Question

Eight charged water drops each of radius \[1mm\] & charge ${10^{ - 10}}C$. Coalesce to form a single drop. Calculate the potential of the single drop.

Answer 1

Hint: The volume of the single drop will be equal to the volume of the eight drops. Use this to find the radius of the single drop. Also, the charge on the single drop will be equal to the charge on the eight drops. We know that the ratio of charge and capacitance is equal to the potential. Find the capacitance of the single drop for finding the required potential.

Complete answer:
Given that the radius of the one water droplet $r = 1mm$
And it has a charge of $q = {10^{ - 10}}C$
Let the radius of the drop formed after coalesce be $R$
Since the single drop is formed from the eight water droplets therefore the volume of the single drop will be equal to the volume of eight drops
$\dfrac{4}{3}\pi {R^3} = 8 \times \dfrac{4}{3}\pi {r^3}$
$ \Rightarrow R = 2r$
$ \Rightarrow R = 2mm = 2 \times {10^{ - 3}}m$
Charge on the single drop will be $Q = 8 \times q = 8 \times {10^{ - 10}}C$
The capacitance of the single drop is $C = 4\pi {\varepsilon _0}R$
$ \Rightarrow C = \dfrac{{2 \times {{10}^{ - 3}}}}{k}$
The potential of the single drop will be $V = \dfrac{Q}{C}$
$ \Rightarrow V = \dfrac{{8 \times {{10}^{ - 10}} \times k}}{{2 \times {{10}^{ - 3}}}}$
$\because k = 9 \times {10^9}$
$ \Rightarrow V = 3600V$
Hence the potential of the single drop is found to be $V = 3600V$

Note:
A drop will be in a form of a sphere due to surface tension hence the volume of the drop is taken as $\dfrac{4}{3}\pi {r^3}$ . Charges are known to diffuse equally across the surface of a conductor. In the case of a charged spherical conductor, all of the charges may be regarded concentrated at the center so capacitance of a charged sphere drop is taken as $C = 4\pi {\varepsilon _0}R$