To solve the problem, we need to determine the potential of a soap bubble after it bursts and becomes a spherical drop. We will use the principles of electrostatics and the conservation of charge.
### Step-by-Step Solution:
1. **Identify Given Values:**
- Radius of the soap bubble, \( R = 3 \, \text{cm} = 0.03 \, \text{m} \)
- Thickness of the soap bubble, \( t = 10^{-2} \, \text{mm} = 10^{-5} \, \text{m} \)
- Initial potential of the bubble, \( V = 0.3 \, \text{V} \)
2. **Calculate the Charge on the Bubble:**
The charge \( Q \) on the soap bubble can be calculated using the formula for the potential of a spherical conductor:
\[
V = \frac{Q}{4 \pi \epsilon_0 R}
\]
Rearranging this gives:
\[
Q = V \cdot 4 \pi \epsilon_0 R
\]
Substituting the known values:
\[
Q = 0.3 \cdot 4 \pi \cdot (8.854 \times 10^{-12}) \cdot 0.03
\]
3. **Calculate the Charge \( Q \):**
First, calculate \( 4 \pi \epsilon_0 \):
\[
4 \pi \epsilon_0 \approx 4 \cdot 3.14 \cdot (8.854 \times 10^{-12}) \approx 1.112 \times 10^{-10} \, \text{C/V}
\]
Now substituting this into the charge equation:
\[
Q = 0.3 \cdot 1.112 \times 10^{-10} \cdot 0.03 \approx 1.003 \times 10^{-12} \, \text{C}
\]
4. **Determine the Radius of the Droplet:**
When the bubble bursts, it becomes a solid spherical drop with the same volume of soap. The volume of the soap bubble is given by:
\[
V_{\text{bubble}} = \frac{4}{3} \pi (R^3 - (R - t)^3)
\]
The volume of the drop is:
\[
V_{\text{drop}} = \frac{4}{3} \pi r^3
\]
Since the volume of the soap remains constant, we can set these equal:
\[
\frac{4}{3} \pi (R^3 - (R - t)^3) = \frac{4}{3} \pi r^3
\]
Simplifying gives:
\[
R^3 - (R - t)^3 = r^3
\]
5. **Calculate \( r \):**
Expanding \( (R - t)^3 \):
\[
(R - t)^3 = R^3 - 3R^2t + 3Rt^2 - t^3
\]
Therefore:
\[
R^3 - (R^3 - 3R^2t + 3Rt^2 - t^3) = r^3
\]
Simplifying gives:
\[
3R^2t - 3Rt^2 + t^3 = r^3
\]
For small \( t \), we can neglect \( t^3 \):
\[
r^3 \approx 3R^2t
\]
Substituting \( R = 0.03 \, \text{m} \) and \( t = 10^{-5} \, \text{m} \):
\[
r^3 \approx 3(0.03)^2(10^{-5}) = 3 \cdot 0.0009 \cdot 10^{-5} = 2.7 \times 10^{-8}
\]
Thus:
\[
r \approx (2.7 \times 10^{-8})^{1/3} \approx 0.029 \, \text{m} \approx 2.9 \, \text{cm}
\]
6. **Calculate the Potential of the Droplet:**
Using the formula for potential again:
\[
V' = \frac{Q}{4 \pi \epsilon_0 r}
\]
Substituting \( Q \) and \( r \):
\[
V' = \frac{1.003 \times 10^{-12}}{4 \pi (8.854 \times 10^{-12}) (0.029)}
\]
Calculate \( V' \):
\[
V' \approx \frac{1.003 \times 10^{-12}}{3.27 \times 10^{-12}} \approx 0.306 \, \text{V}
\]
This is approximately \( 3 \, \text{V} \).
### Final Answer:
The potential of the drop is approximately **3 volts**.
