Solution:
By volume conservation, \(R = N^{1/3} r = 27^{1/3} r = 3r\). The total charge of the combined drop is \(Q = 27q\). The potential of the bigger drop is \(V' = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9 V = 9 \times 220 = 1980\text{ V}\).
Twenty seven drops of same size are charged at \(220\text{ V}\) each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
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