Consider a sphere of radius R with unform charged density and total charge Q. The electrostatic potential distribution inside the sphere is given by ` phi (r ) =(Q )/( 4 pi epsi_0 R) (a + b ( r//R)^c )` Note that the zero of potential is at infinity. The values of (a, b, c) are :-

Options

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

Explanation

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) in the given electrostatic potential distribution formula for a uniformly charged sphere. The potential inside a uniformly charged sphere can be derived from the principles of electrostatics.

### Step-by-Step Solution:

1. **Understanding the Potential Inside a Uniformly Charged Sphere**:
The electrostatic potential \( \phi(r) \) inside a uniformly charged sphere of radius \( R \) and total charge \( Q \) is given by the formula:
\[ \phi(r) = \frac{Q}{4 \pi \epsilon_0 R} \left( \frac{3}{2} - \frac{r^2}{2R^2} \right) \]
This formula is derived from integrating the electric field inside the sphere.

2. **Rearranging the Formula**:
We can rearrange the potential formula to match the form given in the question:
\[ \phi(r) = \frac{Q}{4 \pi \epsilon_0 R} \left( \frac{3}{2} - \frac{1}{2} \frac{r^2}{R^2} \right) \]
This can be expressed as:
\[ \phi(r) = \frac{Q}{4 \pi \epsilon_0 R} \left( \frac{3}{2} + \left(-\frac{1}{2}\right) \left(\frac{r}{R}\right)^2 \right) \]

3. **Identifying Coefficients**:
Now, we can compare this expression with the given potential distribution:
\[ \phi(r) = \frac{Q}{4 \pi \epsilon_0 R} \left( a + b \left(\frac{r}{R}\right)^c \right) \]
From our rearranged formula, we can identify:
- \( a = \frac{3}{2} \)
- \( b = -\frac{1}{2} \)
- \( c = 2 \)

4. **Final Values**:
Thus, the values of \( a \), \( b \), and \( c \) are:
- \( a = \frac{3}{2} \)
- \( b = -\frac{1}{2} \)
- \( c = 2 \)

### Conclusion:
The values of \( (a, b, c) \) are \( \left( \frac{3}{2}, -\frac{1}{2}, 2 \right) \).

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