A conducting liquid bubble of radius a and thickness ` t ( tltlt a)` is charged to potential `V`. If the bubble collapses to a droplet , find the potential on the droplet .

Answer

`V'=V(q/(3t))^(1//3)`

Explanation

Let q be the charge on the bubble, then
`V=(Kq)/a("Here" K=1/(4piepsilon_(0)))`
`:. Q=(Va)/K`
Let after collapsing the radius of droplet becomes R, then equating the volume, we have
`(4pia^(2))t=4/3 pi R^(3)`
`:. R=(3a^(2)t)^(1//3)`
Now, potential of proplet will be `V'=(Kq)/R`
Substituting the values, we have
`V'=((K)((Va)/K))/((3a^(2)t)^(1//3))rArr V'=V(a/(3t))^(1//3)`