Q. A soap bubble is charged to a potential of 16V. Its radius is, then doubled. The potential of the bubble now will be
A) 16V
B) 8V
C) 4V
D) 2V
Explanation
(b) Potential on bubble,
$V = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r}$
$\therefore\frac{V_{1}}{V_{2}} = \frac{r_{2}}{r_{1}}$
$\Longrightarrow \frac{16}{V_{2}} = \frac{2}{1}\ \Longrightarrow V_{2} = 8\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.