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))

A soap bubble of radius r and surface tension constant T is given a ch

1.	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))
Answer» Solution :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)`.