# Electric potential of water droplets

1. Jun 19, 2013

### tsgkl

1. The problem statement, all variables and given/known data
Twenty seven charged water droplets each with a diameter of 2 mm and a charge of 10-12 C coalesce to form a single drop .Calculate the potential of the bigger drop.

2. Relevant equations
V(potential)=$\frac{q}{4∏εr}$

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 19, 2013

### voko

Calculate the potential where? Or do you need to find it as a function?

3. Jun 19, 2013

### dreamLord

Well we know that the 27 drops are forming into a bigger drop. So now you have a big drop who's charge and radius is unknown - you have to calculate the potential of that. Try to find out the radius and total charge of the new big drop.

4. Jun 19, 2013

### royalhacker

The volume of bigger drop will be equal to volume of 27 small drops
and charge is conserved , so add the total charge

5. Jun 19, 2013

### tsgkl

radius came out to be 10-3 m and finally potential came out to be 81V.....thanks for helping guys....

6. Aug 27, 2013

### andyrk

How? The radius comes out to be 6x10-3m and not 10-3m. And the potential comes out to be 81/2 Volts and not 81 Volts. Can somebody please verify?

27x4/3x∏x(2)3=4/3x∏x(R)3
=> R3=23.33
=> R=6mm=6x10-3m

=>Vbig drop=1/(4∏εo)xQ/R=9x109x27x10-12/6x10-3=81/2V

??

Last edited: Aug 27, 2013
7. Aug 27, 2013

### Staff: Mentor

The problem specifies that the DIAMETER of the small drops is 2mm. The radius of the large drop should turn out to be 3 x 10-3 m , or 3 mm.