216 charged droplets, each of radius r and charge q, are combined to form a big drop. The ratio of potential of the big drop to the small droplet will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 36
Potential of a drop of radius r and charge q is:
$V_{small} = \frac{1}{4 \pi \epsilon_0}\frac{q}{r}$
When 216 such droplets combine:
Volume conservation: $R^3 = 216 r^3 \;\;\Rightarrow\;\; R = 6r$
Total charge: $Q = 216q$
Potential of big drop:
$V_{big} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R} = \frac{1}{4 \pi \epsilon_0}\frac{216q}{6r} = \frac{1}{4 \pi \epsilon_0}\frac{36q}{r}$
Ratio:
$\frac{V_{big}}{V_{small}} = \frac{36q/r}{q/r} = 36$
Final Answer: The ratio is 36.