A conducting liquid bubble of radius a and thickness ` t ( tltlt a)` is charged to potential `V`. If the bubble collapses to a droplet , find the potential on the droplet .

Options

A. `V((2a)/(3t))^(1//3)`
B. `v((a)/(3t))^(2//3)`
C. `v((a)/(3t))^(1//3)`
D. `v((a)/(6t))^(1//3)`

Explanation

To find the potential on the droplet after the collapse of a conducting liquid bubble, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- We have a conducting liquid bubble with an outer radius \( a \) and thickness \( t \). The bubble is charged to a potential \( V \).
- The volume of the bubble can be approximated since \( t \ll a \).

2. **Calculating the Charge on the Bubble**:
- The potential \( V \) of the bubble can be related to the charge \( Q \) and radius \( a \) using the formula:
\[ V = \frac{kQ}{a} \]
- Rearranging gives us the charge:
\[ Q = \frac{Va}{k} \]
- Here, \( k \) is Coulomb's constant.

3. **Volume of the Bubble**:
- The volume of the bubble can be approximated as:
\[ \text{Volume of the bubble} = \text{Surface Area} \times \text{Thickness} = 4\pi a^2 t \]

4. **Volume of the Droplet**:
- When the bubble collapses into a droplet, the volume of the droplet is given by:
\[ \text{Volume of the droplet} = \frac{4}{3} \pi r^3 \]
- Setting the volumes equal gives:
\[ 4\pi a^2 t = \frac{4}{3} \pi r^3 \]

5. **Solving for the Radius of the Droplet**:
- Canceling \( 4\pi \) from both sides:
\[ a^2 t = \frac{1}{3} r^3 \]
- Rearranging gives:
\[ r^3 = 3a^2 t \]
- Taking the cube root:
\[ r = (3a^2 t)^{1/3} \]

6. **Finding the Potential of the Droplet**:
- The potential \( V' \) of the droplet can be calculated using the charge \( Q \) and the new radius \( r \):
\[ V' = \frac{kQ}{r} \]
- Substituting for \( Q \):
\[ V' = \frac{k \left(\frac{Va}{k}\right)}{(3a^2 t)^{1/3}} \]
- Simplifying:
\[ V' = \frac{Va}{(3a^2 t)^{1/3}} \]
- This can be rewritten as:
\[ V' = V \cdot \frac{a}{(3^{1/3} a^{2/3} t^{1/3})} = V \cdot \frac{1}{3^{1/3} t^{1/3} a^{1/3}} \]

7. **Final Expression for the Potential**:
- Thus, the potential on the droplet is:
\[ V' = \frac{V a}{3^{1/3} t^{1/3}} \]

### Final Answer:
\[ V' = \frac{V a}{3^{1/3} t^{1/3}} \]