Electric potential of water droplets

AI Thread Summary
The discussion revolves around calculating the electric potential of a larger water droplet formed by the coalescence of 27 smaller droplets, each with a diameter of 2 mm and a charge of 10^-12 C. Participants clarify that the volume of the larger droplet equals the combined volume of the smaller droplets, leading to a radius calculation of 6 mm, not 1 mm as initially suggested. The total charge is conserved, resulting in a potential of 40.5 V for the larger droplet. There is confusion regarding the correct radius and potential values, with some participants asserting the radius should be 3 mm based on the given diameter of the smaller droplets. The final consensus indicates the potential of the larger droplet is 40.5 V, correcting earlier miscalculations.
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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...
 
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Calculate the potential where? Or do you need to find it as a function?
 
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.
 
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
 
radius came out to be 10-3 m and finally potential came out to be 81V...thanks for helping guys...
 
tsgkl said:
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

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

??
 
Last edited:
andyrk said:
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.
 
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