# Homework Help: Potential of the droplet.

1. Dec 10, 2015

### gracy

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Potential at the surface of bubble=$V$=$\frac{Kq}{a}$

3. 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.

2. Dec 10, 2015

### ehild

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

3. Dec 10, 2015

### gracy

Volume of bubble=$(4πa^2)t$?

4. Dec 10, 2015

### ehild

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

5. Dec 10, 2015

### gracy

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

6. Dec 10, 2015

### Staff: Mentor

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

7. Dec 10, 2015

### gracy

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

8. Dec 10, 2015

### Staff: Mentor

That sounds like a good way to approximate it.

9. Dec 10, 2015

### gracy

Is my post #7correct?

10. Dec 10, 2015

### gracy

$\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

11. Dec 10, 2015

### ehild

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

12. Dec 10, 2015

### gracy

Yes.I don't know how to do that

13. Dec 10, 2015

### ehild

expand the denominator

14. Dec 10, 2015

### gracy

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 "

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

15. Dec 10, 2015

### Mister T

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

16. Dec 10, 2015

### gracy

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

17. Dec 10, 2015

### 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?

18. Dec 10, 2015

### Mister T

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

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

19. Dec 10, 2015

### gracy

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

20. Dec 10, 2015

### ehild

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

21. Dec 10, 2015

### gracy

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

=$a$×${a^{-2/3}}$

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

22. Dec 10, 2015

### Staff: Mentor

23. Dec 11, 2015

### 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?

24. Dec 11, 2015