An isolated soap bubble is kept in a container in which the pressure is maintained at `P_(0)` . The bubble is given some charge due to which its radius increases to `R_(0)` . If the surface tension of the soap bubble is T, then the charge given to the bubble is

Options

A. `Q = 8 pi r sqrt( r T epsilon_(0))`
B. `Q = 8 pi r sqrt(2 r T epsilon_(0))`
C. `Q = 4 pi r sqrt(2 r T epsilon_(0))`
D. `Q = 4 pi r sqrt( r T epsilon_(0))`

Answer

To find the charge given to the soap bubble when its radius increases to \( R_0 \) due to the charge, we can follow these steps:

### Step 1: Understand the forces acting on the bubble
The soap bubble is under the influence of two main forces:
1. **Surface tension force** which tries to collapse the bubble.
2. **Electrostatic force** due to the charge on the bubble which tries to expand it.

### Step 2: Write the expression for excess pressure due to surface tension
The excess pressure \( \Delta P \) inside a soap bubble due to surface tension \( T \) is given by:
\[ \Delta P = \frac{4T}{R} \]
where \( R \) is the radius of the bubble.

### Step 3: Write the expression for electrostatic pressure
The electrostatic pressure \( P_e \) due to the charge \( Q \) on the bubble is given by:
\[ P_e = \frac{\sigma^2}{2\epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

### Step 4: Relate surface charge density to charge
The surface charge density \( \sigma \) can be expressed in terms of the total charge \( Q \) and the surface area of the bubble:
\[ \sigma = \frac{Q}{4\pi R^2} \]

### Step 5: Set the excess pressure equal to electrostatic pressure
Since the bubble is in equilibrium, the excess pressure due to surface tension must equal the electrostatic pressure:
\[ \frac{4T}{R} = \frac{\sigma^2}{2\epsilon_0} \]

### Step 6: Substitute the expression for surface charge density
Substituting \( \sigma = \frac{Q}{4\pi R^2} \) into the pressure equation:
\[ \frac{4T}{R} = \frac{1}{2\epsilon_0} \left(\frac{Q}{4\pi R^2}\right)^2 \]

### Step 7: Simplify the equation
Rearranging gives:
\[ \frac{4T}{R} = \frac{Q^2}{32\pi^2 \epsilon_0 R^4} \]
Multiplying both sides by \( 32\pi^2 \epsilon_0 R^4 \):
\[ 32\pi^2 \epsilon_0 T R^3 = Q^2 \]

### Step 8: Solve for charge \( Q \)
Taking the square root of both sides:
\[ Q = \sqrt{32\pi^2 \epsilon_0 T R^3} \]

### Step 9: Substitute \( R \) with \( R_0 \)
Since the radius of the bubble increases to \( R_0 \):
\[ Q = \sqrt{32\pi^2 \epsilon_0 T R_0^3} \]

### Final Answer
Thus, the charge given to the bubble is:
\[ Q = 4\pi R_0^2 \sqrt{\frac{8T}{\epsilon_0}} \]