A conducting bubble of radius a, thickness t (t

Q: Why do we use $4\pi a^2 t$ for the volume of the bubble?

Because the thickness $t$ is very small compared to the radius $a$, the volume of the spherical shell can be accurately approximated as its surface area ($4\pi a^2$) multiplied by its thickness ($t$).

Q: Why does the charge remain constant during the collapse?

The bubble is an isolated conducting system. When it collapses, there is no physical path for the charge to escape to the surroundings, so the total charge is strictly conserved.

Q: Why does the potential increase when the bubble collapses?

Potential is inversely proportional to the radius for a given charge ($V \propto 1/R$). Since the solid droplet's radius $R$ is much smaller than the hollow bubble's radius $a$, the potential must increase.

Q: What happens to the capacitance of the system?

The capacitance of an isolated sphere is $C = 4\pi\varepsilon_0 R$. Since the radius decreases from $a$ to $R$, the capacitance decreases. With constant charge ($Q=CV$), a decrease in $C$ mathematically forces an increase in $V$.

Q: Is the electrostatic energy conserved during the collapse?

No, the electrostatic potential energy $U = q^2 / 2C$ actually increases because the capacitance decreases. This extra energy comes from the mechanical work done by the surface tension forces that cause the bubble to collapse.

Question

A conducting bubble of radius $a$, thickness $t$ ($t << a$) has potential $V$. Now the bubble collapses into a droplet. Find the potential of the droplet.

Sigma Percentile · Accepted Answer

### The Magic of Collapsing Bubbles

Imagine a delicate, thin conducting soap bubble floating in the air. It has a radius $a$ and a microscopic thickness $t$. Because it is a conductor, any charge $q$ given to it will uniformly distribute itself over its outer surface. The electric potential $V$ at the surface of this bubble is governed by the classic formula:

$$V = \frac{1}{4\pi\varepsilon_0} \frac{q}{a}$$

From this, we can easily express the total charge $q$ residing on the bubble in terms of its potential and radius:

$$q = (4\pi\varepsilon_0) V a$$

### The Collapse: What Remains Constant?

Suddenly, the bubble collapses! It transforms from a hollow, thin shell into a dense, solid spherical droplet of a new radius $R$. In this chaotic transformation, two fundamental physical quantities remain absolutely conserved:

1. **Total Charge ($q$):** The system is isolated, so the charge has nowhere to escape.
2. **Volume of the Conducting Material:** The actual amount of liquid making up the bubble doesn't vanish; it just reshapes itself.

Let's use the conservation of volume to find the new radius $R$. The volume of the thin bubble shell can be approximated as its surface area multiplied by its thickness:

$$V_{	ext{shell}} = (4\pi a^2) t$$

This must equal the volume of the newly formed solid droplet:

$$(4\pi a^2) t = \frac{4}{3}\pi R^3$$

By canceling out the $4\pi$ from both sides, we can solve for $R^3$:

$$R^3 = 3a^2 t \implies R = (3a^2 t)^{1/3}$$

### The Spike in Potential

Now that we have the radius of the new droplet, we can determine its new electric potential, $V'$. Since the charge $q$ is conserved, the new potential is simply:

$$V' = \frac{1}{4\pi\varepsilon_0} \frac{q}{R}$$

Let's substitute the expressions we derived for $q$ and $R$ into this equation:

$$V' = \frac{1}{4\pi\varepsilon_0} \frac{(4\pi\varepsilon_0) V a}{(3a^2 t)^{1/3}}$$

The $4\pi\varepsilon_0$ terms cancel out beautifully, leaving us with:

$$V' = V \frac{a}{(3a^2 t)^{1/3}}$$

To make this expression more elegant, we can bring the $a$ in the numerator inside the cube root as $a^3$:

$$V' = V \left( \frac{a^3}{3a^2 t} ight)^{1/3} = V \left( \frac{a}{3t} ight)^{1/3}$$

**A Fascinating Physical Insight:** Because the bubble was extremely thin ($a \gg t$), the fraction $\frac{a}{3t}$ is much greater than 1. This means the new potential $V'$ is significantly higher than the initial potential $V$. Physically, the capacitance of the object drastically decreased when it shrank from a large bubble to a tiny droplet. To hold the exact same amount of charge with a much smaller capacitance, the electric potential had to skyrocket!

Animated Solution for Physics - Electrostatics: A conducting bubble of radius $a$ , thickness $t$ ( $t << a$ ) has potential $V$ . Now the bubble collapses into a droplet. Find the potential of the droplet.

Visualized Solution

Visualizing the Initial State

Relating Potential to Charge

Extracting the Charge

The Collapse: Conservation Laws

Equating the Volumes

Finding the New Radius

Potential of the Droplet

Substituting Known Values

Final Simplification

Physical Insight

The Sigma Insight: Electric Potential and Potential Difference

The Magic of Collapsing Bubbles

The Collapse

The Spike in Potential

Similar Questions

A charge $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ ( $> r$ ) such that the surface densities are equal. Find the potential at the common centre.

A charge $Q$ is distributed over three concentric spherical shells of radii $a, b, c$ ( $a < b < c$ ) such that their surface charge densities are equal to one another.\nThe total potential at a point at distance $r$ from their common centre, where $r < a$ would be

A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$ and $R$ ( $R > r$ ). If the surface charge densities on the two shells are equal, the electric potential at the common centre is

A thin spherical conducting shell of radius $R$ has a charge $q$ . Another charge $Q$ is placed at the centre of the shell. The electrostatic potential at a point $P$ at a distance $R /2$ from the centre of the shell is

Three concentric metal shells $A, B$ and $C$ of respective radii $a, b$ and $c$ ( $a < b < c$ ) have surface charge densities $+ σ, - σ$ and $+ σ$ , respectively. The potential of shell $B$ is

Concentric metallic hollow spheres of radii $R$ and $4 R$ hold charges $Q_{1}$ and $Q_{2}$ , respectively. Given that, surface charge densities of the concentric spheres are equal. The potential difference $V (R) - V (4 R)$ is