Question

In millinkan's oil drop experiment charged oil drop when subjected to potential difference of V falls with some acelleration if the potential difference between the plates is increased to 2V the drops moves up with same acceleration how much potential is required to balance the drop

Answer 1

Let potential f 1 drop = Kq/r =V

Volume of n drops = volume of bigger drop

n 4/3 pi (r)^3 = 4/3 pi (R)^3

Therefore , R = ( (n)^1/3) r

also charge remains conserved

charge on 1 drop = q

charge on big drop containing n drops = (nq)

Therefore potential on big drop = K (nq)/ ((n)^1/3)r

=(( n)^2/3) Kq/r

= (( n)^2/3) V