Question

N drops of mercury of equal radii and possesing equal charges combine to form a big drop. What is the charge capacitance and potential of the drop?

Answer 1

Volume of the big drop=Nr(volume of small drop)

If radius of N small drops are 'r',

$\frac{4}{3}$(radius of the big drop)$\times \pi =N\times \frac{4}{3}\pi r^3$

Radius of the big drop $=N^{1/3}r$

Capacity (change) of the big spherical drop is,$c=4\pi {\epsilon }_{0}R=4\pi {\epsilon }_0\times N^{1/3}\times r$

Total change of big drop =$N_q$(change of each drop)

(capacity)(potential)=change

$\left(4\pi {\epsilon }_0{N}^{1/3}r\right)\left(v\right)=N_q$

$v=\frac{N^{2/3}q}{4\pi {\epsilon }_{0}r}$