1/3); potential = kQ/R scales as n/n^(1/3) = n^(2/3).
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