A soap bubble is charged to a potential of 16V. Its radius is then doubled. The potential of the bubble now will be

Options

A. 16V
B. 8V
C. 4V
D. 2V

Explanation

To solve the problem of determining the potential of a soap bubble after its radius is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The soap bubble is initially charged to a potential \( V = 16 \, \text{V} \).
- Let the initial radius of the bubble be \( R \).

2. **Relating Charge and Potential**:
- The potential \( V \) of a charged conducting sphere is given by the formula:
\[ V = \frac{Q}{4 \pi \epsilon_0 R} \]
- Here, \( Q \) is the charge on the bubble, and \( \epsilon_0 \) is the permittivity of free space.

3. **Calculating the Initial Charge**:
- Rearranging the formula for charge \( Q \):
\[ Q = V \cdot 4 \pi \epsilon_0 R \]
- Substituting the initial potential:
\[ Q = 16 \cdot 4 \pi \epsilon_0 R \]

4. **Doubling the Radius**:
- The radius of the bubble is then doubled, so the new radius \( R' = 2R \).

5. **Finding the New Potential**:
- The potential \( V' \) at the new radius can be calculated using the same formula for potential:
\[ V' = \frac{Q}{4 \pi \epsilon_0 R'} \]
- Substituting \( R' = 2R \):
\[ V' = \frac{Q}{4 \pi \epsilon_0 (2R)} = \frac{Q}{2 \cdot 4 \pi \epsilon_0 R} \]
- Now substituting \( Q \) from the earlier calculation:
\[ V' = \frac{16 \cdot 4 \pi \epsilon_0 R}{2 \cdot 4 \pi \epsilon_0 R} \]
- Simplifying this gives:
\[ V' = \frac{16}{2} = 8 \, \text{V} \]

6. **Conclusion**:
- The potential of the bubble after its radius is doubled is \( 8 \, \text{V} \).

### Final Answer:
The potential of the bubble now will be **8 V**.