A soap bubble of radius 3 cm is charged with 9.0nC. The excess pressure inside the bubble, if surface tension of soap solution = 3×10^-3 N/m is

(1)0.26N/m^2

(2)0.36

(3)0.44

(4)0.52

Asked by shahrithik07 | 27th Oct, 2018, 07:53: PM

Expert Answer:

In a charged soap bubble, there are two forces acting on it.
One is surface tension forces (4T/r) that try to reduce the size of bubble, where T is surface tension and r is radius of bubble.
Other one is electrostaic force [ σ2/(2ε0) ] due to the chrage distribution on the bubble that try to expand its size,
where σ is surface charge density and ε0 is permitivity of free space. Both the forces mentioned above are per unit area.
 
Hence the excess pressure P inside bubble is given by
 
P = (4T/r) - [ σ2/(2ε0) ] ................(1)
σ = q/(4πr2) = (9×10-9)/(4π×3×3×10-4) = 7.97×10-7 C/m2 ..................(2)
 
By substituting surface charge density from (2) in (1) along with other data,

we get P = (4×3×10-3)/(3×10-2) - (7.97×7.97×10-14)/(2×8.854×10-12) = 0.36 N/m2

Answered by Thiyagarajan K | 28th Oct, 2018, 02:28: AM

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