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(NONE) Let the charge on the big bubble be $Q$ and its radius be $R$. The charge density is $\sigma = \frac{Q}{4\pi R^2}$.
The mechanical force per u

Vedclass · Accepted Answer

$16: 1$

Vedclass · Answer

(NONE) Let the charge on the big bubble be $Q$ and its radius be $R$. The charge density is $\sigma = \frac{Q}{4\pi R^2}$.
The mechanical force per unit area (electrostatic pressure) on a charged bubble is given by $P = \frac{\sigma^2}{2\epsilon_0} = \frac{Q^2}{2\epsilon_0 (4\pi R^2)^2} = \frac{Q^2}{32\pi^2 \epsilon_0 R^4}$.
When the bubble breaks into $64$ small bubbles,the volume is conserved: $\frac{4}{3}\pi R^3 = 64 	imes \frac{4}{3}\pi r^3$,which gives $R^3 = 64r^3$,so $R = 4r$.
The total charge $Q$ is conserved,so each small bubble has charge $q = \frac{Q}{64}$.
The electrostatic pressure on a small bubble is $p = \frac{q^2}{32\pi^2 \epsilon_0 r^4} = \frac{(Q/64)^2}{32\pi^2 \epsilon_0 (R/4)^4} = \frac{Q^2 / 64^2}{32\pi^2 \epsilon_0 (R^4 / 256)} = \frac{Q^2}{32\pi^2 \epsilon_0 R^4} 	imes \frac{256}{4096} = P 	imes \frac{1}{16}$.
Therefore,the ratio of the mechanical force per unit area of the big bubble to that of a small bubble is $P/p = 16:1$.

In air,a charged soap bubble of radius $R$ breaks into $64$ small soap bubbles of equal radius $r$. The ratio of mechanical force per unit area of the big soap bubble to that of a small bubble is

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