Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Eight charged water droplets, each with a radius of 1 mm and a charge of $10^{–10}\; C$, coalesce to form a single drop. Calculate the potential of the bigger drop.

Can you answer this question?

1 Answer

0 votes
Let R and r be the radii of the bigger drop and one droplet respectively.
Volume of 8 droplets = Volume of bigger drop
$ 8 \times \large\frac{4}{3} $$\pi r^3=\large\frac{4}{3} \pi $$R^3$
=>$R= (8)^{1/3}r=2r$
or $R= 2 \times 1 \times 10^{-3} m= 2 \times 10^{-3} m$
Charge on bigger drop, $Q = 8 × 10^{–10} C$
Potential of the bigger drop $=\large\frac{1}{4 \pi \in_0} \frac{Q}{R}$
=>$3.6 \times 10^{4} volt$
Hence A is the correct answer.
answered Jun 7, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App