The radius of a soap bubble whose potential is `16 V` is doubled. The new potential of the bubble will be

Options

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

Explanation

To solve the problem step by step, we need to understand the relationship between the electric potential of a soap bubble and its radius.

### Step-by-Step Solution:

1. **Understanding Electric Potential**:
The electric potential \( V \) of a soap bubble is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is a constant, \( Q \) is the charge on the bubble, and \( R \) is the radius of the bubble.

2. **Initial Conditions**:
We are given that the initial potential \( V_1 \) is \( 16 \, \text{V} \) and the initial radius \( R_1 \) is \( R \).

3. **Doubling the Radius**:
The radius is doubled, so the new radius \( R_2 \) is:
\[ R_2 = 2R \]

4. **Finding New Potential**:
Since the charge \( Q \) remains constant when the radius is changed, we can express the new potential \( V_2 \) using the new radius:
\[ V_2 = \frac{kQ}{R_2} = \frac{kQ}{2R} \]

5. **Relating New and Old Potentials**:
We can relate the new potential \( V_2 \) to the old potential \( V_1 \):
\[ V_2 = \frac{kQ}{2R} = \frac{1}{2} \cdot \frac{kQ}{R} = \frac{1}{2} V_1 \]
Substituting \( V_1 = 16 \, \text{V} \):
\[ V_2 = \frac{1}{2} \cdot 16 \, \text{V} = 8 \, \text{V} \]

6. **Conclusion**:
The new potential of the bubble after doubling the radius is:
\[ V_2 = 8 \, \text{V} \]

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