Electric potential of water droplets

[tsgkl]

Homework Statement

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.

Homework Equations

V(potential)=$\frac{q}{4∏εr}$

The Attempt at a Solution

i don't really have an idea from where to start...please help...

 

[voko]

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

 

[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.

 

[royalhacker]

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

 

[tsgkl]

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

 

[andyrk]

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

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

=>V_{big drop}=1/(4∏ε_o)xQ/R=9x10⁹x27x10^-12/6x10^-3=81/2V

??

 

[gneill]

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

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.