Soap bubble

A conductive soap bubble with a radius of r=2 cm and charged to a potential of φ= 10000 V will burst into a drop of water with a radius of r1= 0.05 cm. What is the potential φ1 of the drop?

Final Answer:

φ1 =  400000 V

Step-by-step explanation:

$$\begin{gathered} r=2\text{cm}\to \text{m}=2:100\text{m}=0.02\text{m} \\ \phi =10000\text{V} \\ {r}_1=0.05\text{cm}\to \text{m}=0.05:100\text{m}=0.0005\text{m} \\ \phi =k/r \\ {\phi }_1=k/r_1 \\ \phi \cdot r={\phi }_1\cdot r_1 \\ {\phi }_1=\phi \cdot \frac{r}{r_1}=10000\cdot \frac{0.02}{0.0005}=400000\text{V} \end{gathered}$$

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