I was thinking about what'd happen to the radius of a bubble, if a charge is uniformly distributed over it.

At the beginning, I thought that the bubble would expand, for sure, because of the like-charge repulsion over the surface.

But then I realized that the Pressure difference between the atmosphere and the interior of the bubble is given by $$P-P_o=\frac{4T}{R}$$ where $R$ is the radius and $T$ the surface tension of the liquid which forms the bubble. $P-P_o$ would remain the same after charging the bubble, but the repulsive forces in the bubble would oppose surface tension and reduce it, and hence, it follows that $R$ should decrease, in order to keep the whole quantity constant.

(I can't decide between which of my opposing views is correct. I think it might expand, because I'm probably neglecting a charge generated pressure in the expression I gave above.)

What would actually happen?


1 Answer 1


You are confusing yourself. The statement

$P - P_0$ would remain the same

is false. Why would it remain the same? There is a certain amount of compressed gas inside the bubble, and there is a force that maintains it compressed. In the first case, this force is just the surface tension. In the second case, it is the surface tension reduced by the surface charge. The gas will expand according to $PV=nRT$ until a new equilibrium is found, with a new pressure that exactly cancels the force/area of the new bubble.


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