When a soap bubble of radius 0.2 mm is charged it experiences an outward electrostatic pressure of magnitude `sigma^(2)/(2epsilon_(0))` , where `sigma=20muCm^(-2)` is the surface charge density. If the excess pressure inside the soap bubble due to the tension is same as this electrostatic pressure then the surface tension of the soap solution is
`(epsi_(0)=8.85xx10^(-12)C^(2)N^(-1)m^(-1))`

Options

A. `8.85 xx10^(-4)Nm^(-1)`
B. `12.4 xx10^(-4) Nm^(-1)`
C. `11.3xx10^(-4)Nm^(-1)`
D. `90xx10^(-4)Nm^(-1)`

Answer

C

Explanation

Excess pressure inside soap bubble due to surface tension
`p= ((4S)/( R))` where S = surface tension and R = radius . `implies ` Electrostatic pressure ` p = (sigma^(2))/(2epsi_(0))`
According to the equestion, `(4S)/( R) = (sigma^(2))/(2epsi_(0))`
`S= (sigma^(2)R)/(deltaepsi_(0)) " " ........(i)`
Putting all values in Eq. (i) we get
`S=((20xx10^(-6))^(2)xx(0.2xx10^(-3)))/((8xx8.85xx10-12))`
`implies S= 11.3xx10^(-4)N//m`