A sphere of radius `R` has a variable charge density `rho(r)=rho_(0)r^(n-1)(rleR)`. If ratio of "potential at centre" and "potential at surface" be `3//2` then find the value of `n`

Explanation

To solve the problem, we need to find the value of \( n \) given the charge density \( \rho(r) = \rho_0 r^{n-1} \) for a sphere of radius \( R \), and the ratio of the potential at the center \( V_c \) to the potential at the surface \( V_s \) is \( \frac{3}{2} \).

### Step-by-Step Solution:

1. **Calculate the Total Charge \( Q \)**:
The total charge \( Q \) inside the sphere can be calculated by integrating the charge density over the volume of the sphere. The volume element in spherical coordinates is \( dV = 4\pi r^2 dr \).

\[ Q = \int_0^R \rho(r) \, dV = \int_0^R \rho_0 r^{n-1} \cdot 4\pi r^2 \, dr = 4\pi \rho_0 \int_0^R r^{n+1} \, dr \]

Evaluating the integral:

\[ Q = 4\pi \rho_0 \left[ \frac{r^{n+2}}{n+2} \right]_0^R = 4\pi \rho_0 \frac{R^{n+2}}{n+2} \]

2. **Calculate the Potential at the Surface \( V_s \)**:
The potential at the surface of the sphere is given by:

\[ V_s = k \frac{Q}{R} = k \frac{4\pi \rho_0 \frac{R^{n+2}}{n+2}}{R} = k \frac{4\pi \rho_0 R^{n+1}}{n+2} \]

3. **Calculate the Potential at the Center \( V_c \)**:
The potential at the center of the sphere can be calculated using the formula for a uniformly charged sphere. For a non-uniform charge distribution, we can integrate the contributions from each shell of charge:

\[ V_c = \int_0^R \frac{Q(r)}{r} \, dQ \]

where \( Q(r) \) is the charge within radius \( r \):

\[ Q(r) = 4\pi \rho_0 \int_0^r r^{n-1} r^2 \, dr = 4\pi \rho_0 \int_0^r r^{n+1} \, dr = 4\pi \rho_0 \frac{r^{n+2}}{n+2} \]

Thus, the potential at the center becomes:

\[ V_c = k \int_0^R \frac{Q(r)}{r^2} \, dr = k \int_0^R \frac{4\pi \rho_0 \frac{r^{n+2}}{n+2}}{r^2} \, dr = k \frac{4\pi \rho_0}{n+2} \int_0^R r^{n} \, dr \]

Evaluating the integral:

\[ V_c = k \frac{4\pi \rho_0}{n+2} \left[ \frac{r^{n+1}}{n+1} \right]_0^R = k \frac{4\pi \rho_0 R^{n+1}}{(n+2)(n+1)} \]

4. **Set Up the Ratio**:
Given that \( \frac{V_c}{V_s} = \frac{3}{2} \):

\[ \frac{k \frac{4\pi \rho_0 R^{n+1}}{(n+2)(n+1)}}{k \frac{4\pi \rho_0 R^{n+1}}{n+2}} = \frac{3}{2} \]

Simplifying this gives:

\[ \frac{1}{(n+1)} = \frac{3}{2} \]

5. **Solve for \( n \)**:
Rearranging the equation:

\[ 2 = 3(n + 1) \implies 2 = 3n + 3 \implies 3n = -1 \implies n = -\frac{1}{3} \]

### Final Answer:
The value of \( n \) is \( -\frac{1}{3} \).