A soap bubble is charged to a potential 12 V. If its radius is doubled, then the potential of the bubble
becomes 

We know that, potential, $\mathrm{V}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

Here,              ${\mathrm{r}}_1=\mathrm{r} \text{ and } {\mathrm{r}}_2=2\mathrm{r}$

So,                 ${\mathrm{V}}_1=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}\text{ and }{\mathrm{V}}_2=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\left(2\mathrm{r}\right)}$

Hence,           $\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}\times \frac{4{\mathrm{\pi \epsilon }}_0\left(2\mathrm{r}\right)}{\mathrm{q}}$

$\Rightarrow \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{2}{1};{\mathrm{V}}_2=\frac{{\mathrm{V}}_1}{2}$

In question, V₁ = 12 V                                                           (given)

So,              ${\mathrm{V}}_2=\frac{12}{2}=6 \mathrm{V}$