Coalesce of mercury dropssix at different potential

AI Thread Summary
The discussion revolves around calculating the final potential of six mercury drops initially at +3V and two drops at -3V upon coalescence. Participants explore the challenge of applying the formula for equal potentials to a scenario with differing potentials. A method involving the relationship between charge, potential, and radius is suggested for deriving the final potential. The final calculation leads to a conclusion of 6 volts after considering the total charge from the drops. The solution highlights the importance of understanding the underlying principles of electrostatics in such problems.
AMRIT GAUTAM
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Six mercury drops of equal size given potential of +3v and two other drops are given -3v potential. If they coalesce what is final potential

This question had been eating my brain i know solution if they are given same potential but what to do when they have different potential
 
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Hello and welcome to PF!

If you can work the problem for the case where all 8 drops initially have the same potential, then you should find that it's not much harder to deal with the case of different initial potentials.

Can you show how you get the answer when all 8 drops have the same initial potential?
 
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TSny said:
Hello and welcome to PF!

If you can work the problem for the case where all 6 drops initially have the same potential, then you should find that it's not much harder to deal with the case of different initial potentials.

Can you show how you get the answer when all 6 drops have the same initial potential?
Um i solve it using formula
V=n^(2/3) vs where n is no of drops and vs is potential of each drop but it works for condition when all are provided same potential
 
Do you understand how to derive the formula V = n2/3 vs for equal initial potentials?

If so, then you can use essentially the same method of derivation to get the result for unequal initial potentials.

If not, my hint would be to consider how the charge on a spherical drop is related to the potential and radius of the drop.
 
TSny said:
Do you understand how to derive the formula V = n2/3 vs for equal initial potentials?

If so, then you can use essentially the same method of derivation to get the result for unequal initial potentials.

If not, my hint would be to consider how the charge on a spherical drop is related to the potential and radius of the drop.
I know but this methode only useful for similar charge only searched whole internet can't find right answer. All links leads to same potential problem only i tried it myself bt couldnt
 
Thank you guys for idea
Finally solved it first used formula 4/3 pi R^3= n 4/3 pi r ^3
Which gave R= n^(1/3)r ...(1)
And potential V= k Q/R ...(2)
where Q = 6q-2q since they are provided potential of different sign
And gave
Combining 1 and 2 and puting value of Q and n =8
I came up with answer 6 volts
 
Looks good!
 
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