In the previous question, if `V` is the electric potential of the first sphere, what would be the electric potential of the second sphere ?

Options

A. 2 V
B. `V//2`
C. `V//4`
D. V

Explanation

To solve the problem of finding the electric potential of the second sphere given that the electric potential of the first sphere is \( V \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Electric Potential**:
The electric potential \( V \) at a distance \( r \) from a charged sphere is given by the formula:
\[ V = \frac{Q}{4\pi \epsilon_0 r} \]
where \( Q \) is the charge of the sphere and \( \epsilon_0 \) is the permittivity of free space.

2. **Identify the Radii of the Spheres**:
Let the radius of the first sphere be \( r \) and the radius of the second sphere be \( 2r \) (as mentioned in the transcript).

3. **Relate the Electric Potentials**:
If the electric potential of the first sphere is \( V \), then for the second sphere, which has double the radius, we can express its potential as:
\[ V' = \frac{Q}{4\pi \epsilon_0 (2r)} \]
This can be simplified to:
\[ V' = \frac{Q}{8\pi \epsilon_0 r} \]

4. **Express \( V' \) in Terms of \( V \)**:
Since we know that:
\[ V = \frac{Q}{4\pi \epsilon_0 r} \]
We can relate \( V' \) to \( V \):
\[ V' = \frac{1}{2} \cdot \frac{Q}{4\pi \epsilon_0 r} = \frac{1}{2} V \]

5. **Conclusion**:
Therefore, the electric potential of the second sphere is:
\[ V' = \frac{1}{2} V \]

### Final Answer:
The electric potential of the second sphere is \( \frac{1}{2} V \).

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