Solution:
When \(N = 64\) drops combine, the new radius is \(R = N^{1/3}r = 4r\). The new charge is \(Q = Nq = 64q\). The potential of the bigger drop is \(V' = \frac{kQ}{R} = \frac{k(64q)}{4r} = 16 \left(\frac{kq}{r}\right) = 16V\).
Sixty four identical drops of water having equal charge combine to form a bigger drop. The factor by which potential of bigger drop change in comparison to a small drop is
Leave a Reply