If a charged spherical conductor of radius 5 cm has potential V at a point distant 5 cm its centre, then the potential at a point distant 30 cm from the centre will be

Options

A. `(1)/(3)V`
B. `(1)/(6)V`
C. `(3)/(2)V`
D. `3 V`

Answer

To solve the problem, we need to determine the potential at a point 30 cm from the center of a charged spherical conductor with a radius of 5 cm, given that the potential at a point 5 cm from the center is V.

### Step-by-Step Solution:

1. **Understand the Concept of Electric Potential**:
The electric potential \( V \) at a distance \( r \) from a charged spherical conductor can be determined using the following formulas:
- For points inside or on the surface of the conductor (i.e., \( r \leq R \)):
\[ V = \frac{kQ}{R} \]
- For points outside the conductor (i.e., \( r > R \)):
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the conductor, and \( R \) is the radius of the conductor.

2. **Identify the Given Values**:
- Radius of the conductor \( R = 5 \) cm
- Distance from the center for the first case \( a = 5 \) cm (which is equal to \( R \))
- Distance from the center for the second case \( a' = 30 \) cm

3. **Calculate the Potential at 5 cm**:
Since \( a = R \), we can use the formula for the potential at the surface:
\[ V = \frac{kQ}{R} = \frac{kQ}{5 \, \text{cm}} \]

4. **Calculate the Potential at 30 cm**:
Since \( a' > R \), we use the formula for points outside the conductor:
\[ V' = \frac{kQ}{a'} = \frac{kQ}{30 \, \text{cm}} \]

5. **Relate the Two Potentials**:
To find the relationship between \( V \) and \( V' \):
\[ V' = \frac{kQ}{30} \quad \text{and} \quad V = \frac{kQ}{5} \]
We can express \( V' \) in terms of \( V \):
\[ V' = V \cdot \frac{5}{30} = V \cdot \frac{1}{6} \]

6. **Conclusion**:
Therefore, the potential at a point 30 cm from the center of the conductor is:
\[ V' = \frac{V}{6} \]

### Final Answer:
The potential at a point distant 30 cm from the center will be \( \frac{V}{6} \).