# Spherical droplet - potential

1. Sep 12, 2009

### exitwound

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

A spherical drop of water carrying a charge of 24 pC has a potential of 490 V at its surface (with V = 0 at infinity).

(a) What is the radius of the drop?

(b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

2. Relevant equations

V=kQ/r

3. The attempt at a solution

If V=kQ/r then if the radius is doubled and the charge is doubled, why isn't the answer the same? V=2kQ/2r = kQ/r

2. Sep 12, 2009

### kuruman

Because two drops of radius r when added together do not give a drop of twice the radius. Think about it, what is it that doubles?

3. Sep 12, 2009

### exitwound

Ok The volume doubles. I'm a moron.

if
$Vol=(4/3)\pi R^3$
$Vol=(4/3)\pi (4.41e-4)^3$
$Vol=3.59e-10$

3.59*2=7.18e-10

$7.18e-10=(4/3)\pi (R)^3$
$R=5.55e-4$

$V=kQ/r$
$V=(9e9)(2*24e-12)/(5.55e-4)$
$V=778 Volts$

Correct? don't wanna lose any more points on these problems.

4. Sep 12, 2009

### kuruman

I didn't check your calculations of the potential but the new radius is correct. For future reference, since the volume depends on the cube of the radius, if you double the volume, the radius increases by a factor of 21/3.