In air, a charged soap bubble of radius 'r' is in equilibrium having outside and inside pressures being equal. The charge on the drop is (`in_(0)` = permittivity of free space, T = surface tension of soap solution)

Options

A. `4pir^(2)sqrt((2Tin_(0))/r)`
B. `4pir^(2)sqrt((4Tin_(0))/r)`
C. `4pir^(2)sqrt((6Tin_(0))/r)`
D. `4pir^(2)sqrt((8Tin_(0))/r)`

Explanation

To solve the problem, we need to find the charge on a charged soap bubble of radius 'r' that is in equilibrium, with the outside and inside pressures being equal.

### Step-by-Step Solution:

1. **Understand the Excess Pressure in the Soap Bubble:**
The excess pressure \( P \) inside a soap bubble is given by the formula:
\[ P = \frac{4T}{r} \]
where \( T \) is the surface tension of the soap solution and \( r \) is the radius of the bubble.

2. **Equate Inside and Outside Pressure:**
Since the bubble is in equilibrium, the excess pressure inside the bubble must equal the electrostatic pressure due to the charge on the bubble. Thus, we have:
\[ P = \frac{\sigma^2}{2\epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

3. **Set the Two Pressures Equal:**
Setting the two expressions for pressure equal gives:
\[ \frac{4T}{r} = \frac{\sigma^2}{2\epsilon_0} \]

4. **Solve for Surface Charge Density \( \sigma \):**
Rearranging the equation to solve for \( \sigma^2 \):
\[ \sigma^2 = \frac{8T\epsilon_0}{r} \]

5. **Relate Surface Charge Density to Total Charge:**
The surface charge density \( \sigma \) is related to the total charge \( Q \) on the bubble by the formula:
\[ \sigma = \frac{Q}{A} \]
where \( A \) is the surface area of the bubble. For a sphere, the surface area \( A \) is:
\[ A = 4\pi r^2 \]
Therefore, we can express \( \sigma \) as:
\[ \sigma = \frac{Q}{4\pi r^2} \]

6. **Substitute \( \sigma \) into the Pressure Equation:**
Substitute \( \sigma \) back into the equation for \( \sigma^2 \):
\[ \left(\frac{Q}{4\pi r^2}\right)^2 = \frac{8T\epsilon_0}{r} \]

7. **Solve for \( Q \):**
Squaring the left side gives:
\[ \frac{Q^2}{16\pi^2 r^4} = \frac{8T\epsilon_0}{r} \]
Multiplying both sides by \( 16\pi^2 r^4 \):
\[ Q^2 = 16\pi^2 r^4 \cdot \frac{8T\epsilon_0}{r} \]
Simplifying gives:
\[ Q^2 = 128\pi^2 T\epsilon_0 r^3 \]
Taking the square root:
\[ Q = \sqrt{128\pi^2 T\epsilon_0 r^3} \]

8. **Final Expression:**
Thus, the charge \( Q \) on the soap bubble is:
\[ Q = 8\pi \sqrt{2T\epsilon_0 r^3} \]