Eight charged water droplets, each of radius 1 mm and carrying a charge \( 5 \times 10^{-10}\,\mathrm{C} \), coalesce to form a single drop. The potential of the bigger drop is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 18000 V
When the drops coalesce, both the total charge and the total volume are conserved. The bigger drop is a charged sphere, so its potential is \( V = Q/(4\pi\varepsilon_0 R) \).
\( \displaystyle 8 \times \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi R^{3} \;\Rightarrow\; R = 2r = 2\,\mathrm{mm} = 2\times10^{-3}\,\mathrm{m} \)
\( \displaystyle Q = 8q = 8 \times 5\times10^{-10}\,\mathrm{C} = 4\times10^{-9}\,\mathrm{C} \)
\( \displaystyle V = \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{R} = \frac{9\times10^{9} \times 4\times10^{-9}}{2\times10^{-3}}\,\mathrm{V} = 18000\,\mathrm{V} \)
Shortcut: For \( n \) identical drops merging, \( V_{\mathrm{big}} = n^{2/3}\,V_{\mathrm{small}} \). Here \( V_{\mathrm{small}} = 9\times10^{9} \times 5\times10^{-10}/10^{-3} = 4500\,\mathrm{V} \) and \( 8^{2/3} = 4 \), so \( V_{\mathrm{big}} = 18000\,\mathrm{V} \).