Question

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!!

Answer 1

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.