A soap bubble of radius `r` and surface tension constant `T` is given a charge, so that its surface charge density is `sigma`. Due to charge, the radius os the soap bubble becomes double then find `'sigma'`. (atmospheric pressure `= P_(0)`)

Explanation

initial pressure inside the bubble
`P_(i) = P_(0) + (4T)/(r)`
Now a uniform surface charge in given to the bubble
The surface tension is a pulling a force, which increases pressure inside the bubble (by `(4T)/(r))`
but the charge given to the surface will repel each other. So due to the charge given. pressure inside the bubble will decrease (by `(sigma^(2))/(2epsilon_(0))`)
So, final pressure inside the bubble
`P_(1) = P_(0) + (4T)/(r_(f)) - (sigma^(2))/(2epsilon)`
As the temperature of the gas inside the bubble if costant so,
`P_(1)V_(1) = P_(1)V_(1)`
`(P_(0) + (2T)/(r))((4)/(3)pir^(3)) = (P_(0) + (4T)/(r_(f))) = (P_(0) + (4T)/(r_(f)) - (sigma^(2))/(2epsilon_(0)))((4)/(3)pir_(f)^(3))`
Here Put`r_(1) = 2r`
So, get `sigma = sqrt((7P_(0) - (12T)/(r))2epsilon)`.