Radially oscillating charged soap bubble

Since the charge $q$ is small, the amplitude of oscillations will be small. Let $\sigma_{0} = \dfrac{q}{4 \pi r^2}$ be the surface charge density.

Consider the instant when the radius is $r = R + x$ , where $x << R$

Initial pressure = $P_{1} = P_{A} + 4 \dfrac{\sigma}{R}$

Final pressure = $P_{2} = P_{1} \bigg(\dfrac{V_{1}}{V_{2}} \bigg)^{5/3}$

Hence, $P_{2} = \bigg(P_{A}+4 \dfrac{\sigma}{R}\bigg)\bigg(\dfrac{R}{R+x}\bigg)^5 \approx \bigg(P_{A}+\dfrac{4\sigma}{R} \bigg)\bigg(1 - 5 \dfrac{x}{R}\bigg)$

Now, consider a small area element $dA$ with mass $dM = \dfrac{M}{4 \pi r^2} dA$ on the surface.

The electrostatic pressure is $P_{E} = \dfrac{\sigma_{0}^2}{2 \epsilon_{0}}$

Now, clearly, the net pressure is :

$P_{net} = P_{2} + P_{E} - \dfrac{4 \sigma}{r} - P_{A}$

= $-5P_{A} \dfrac{x}{R} - + \bigg(\dfrac{4 \sigma}{R}\bigg(1 - \dfrac{5x}{R} \bigg) - \dfrac{4 \sigma}{R} \bigg(1+ \dfrac{x}{R}\bigg) + \dfrac{\sigma_{0}^2}{2 \epsilon_{0}}$

= $-5P_{A} \dfrac{x}{R} - 16\sigma \dfrac{x}{R^2} + \dfrac{q^2}{32 \pi^2 \epsilon_{0} r^4}$

$dF = P_{net} dA = dM \ddot{x} = \dfrac{M}{4 \pi r^2} dA \ddot{x}$

$\Rightarrow M \ddot{x} = 4 \pi r^2 P_{net}$

Hence,

$M \ddot{x} \approx -4 \pi (5P_{A}R +16 \sigma)x + \dfrac{q^2}{8 \pi \epsilon_{0} R^2} \bigg(1 - 2 \dfrac{x}{R}\bigg)$

Now , we neglect $\dfrac{q^2 x}{4 \pi \epsilon_{0} R^3}$

Clearly, this represents SHM with a time period of $T_{0} = \sqrt{\dfrac{\pi M}{5P_{A}R+16 \sigma}}$

Hence, $a+b+c = 22$

Note: Another method can be energy conservation, where electrostatic energy is $\dfrac{q^2}{8 \pi \epsilon_{0} r}$ .

The question and the given answer seems correct to me, wonder why some have reported it,

Here is my way,

At any instant the forces acting on an elementary surface of the bubble are

1) The pressure of the air trapped inside (it is important to mention here that the process is adiabatic as mentioned , so P1V1=P2V2 is not valid) Instead use adiabatic equation

2) The atmospheric pressure which i assumed to be constant

3) The electrostatic repulsion among the chosen surface and other surfaces( here we will neglect it at the end as q is very small so q^2 is smaller )( and it is logical since SHM approximation only holds if such is the case)

4) The inward pull of the surface tension trying to contract the bubble into a point , (again note the pressure corresponding to this is 4S/R)

Now let us choose an elementary surface area to operate with say $\Delta S$

Now at any instant,, if the radius of the bubble is R, the force analsys will give

$\Delta S({ P }_{ i }-\frac { 4S }{ R } -{ P }_{ o }+E)=(\Delta M)\ddot { r } =\rho \Delta S\ddot { r } =\frac { M }{ 4\pi { R }^{ 2 } } \Delta S\ddot { r } \\$

It is important to know that you cannot ignore the changing density of the bubble, it matters

Now using

$P{ V }^{ \frac { 5 }{ 3 } }=k$

and also that initial pressure inside bubble (before giving charge) was

${ P }_{ o }+\frac { 4S }{ R }$

we have

$({ (P }_{ o }+\frac { 4S }{ { R }_{ o } } )\frac { { { R }_{ 0 } }^{ 5 } }{ { R }^{ 5 } } -\frac { 4S }{ R } -{ P }_{ o }+E)=\frac { M }{ 4\pi { R }^{ 2 } } \ddot { r } \\$

Now rearrangeing we get

$({ (P }_{ o }+\frac { 4S }{ { R }_{ o } } )\frac { { { R }_{ 0 } }^{ 5 } }{ { R }^{ 3 } } -\frac { 4SR }{ 1 } -{ P }_{ o }{ R }^{ 2 }+E{ R }^{ 2 })=\frac { M }{ 4\pi } \ddot { r } \\ \\$

Now consider the bubble in equillibrium,

ALso you can check that for small charge the original radius of the bubble is almost same as it was in equillibirum, we will use that later, for now let us use ${ R }_{ e }\\$ to denote it

so we have $\\ \\ ({ (P }_{ o }+\frac { 4S }{ { R }_{ o } } )\frac { { { R }_{ 0 } }^{ 5 } }{ { { R }_{ e } }^{ 3 } } -\frac { 4SR_{ e } }{ 1 } -{ P }_{ o }{ { R }_{ e } }^{ 2 }+{ E }_{ e }{ { R }_{ e } }^{ 2 })=0$

Now subtracting both and using differential approximation

or $\\ f(x+\Delta x)-f(x)=f^{ 1 }\left( x \right) \Delta x$

(it is valid since subtracting something by 0 doesnt change it)

we have

$({ -3(P }_{ o }+\frac { 4S }{ { R }_{ o } } )\frac { { { R }_{ 0 } }^{ 5 } }{ { { R }_{ e } }^{ 4 } } -\frac { 4S }{ 1 } -{ 2P }_{ o }{ { R }_{ e } })\Delta R+\Delta (E{ R }^{ 2 })=\frac { M }{ 4\pi } \ddot { r } \\ \\ \\$

Now using the fact that the electrostatic repulsion expression contains a q^2 term which is negligible (you can check it easily yourself, i am not doing it since is is bulky and the equations are already bulky enough)

Also taking

${ R }_{ 0 }\simeq { R }_{ e }\\$

we have a simple SHM type equation

$-({ 3(P }_{ o }+\frac { 4S }{ { R }_{ o } } ){ R }_{ 0 }+4S+{ 2P }_{ o }{ { R }_{ 0 } })\Delta R=\frac { M }{ 4\pi } \ddot { r } \\$

Now collecting like terms and using 2pi/w=T we get

$T=\sqrt { \frac { M\pi }{ 5{ P }_{ 0 }{ R }_{ 0 }+16S } }$

thus answer 22,

hopefully i am correct

Not difficult at all but it sure tests ur basics :) the given condition r adiabatic ..... use PV^(gamma)=constant . and proceed as the standard process to evaluate time period :)