27 identical drops of water are equally and simillarly charged to potential V.They are then united to form a bigger drop.The potential of the bigger drop is??Thank u!!

1 Answer
Feb 1, 2016

Let me derive the general expressions for this condition.

Let there be #n# small drops each having a charge #q# on it and the radius #r#, #V# be its potential and let the volume of each be denoted by #B#.

When these #n# small drops are coalesced there is a new bigger drop formed.

Let the radius of the bigger drop be #R#, #Q# be charge on it, #V'# be its potential and its volume be #B'#

The volume of the bigger drop must be equal to the sum of volumes of #n# individual drops.

#implies B'=B+B+B+......+B#

There are total #n# small drops therefore the sum of volumes of all the individual drops must be #nB#.

#implies B'=nB#

A drop is spherical in shape. Volume of a sphere is given by #4/3pir^3# where #r# is its radius.

#implies 4/3piR^3=n4/3pir^3#

#implies R^3=nr^3#

Taking third root on both sides.

#implies R=n^(1/3)r#

Also the charge of the bigger drop must be equal to the sum of charges on the individual drops.

#implies Q=nq#

The potential of the bigger drop can be given by

#V'=(kQ)/R#

#implies V'=(knq)/(n^(1/3)r)#

#implies V'=n^(1-1/3)(kq)/r#

#implies V'=n^(2/3)(kq)/r#

Since, #kq/r# represents the potential of small drop which we have symbolized by #V#.

Therefore, #V'=n^(2/3)V#

Now we have found a general equation for this case.

In this case there are #27# identical drops.

#implies V'=27^(2/3)V#

#implies V'=9V#

This shows that in your case the potential of the bigger drop is #9# times the potential of the smaller drop.