Q131 / 160 · WBJEE 2010 · Physics

$n$ identical droplets are charged to $v$ volt each. If they coalesce to form a single drop, then its potential will be

  • Correct answer
    A
    $n^{2 / 3} v$
  • B
    $n^{1 / 3} v$
  • C
    $n v$
  • D
    $v / n$

Solution

$$ \begin{aligned} & \text { Hints: } n \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi \mathrm{R}^3 \\ & \Rightarrow \mathrm{R}=r n^{1 / 3} \\ & \mathrm{C}_0=4 \pi \varepsilon_0 r \\ & q_0=\mathrm{C}_0 \mathrm{~V}=\left(4 \pi \varepsilon_0 r\right) \mathrm{V} \end{aligned} $$ Capacitance of Bigger drop, $$ \mathrm{C}=4 \pi \varepsilon_0 \mathrm{R} $$ So, $\mathrm{V}=\frac{n q_0}{\mathrm{C}}=\frac{n\left(4 \pi \varepsilon_0 r \mathrm{~V}\right)}{4 \pi \varepsilon_0 \mathrm{R}}=n\left(\frac{r}{\mathrm{R}}\right) \mathrm{V}=n\left(\frac{1}{n^{1 / 3}}\right) \mathrm{V}=n^{2 / 3} \mathrm{~V}$

Hence, the correct answer is option: 'A'.

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