Potential of two surfaces coming together

[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

 

[dx]

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

 

[terminator88]

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

 

[dx]

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

 

[terminator88]

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

 

[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 $$2kr^3$$.

 

[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

 

[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 $$2^{2/3}$$.

 

[terminator88]

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