A conducting bubble of radius a, thickness t ( lt lt a) has a potential V. Now the bubble transforms into a droplet. Find the potential on the surface of the droplet.

Options

A. `v'=v((a)/(3t))^((1)/(3))`
B. `v'=v((a)/(3t))^((1)/(2))`
C. `v'=v((a)/(3t))`
D. None of these

Answer

To solve the problem of finding the potential on the surface of a droplet formed from a conducting bubble, we can follow these steps:

### Step 1: Understand the Initial Conditions
We have a conducting bubble with:
- Radius = \( a \)
- Thickness = \( t \) (where \( t \ll a \))
- Potential = \( V \)

### Step 2: Relate Charge to Potential
The potential \( V \) at the surface of the bubble can be expressed using the formula for electric potential due to a charged sphere:
\[ V = \frac{kQ}{a} \]
where \( k \) is Coulomb's constant and \( Q \) is the charge on the bubble.

### Step 3: Express Charge in Terms of Potential
From the equation above, we can express the charge \( Q \) as:
\[ Q = \frac{V a}{k} \]

### Step 4: Volume Conservation During Transformation
When the bubble collapses into a droplet, the volume of the conducting bubble must equal the volume of the droplet. The volume of the bubble can be approximated as:
\[ \text{Volume of bubble} \approx 4\pi a^2 t \]
The volume of the droplet (assuming it is a sphere of radius \( R \)) is:
\[ \text{Volume of droplet} = \frac{4}{3} \pi R^3 \]

Setting these two volumes equal gives:
\[ 4\pi a^2 t = \frac{4}{3} \pi R^3 \]

### Step 5: Solve for the Radius of the Droplet
Cancelling \( 4\pi \) from both sides and rearranging gives:
\[ R^3 = 12 a^2 t \]
Taking the cube root, we find:
\[ R = (12 a^2 t)^{1/3} \]

### Step 6: Calculate the Potential of the Droplet
Now we can find the potential \( V' \) at the surface of the droplet using the same potential formula:
\[ V' = \frac{kQ}{R} \]
Substituting for \( Q \) from Step 3:
\[ V' = \frac{k \left( \frac{V a}{k} \right)}{R} = \frac{V a}{R} \]

### Step 7: Substitute for \( R \)
Substituting the expression for \( R \) from Step 5:
\[ V' = \frac{V a}{(12 a^2 t)^{1/3}} = V \cdot \frac{a}{(12^{1/3} a^{2/3} t^{1/3})} \]
This simplifies to:
\[ V' = V \cdot \frac{a^{1/3}}{12^{1/3} t^{1/3}} \]

### Final Result
Thus, the potential on the surface of the droplet is:
\[ V' = V \cdot \frac{a^{1/3}}{(3t)^{1/3}} \]