Considering a uniformly charged spherical shell, it is easy to show with a virtual work argument that the pressure will be given by

$$ p = \frac{Q^2}{32\pi^2 \epsilon_0 R^4} $$

On the other side, the excess pressure in a soap bubble is given by the Laplace pressure $ p_L = 2 \gamma / R $ and is needed to balance the tension from the walls.

My question is, what happens if a soap bubble is charged? The charges will spread evenly and expand the bubble, which is ultimately limited by the balance from the tension(?) I.e. can we "exchange" the Laplace pressure to the electrical one? Am I right in thinking that that would give equilibrium at: $$ \frac{Q^2}{32\pi^2 \epsilon_0 R^4} = 2 \frac{\gamma}{R} $$? But is this an unstable equilibrium due to the different R dependence?

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    $\begingroup$ Both the electrostatic pressure and the excess fluid pressure are pushing outward, so in a charged soap bubble, you're adding one to the other, not replacing one with the other. As such, you'll find that the excess fluid pressure required to balance tension from the walls is less than the usual Laplace pressure. $\endgroup$ May 21, 2018 at 14:05
  • $\begingroup$ @probably_someone if the fluid pressure decreases due to the electric pressure, doesn't that mean that it is "replaced" partly by the electric pressure? Will the equilibrium be stable? $\endgroup$
    – Jhonny
    May 26, 2018 at 13:20
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    $\begingroup$ I was referring to the static case here, in which you're comparing the internal pressure inside an uncharged bubble that has existed forever to the internal pressure inside a charged bubble that has existed forever. The equations you outline do not apply in non-equilibrium situations (like if you were to put charge on an uncharged bubble), so I'm not sure how you're making any stability claims for equations that manifestly only hold in equilibrium. $\endgroup$ May 26, 2018 at 22:58
  • $\begingroup$ @probably_someone I apologise, I didn't mean to missuse the equations. What I am in essence trying to figure out is if we start with an uncharged bubble and then add on charge if this will be stable or not. I am not sure what equations to apply to work this out. $\endgroup$
    – Jhonny
    May 27, 2018 at 16:06

1 Answer 1


When a bubble is charged it tends to expand because the charges prefer to stay as far away from each other as possible. Lord Rayleigh, in a paper published in 1882, showed that the maximum charge that a bubble of radius $r$ can hold without disintegrating is $q = 4\sqrt{\pi\gamma} r^{3/2}$, where $\gamma$ is the interfacial tension between the drop and the surrounding medium. Please note that the result is in CGS (gaussian) units.

Thus, a soap bubble is more unstable when it is charged than when it is not.


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