question_answer

27 drops of same size are charged at 220 V each. They coalesce to form a bigger drop. Calculate the potential of bigger drop.

Answer:

Here,     n = 27, V = 220 V, V? = ? Let R be the radius of big drop and r be the radius of each small drop. As, mass remains unchanged, therefore             \[\frac{4}{3}\pi {{R}^{3}}=27\times \frac{4}{3}\pi {{r}^{3}}\] or         R = 3r If q is charge on each small drop, then charge on one big drop,         q' = 27q Capacity of big drop, \[C'=4\pi {{\varepsilon }_{0}}R\Rightarrow C'=4\pi {{\varepsilon }_{0}}(3r)\] Potential of big drop, \[V=\frac{q'}{C}\] or         \[V'=\frac{27q}{4\pi {{\varepsilon }_{0}}(3r)}\]             \[V'=9\times V=9\times 220=1980\,V\]