Question

spherical drop of mercury of radius R has a Capacitance given by C = 4πϵ0 R. If two such drops combine to form a single larger drop, what is its capacitance?

Answer 1

Given :

Capacitance of spherical drop of radius R, C = 4πϵ0 R

Two drops combined together.

To find :

Capacitance of combined drop.

Solution :

It is given that the capacitance of spherical drop of mercury of radius R is :

$C = 4\pi \epsilon _{0} R$

(equation 1)

If the raduis of 1 drop is R,

when drops are combined together then the volume of combined drop will be double of the volume of single drop.

volume of 1 drop :

$\frac{4}{3} \pi {R}^{3} \:$

On adding to drops, the volume of new drop will be double of 1st drop so,

volume of new drop is :

$\frac{4}{3} \pi {R}^{3} \: + \frac{4}{3} \pi {R}^{3} \: = \frac{8}{3} \pi {R}^{3} \:$

(equation 2)

Let the radius of new droplet is r,

so volume of new drop with respect to radius r :

$\frac{4}{3} \pi {r}^{3} \:$

(equation 3)

By equation 2 and 3:

$\frac{4}{3} \pi {r}^{3} \: = \frac{8}{3} \pi {R}^{3} \: \\ \\ {r}^{3} = 2{R}^{3} \: so \\ r \: = \sqrt[3]{2} R \: = 1.26R \:$

So

Radius of new drop is r = 1.26 R

Putting the value of r in equation 1.

The Capacitance of new drop is C':

$C'= 4\pi \epsilon _{0} r = 4\pi \epsilon _{0} \times (1.26R) = 1.26 \times (4\pi \epsilon _{0} R) = 1.26C \:$

So Capacitance of new drop :

C' = 1.26 C