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There are 27 drops of a conducting fluid...

There are `27` drops of a conducting fluid. Each drop has radius `r`, and each of them is charged to the same potential `V_1`. They are then combined to from a bigger drop. The potential of the bigger drop is `V_2`. Find the ratio `V_2//V_1`. Ignore the change in density of the fluid on combining the drops.

Text Solution

Verified by Experts

The correct Answer is:
9
Let `q` be the charge on each drop, then `V_1 = kq//r`. Let `R` be radius of single drop. Then
`(4)/(3) pi R^3 = 27 (4)/(3) pi r^3 or R = 3 r`
`V_2 = (k(27 q))/(R) = (k27 q)/(3 r) = 9 V_1 or (V_2)/(V_1) = 9`.
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Knowledge Check

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