If a charged spherical conductor of radius `10 cm` has potential `V` at a point distant `5 cm` from its centre, then the potential at a point distant `15 cm` from the centre will be

Options

A. `(1)/(3) V`
B. `(2)/(3) V`
C. `(3)/(2) V`
D. `3 V`

Explanation

To solve the problem, we need to find the potential at a point distant `15 cm` from the center of a charged spherical conductor with a radius of `10 cm`. We know that the potential at a point `5 cm` from the center is `V`.

### Step-by-Step Solution:

1. **Understanding the Spherical Conductor:**
- A charged spherical conductor has a uniform charge distribution. The electric potential inside the conductor is constant and equal to the potential on its surface.

2. **Identify the Given Values:**
- Radius of the spherical conductor, \( R = 10 \, \text{cm} \)
- Distance from the center for point P, \( r_P = 5 \, \text{cm} \)
- Potential at point P, \( V_P = V \)
- Distance from the center for point Q, \( r_Q = 15 \, \text{cm} \)

3. **Potential Inside the Spherical Conductor:**
- The potential at any point inside the conductor (for \( r < R \)) is equal to the potential on the surface.
- Therefore, the potential at point P (inside the conductor) is equal to the potential on the surface of the conductor.
- Thus, \( V_P = V_S \), where \( V_S \) is the potential on the surface.

4. **Calculating the Potential on the Surface:**
- The potential on the surface of the spherical conductor can be expressed as:
\[ V_S = \frac{kQ}{R} \]
- Here, \( k \) is Coulomb's constant, and \( Q \) is the total charge on the conductor.

5. **Potential Outside the Spherical Conductor:**
- For a point outside the conductor (for \( r > R \)), the potential can be calculated using:
\[ V_Q = \frac{kQ}{r_Q} \]
- Substituting \( r_Q = 15 \, \text{cm} \):
\[ V_Q = \frac{kQ}{15} \]

6. **Relating the Potentials:**
- We already established that \( V_S = V \) (the potential at the surface).
- Therefore, we can express \( V_S \) in terms of \( V \):
\[ V = \frac{kQ}{10} \]
- Now, we can relate \( V_Q \) to \( V \):
\[ V_Q = \frac{kQ}{15} = V \cdot \frac{10}{15} = V \cdot \frac{2}{3} \]

7. **Final Result:**
- Thus, the potential at a point distant `15 cm` from the center is:
\[ V_Q = \frac{2V}{3} \]

### Conclusion:
The potential at a point distant `15 cm` from the center of the spherical conductor is \( \frac{2V}{3} \).