Potential of two surfaces coming together

1. May 10, 2008

terminator88

[SOLVED] potential of two surfaces coming together

lets say we have two water droplets(assume them to be spherical shape) with a each having charge q and each has the potential at its surface as kq/r.What is the potential if the two water droplets combine?
I thought tht the radius and charge wud double...and so the potential wud double...but its wrong.Why?Thx in advance

2. May 11, 2008

dx

Radius will not double. The correct factor is the cube root of 2. Why do you think the radius will double?

3. May 11, 2008

terminator88

why is it cube root of two???I thought the radius will double cuz they come together....

4. May 11, 2008

dx

No, the volume will double if you put them together. Can you work it out now?

5. May 11, 2008

terminator88

nope still can figure it out..how can I get the new radius from the doubling of the volume and the charge??

6. May 11, 2008

dx

The radius has nothing to do with the charge. Initially you have two spherical droplets of radius r. They each have volume $$k r^3$$. Put together they have total volume $$2kr^3$$. Now what would the radius of a sphere have to be if its volume is $$2kr^3$$? Call the radius of the combined sphere $$r_{1}$$. Write the expression for its volume interms of $$r_{1}$$. Equate that to $$2kr^3$$.

Last edited: May 11, 2008
7. May 11, 2008

terminator88

ohhh..I c.So r is radius of the original drop and r1 is the radius of the new combined drop.
So equating gives kr^3=2kr1^3 which gives r1=2^(1/3)r
then substituting in the original eqtn gives 2kq/{[2^(1/3)]r}...then u get 2^(2/3) not 2^(1/3)...so wht did I do wrong this time?thank you

8. May 11, 2008

dx

$$2^{1/3}$$ is the factor by which the radius changes. The potential is proportional to the charge, and inversely proportional to the radius. So if the charge is doubled, and the radius is increased by a factor of $$2^{1/3}$$, the potential will increase by a factor of $$2^{2/3}$$.

9. May 11, 2008

terminator88

ohk thank you so much...you r a genius!