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. 3V

Answer

To solve the problem, we need to understand the behavior of electric potential around a charged spherical conductor.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
We have a charged spherical conductor with a radius of \( R = 10 \, \text{cm} \). The potential \( V \) is given at a point \( P_1 \) which is \( 5 \, \text{cm} \) from the center of the sphere.

2. **Potential Inside the Conductor**:
Inside a charged conductor, the electric field is zero. Therefore, the potential remains constant throughout the conductor and is equal to the potential at its surface. Since \( P_1 \) is at \( 5 \, \text{cm} \) which is inside the conductor, the potential at this point is the same as the potential at the surface of the conductor.

3. **Potential at the Surface**:
The potential at the surface of the conductor (at \( R = 10 \, \text{cm} \)) can be expressed as:
\[ V_{\text{surface}} = \frac{KQ}{R} \]
where \( K \) is Coulomb's constant and \( Q \) is the charge on the sphere.

4. **Potential at Point \( P_2 \)**:
Now, we need to find the potential at a point \( P_2 \) which is \( 15 \, \text{cm} \) from the center of the sphere. Since \( P_2 \) is outside the conductor, we can use the formula for the potential outside a charged sphere:
\[ V_{P_2} = \frac{KQ}{r} \]
where \( r = 15 \, \text{cm} \).

5. **Relating the Potentials**:
We know that the potential at point \( P_1 \) (which is \( 5 \, \text{cm} \)) is equal to \( V \). Thus,
\[ V = \frac{KQ}{10} \]
Now substituting \( KQ \) from this equation into the equation for \( V_{P_2} \):
\[ V_{P_2} = \frac{KQ}{15} = \frac{10V}{15} = \frac{2V}{3} \]

6. **Final Answer**:
Therefore, the potential at a point distant \( 15 \, \text{cm} \) from the center of the sphere is:
\[ V_{P_2} = \frac{2V}{3} \]

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

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