Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/196

If a spherical soap bubble is electrified to a potential $A$, then, if its radius is $a$, the charge will be $Aa$, and the surface-density will be

$\sigma ={\frac {1}{4\pi }}{\frac {A}{a}}.$

The resultant electrical force just outside the surface will be $4\pi \sigma$, and inside the bubble it is zero, so that by Art. 79 the electrical force on unit of area of the surface will be $2\pi \sigma ^{2}$, acting outwards. Hence the electrification will diminish the pressure of the air within the bubble by $2\pi \sigma ^{2}$, or

${\frac {1}{8\pi }}{\frac {A^{2}}{a^{2}}}.$

But it may be shewn that if $T$ is the tension which the liquid film exerts across a line of unit length, then the pressure from $T$ within required to keep the bubble from collapsing is $2{\tfrac {T}{a}}$. If the electrical force is just sufficient to keep the bubble in equilibrium when the air within and without is at the same pressure

$A^{2}=16\pi aT.$

Two Infinite Coaxal Cylindric Surfaces.

126.] Let the radius of the outer surface of a conducting cylinder be $a$, and let the radius of the inner surface of a hollow cylinder, having the same axis with the first, be $b$. Let their potentials be $A$ and $B$ respectively. Then, since the potential $V$ is in this case a function of $r$, the distance from the axis, Laplace’s equation becomes

${\frac {d^{2}V}{dr^{2}}}+{\frac {1}{r}}{\frac {dV}{dr}}=0,$

whence

$V=C_{1}+C_{2}\log r.$

Since $V=A$ when $r=a$, and $V=B$ when $r=b$,

$V={\frac {A\log {\frac {b}{r}}+B\log {\frac {r}{a}}}{\log {\frac {b}{a}}}}.$

If $\sigma _{1},\ \sigma _{2}$ are the surface-densities on the inner and outer surfaces,

$4\pi \sigma _{1}={\frac {A-B}{a\log {\frac {b}{a}}}},\ 4\pi \sigma _{2}={\frac {B-A}{b\log {\frac {b}{a}}}}.$