Question

Assume that a drop of mercury is an isolated sphere. What is the capacitance of a drop that results when two drops each of radius R = 1.89 mm merge?

Answer 1

Here we have given that,

R = 1.89 mm

And we know that capacitance of a conducting sphere of radius R is given as,

C= 4πe0 R = 4×3.14×8.85×10^-12×1.89×10^-3

C = 21.02 ×10^-14 F

Now Here we can that capacitance is proportional to
the radius of the sphere so that,

When we merge two droplets then the volume of the combined droplet is twice that of the smaller
droplets,

and

in case of the sphere volume is proportional to the radius cubed,

So that the radius of the merged droplet will becomes

R' = 2¹/³ = 1.2599 times that of the smaller droplets,

so the capacitance increases as,

C' = 1.2599 C

C' = 26.49 ×10^-14 F

Which is our required increased or new capacitance.