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Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 μC are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:

1. 4: 1
2. 1: 4
3. 1: 8
4. 8: 1

Correct answer: 4: 1

Solution

Volume conservation: 64*(4/3)pi r^3 = (4/3)pi R^3 -> R = 4r. Total charge Q = 64q. Surface density ratio = (64q/R^2)/(q/r^2) = 64*(r^2/R^2) = 64/16 = 4, so 4:1.

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