A soap bubble is charged to a potential 12 V. If its radius is doubled, then the potential of the bubble becomes

Options

A. 12 V
B. 24 V
C. 3 V
D. 6 V

Answer

To solve the problem, we need to understand the relationship between the electric potential (V) of a charged sphere (like a soap bubble) and its radius (R). The electric potential V at the surface of a charged sphere is given by the formula:

\[ V = \frac{kQ}{R} \]

where:
- \( V \) is the electric potential,
- \( k \) is Coulomb's constant,
- \( Q \) is the charge of the sphere,
- \( R \) is the radius of the sphere.

### Step-by-step Solution:

1. **Initial Potential Calculation**:
Given that the initial potential \( V_1 \) of the soap bubble is 12 V and the radius is \( R \), we can express this as:
\[ V_1 = \frac{kQ}{R} = 12 \, \text{V} \]

2. **Doubling the Radius**:
Now, if the radius of the bubble is doubled, the new radius \( R_2 \) becomes:
\[ R_2 = 2R \]

3. **New Potential Calculation**:
The new potential \( V_2 \) when the radius is doubled can be calculated using the same formula:
\[ V_2 = \frac{kQ}{R_2} = \frac{kQ}{2R} \]

4. **Relating New Potential to Initial Potential**:
We can relate the new potential \( V_2 \) to the initial potential \( V_1 \):
\[ V_2 = \frac{kQ}{2R} = \frac{1}{2} \left(\frac{kQ}{R}\right) = \frac{1}{2} V_1 \]

5. **Substituting the Initial Potential**:
Since we know \( V_1 = 12 \, \text{V} \):
\[ V_2 = \frac{1}{2} \times 12 \, \text{V} = 6 \, \text{V} \]

### Final Answer:
Thus, the potential of the bubble after doubling its radius becomes **6 V**.

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