A soap bubble of radius R = 1 cm is charged with the maximum charge for which breakdown of air on its surface does not occur. Calculate the electrostatic pressure on the surface of the bubble. It is know that dielectric breakdown of air takes place when electric field becomes larger then `E_(0)=3xx10^(6)W//m`

Answer

To solve the problem, we need to calculate the electrostatic pressure on the surface of a soap bubble that has been charged to the maximum limit before the breakdown of air occurs. Here’s how we can approach the solution step by step:

### Step 1: Understand the relationship between electric field and charge
The electric field \( E \) just outside the surface of a charged sphere (or soap bubble) is given by the formula:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R^2} \]
where:
- \( Q \) is the charge on the bubble,
- \( R \) is the radius of the bubble,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 2: Set the electric field equal to the breakdown field
Given that the breakdown of air occurs at \( E_0 = 3 \times 10^6 \, \text{N/C} \), we can set:
\[ \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R^2} = E_0 \]

### Step 3: Solve for the charge \( Q \)
Rearranging the equation to solve for \( Q \):
\[ Q = 4\pi \epsilon_0 E_0 R^2 \]

### Step 4: Substitute the values
Now, substituting the known values:
- \( R = 1 \, \text{cm} = 0.01 \, \text{m} \)
- \( E_0 = 3 \times 10^6 \, \text{N/C} \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

Calculating \( Q \):
\[ Q = 4\pi (8.85 \times 10^{-12}) (3 \times 10^6) (0.01)^2 \]
\[ Q = 4\pi (8.85 \times 10^{-12}) (3 \times 10^6) (0.0001) \]
\[ Q = 4\pi (8.85 \times 10^{-12}) (3 \times 10^2) \]
\[ Q = 4\pi (2.655 \times 10^{-9}) \approx 3.34 \times 10^{-8} \, \text{C} \]

### Step 5: Calculate the electrostatic pressure
The electrostatic pressure \( P \) on the surface of the bubble is given by:
\[ P = \frac{E^2}{2 \cdot \epsilon_0} \]
Substituting \( E = E_0 \):
\[ P = \frac{(3 \times 10^6)^2}{2 \cdot (8.85 \times 10^{-12})} \]
\[ P = \frac{9 \times 10^{12}}{1.77 \times 10^{-11}} \approx 5.08 \times 10^3 \, \text{N/m}^2 \]

### Final Answer
The electrostatic pressure on the surface of the soap bubble is approximately \( 5.08 \times 10^3 \, \text{N/m}^2 \).

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