# Two drops of liquid are charged to the same potential of 100v. They are then merged into one large drop , the potential of the large drop is.

chetan jangir
101 Points
5 years ago
Here we only apply two basic concepts first volume conservation and the other is charge conservation.So applying volume conservation the new radius can be found easily as 2 *4/3πR^3=4/3πr^3 where is new one radius .solving we get r=R2^1/3 . And now we know as the charge is conserved we can write the total charge as 2Q where Q was charge due to which a single drop has potential 100V .As the the potential of a sphere (we have let the drops to be spheres here for solving problems otherwise this question can't be solved)as KQnet/Rfinal and putting values V(final) =K2Q/R2^1/3 and as we know 100=KQ/R.Putting valve V(final)=100(2^2/3) .Here ^ stands for to power
Thanks...  If you found answer to be correct and logic understanding pls approve it and ask further we will be happy to build up your concepts
Khimraj
3007 Points
5 years ago
Assuming spherical drops:
The potential at surface of a sphere is kQ/r where r  is radius of sphere and Q is its charge , k is constant whose value depends on units used (1/(4XpiXepsilon))
Now when 2 drops coalesce to form 1 drop volume remains same.
2(4 pi r^3)/3 = (4 pi R^3)/3   ; R= radius of bigger drop and r = radii of small drops
thus R=(2)1/3r
Applying conservation of charge the total charge of bigger sphere is 2Q where Q is charge of one small drop.
Now potential of bigger sphere is
k(2Q)\R = k(2Q)\(2)1/3r = 9(kQ\r) =10*(2)2/3