JEE Main & Advanced Physics Electrostatics & Capacitance Equilibrium of Charged Soap Bubble

(1) For a charged soap bubble of radius R and surface tension T and charge density \[\sigma .\] The pressure due to surface tension \[4\frac{T}{R}\] and atmospheric pressure \[{{P}_{\text{out}}}\] act radially inwards and the electrical pressure \[({{P}_{el}})\] acts radially outward.

(2) The total pressure inside the soap bubble

\[{{P}_{\text{in}}}={{P}_{\text{out}}}+\frac{4T}{R}-\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}\]

(3) Excess pressure inside the charged soap bubble

\[{{P}_{\text{in}}}-{{P}_{\text{out}}}\,\,={{P}_{\text{excess}}}=\frac{4T}{R}-\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}\].

(4) If air pressure inside and outside are assumed equal then \[{{P}_{\text{in}}}={{P}_{\text{out}}}\]i.e.,\[{{P}_{\text{excess}}}=0\]. So, \[\frac{4T}{R}=\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}\]

(i) Charge density : Since \[\frac{4T}{R}=\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}\]\[\Rightarrow \]\[\sigma \,=\sqrt{\frac{8{{\varepsilon }_{0}}T}{R}}=\sqrt{\frac{2T}{\pi kR}}\]

(ii) Radius of bubble \[R=\frac{8\,{{\varepsilon }_{0}}\,T}{{{\sigma }^{2}}}\]

(iii) Surface tension \[T=\frac{{{\sigma }^{2}}R}{8{{\varepsilon }_{0}}}\]

(iv) Total charge on the bubble \[Q=8\pi R\sqrt{2{{\varepsilon }_{0}}TR}\]

(v) Electric field intensity at the surface of the bubble

\[E=\sqrt{\frac{8T}{{{\varepsilon }_{0}}R}}=\sqrt{\frac{32\pi \,kT}{R}}\]

(vi) Electric potential at the surface \[V=\sqrt{32\pi RTk}=\sqrt{\frac{8RT}{{{\varepsilon }_{0}}}}\]