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`        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.`
one year 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 powerThanks...  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
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one year 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 dropsthus R=(2)1/3rApplying 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)2/3V
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one year ago
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