Calculating the Potential of a Collapsing Droplet: A Scientific Approach

[gracy]

Homework Statement

A conducting bubble of radius a, thickness t(t<<a) has potential V. Now the bubble collapses into a droplet. Find the potential of the droplet.

Homework Equations

Potential at the surface of bubble=$V$=$\frac{Kq}{a}$

The Attempt at a Solution

Potential at the surface of bubble=$V$=$\frac{Kq}{a}$
bubble collapses into droplet.Let the radius of droplet to be R. Volume of droplet should be same as of bubble.
volume of droplet=$\frac{4}{3}$$πR^3$
volume of bubble =$\frac{4}{3}$$πa^3$
$\frac{4}{3}$$πR^3$=$\frac{4}{3}$$πa^3$
Am I right till here?I don't think so,because I have not used thickness t anywhere.

 

[ehild]

Homework Statement

A conducting bubble of radius a, thickness t(t<<a) has potential V. Now the bubble collapses into a droplet. Find the potential of the droplet.

Homework Equations

Potential at the surface of bubble=$V$=$\frac{Kq}{a}$

The Attempt at a Solution

Potential at the surface of bubble=$V$=$\frac{Kq}{a}$
bubble collapses into droplet.Let the radius of droplet to be R. Volume of droplet should be same as of bubble.
volume of droplet=$\frac{4}{3}$$πR^3$
volume of bubble =$\frac{4}{3}$$πa^3$
$\frac{4}{3}$$πR^3$=$\frac{4}{3}$$πa^3$
Am I right till here?I don't think so,because I have not used thickness t anywhere.

No. The bubble is a shell of thickness t. Its volume is not the same as the volume of the enclosed sphere.

 

[gracy]

Its volume is not the same as the volume of the enclosed sphere.

Volume of bubble=$(4πa^2)t$?

 

[ehild]

Volume of bubble=$(4πa^2)t$?

Yes, when t<<a and a is the radius of the bubble.

 

[gracy]

I have just guessed it !Is formula of volume of a shell is (4πradius squared)multiplied by thickness?

 

[NascentOxygen]

To calculate the exact volume of material comprising the bubble wall, you would use "outer sphere's volume minus inner sphere's volume".

 

[gracy]

Whenever thickness <<radius ,I should apply the below formula
(4πradius squared)multiplied by thickness?
And in case of droplet as nothing such is mentioned we will take as usual formula of volume 4/3 πr^3,right?

 

[NascentOxygen]

I have just guessed it !Is formula of volume of a shell is (4πradius squared)multiplied by thickness?

That sounds like a good way to approximate it.

 

[gracy]

Is my post #7correct?

 

[gracy]

Volume of droplet should be same as of bubble.

$\frac{4}{3}$$πR^3$=$(4πa^2)$$t$

$R^3$=$3a^2$$t$

$R$=($3a^2$$t$)^1/3 ...... . hope I have written it correctly

potential of bubble=V(has been given)

potential of droplet =V'(have to find)

= $\frac{V'}{V}$=$\frac{Kq}{R}$ ÷ $\frac{Kq}{a}$= $\frac{Kq}{(3a^2t)^1/3}$ ÷ $\frac{Kq}{a}$

= $\frac{a}{(3a^2t)^1/3}$

I don't know how to cancel these two

 

[ehild]

= $\frac{V'}{V}$=$\frac{Kq}{R}$ ÷ $\frac{Kq}{a}$= $\frac{Kq}{(3a^2t)^1/3}$ ÷ $\frac{Kq}{a}$

= $\frac{a}{(3a^2t)^1/3}$

I don't know how to cancel these two

What do you want to cancel? You can simplify. Expand the power of 1/3

 

[gracy]

You can simplify

Yes.I don't know how to do that

 

[ehild]

Yes.I don't know how to do that

expand the denominator

 

[gracy]

expand the denominator

That is my problem.I don't know how to expand the denominator.Because numerator is only "a "we are not going to do anything with it.
Can not we cancel "a "

$\frac{a}{(3a^2t)^1/3}$=$\frac{1}{(3at)^1/3}$

 

[Mister T]

$(a^2)^{\frac{1}{3}}=(a)^{\frac{2}{3}}$

 

[gracy]

Why Can't we cancel "a see post #14

 

[ehild]

You do not cancel a. It stays in the formula. Post #14 is wrong. How do you rise a product to a power? How do you divide powers of a?

 

[Mister T]

You cancel when $\frac{a}{a}=1$. Not when $\frac{a}{a^{2/3}}\neq 1$.

Try, as an example, $a=8$.

 

[gracy]

$\frac{a}{(3a^2t)^1/3}$=$\frac{a}{(3t)^1/3(a)^2\3}$
Or if it is not clear

 

[ehild]

Simplify $\frac{a}{a^{2/3}}$

 

[gracy]

$\frac{a}{a^{2/3}}$

=$a$×${a^{-2/3}}$

=${a^{1/3}}$

 

[NascentOxygen]

$\frac{a}{a^{2/3}}$

=$a$×${a^{-2/3}}$ =${a^{1/3}}$ $\color{red}{✓}$

 

[gracy]

We were asked to find potential of droplet (V')

$\frac{V'}{V}$=$(\frac{a}{3t})$$^{1/3}$

$V'$=$V$$(\frac{a}{3t})$$^{1/3}$

Right?

 

[NascentOxygen]

We were asked to find potential of droplet (V')

$\frac{V'}{V}$=$(\frac{a}{3t})$$^{1/3}$

$V'$=$V$$(\frac{a}{3t})$$^{1/3}$

Right?

[emoji106]