The potential of the field inside a charged sphere depends upon the distance from its centre to the point under consideration in the following way:
`V=ar^n+b`, where a and b are constants
If the charge is uniformly distributed within the sphere, then the value of n is

Options

A. 2
B. `-2`
C. 1
D. `-1`

Answer

To solve the problem, we need to find the value of \( n \) in the potential equation \( V = ar^n + b \) for a uniformly charged sphere.

### Step-by-Step Solution:

1. **Understanding the Potential Inside a Charged Sphere**:
For a uniformly charged sphere, the potential \( V \) at a distance \( r \) from the center (where \( r \) is less than the radius \( R \) of the sphere) is given by the formula:
\[ V = \frac{KQ}{2R} \left(3 - \frac{r^2}{R^2}\right) \]
where \( K \) is the Coulomb's constant, \( Q \) is the total charge of the sphere, and \( R \) is the radius of the sphere.

2. **Rearranging the Formula**:
We can rewrite the potential equation:
\[ V = \frac{KQ}{2R} \cdot 3 - \frac{KQ}{2R^3} r^2 \]
This can be expressed as:
\[ V = \left(-\frac{KQ}{2R^3}\right) r^2 + \frac{3KQ}{2R} \]

3. **Identifying Constants**:
From the equation \( V = ar^n + b \), we can identify:
- \( a = -\frac{KQ}{2R^3} \)
- \( b = \frac{3KQ}{2R} \)

4. **Comparing Powers of \( r \)**:
In the equation \( V = ar^n + b \), the term involving \( r \) is \( ar^n \). From our rearranged potential equation, we see that the term involving \( r \) is \( -\frac{KQ}{2R^3} r^2 \).

5. **Finding the Value of \( n \)**:
Since the term \( r^2 \) corresponds to \( r^n \), we can conclude that:
\[ n = 2 \]

### Final Answer:
The value of \( n \) is \( 2 \).

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