Question

1000 identical drops are combined to form a big drop.if electric field of each drop is 5 v/m then what would be electric field of big drop?

Answer 1

Let r be the radius of small drops and big drop R.

kq/r = 1  ….......(1)

1000* 4/3pi r^3 = 4/3piR^3  ….......(2)

 

potential of big drop = kQ/R                    

 

Substitute R in terms of r using eq 2 & find the ans

Answer 2

Hi...

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.

⇒B'=B+B+B+......+B

There are total n small drops therefore the sum of volumes of all the individual drops must be n B.
⇒B'=nB
A drop is spherical in shape. Volume of a sphere is given by 4/3πr^3
where r is its radius.
⇒4/3πR^3=n4/3πr^3
⇒R^3=nr^3

Taking third root on both sides.

⇒R=n^1/3r

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

⇒Q=nq

The potential of the bigger drop can be given by

V'=kQ/R
⇒V'=knq / n^1/3r

⇒ V'=n^1−1/3 * kq/ r.

⇒V' = n^2/3 * kq/r
Since,
kq/r represents the potential of small drop which we have symbolized by
V.

Therefore,
V'=n^2/3V

Now we have found a general equation for this case.

In this case there are 1000 identical drops.

⇒V'=1000^2/3 V

⇒V'=10V

This shows that in your case the potential of the bigger drop is
10
times the potential of the smaller drop...

Hope this helps u!!