Within a spherical charge distribution of charge density `rho(r)`, N equipotential surface of potential `V_(0), V_(0)+DeltaV, V_(0)+2DeltaV, … V_(0)+NDeltaV (DeltaV gt 0)`, are drawn and have increasing radii `r_(0), r_(1), r_(2), …. R_(N)`, respectively. If the difference in the radii of the surface is constant for all values of `V_(0)` and `DeltaV` then :-

Options

A. (a)`rho(r )` = constant
B. (b)`rho(r ) prop 1/(r^2)`
C. (c)`rho(r ) prop 1/(r)`
D. (d)`rho(r ) prop r`

Explanation

To solve the problem, we will analyze the spherical charge distribution and the relationship between the potential and the radius of the equipotential surfaces.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a spherical charge distribution with a charge density \( \rho(r) \). There are \( N \) equipotential surfaces with potentials \( V_0, V_0 + \Delta V, V_0 + 2\Delta V, \ldots, V_0 + N\Delta V \) and corresponding radii \( r_0, r_1, r_2, \ldots, r_N \). The difference in the radii of these surfaces is constant.

2. **Applying Gauss's Law**:
To analyze the electric field \( E \) due to a spherical charge distribution, we can use Gauss's Law:
\[ \Phi = \oint E \cdot dA = \frac{Q_{\text{enc}}}{\epsilon_0} \]
For a spherical surface of radius \( r \):
\[ E \cdot 4\pi r^2 = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the charge enclosed within radius \( r \).

3. **Charge Enclosed**:
The charge enclosed can be expressed as:
\[ Q_{\text{enc}} = \int_0^r \rho(r') \cdot 4\pi r'^2 dr' \]

4. **Relating Electric Field and Potential**:
The electric field \( E \) is related to the potential \( V \) by:
\[ E = -\frac{dV}{dr} \]
Given that the potential difference \( \Delta V \) is constant, we can write:
\[ \frac{dV}{dr} = \text{constant} \]

5. **Finding the Relationship**:
From the expression for \( E \) and using Gauss's law:
\[ E = \frac{Q_{\text{enc}}}{4\pi \epsilon_0 r^2} \]
Combining this with the relationship for the potential, we can derive:
\[ \frac{dV}{dr} = -\frac{Q_{\text{enc}}}{4\pi \epsilon_0 r^2} \]

6. **Density and Radius Relationship**:
Since \( \frac{dV}{dr} \) is constant, we can conclude that:
\[ \rho(r) \cdot r^2 \text{ is constant} \]
This leads us to:
\[ \rho(r) \propto \frac{1}{r} \]

7. **Conclusion**:
The relationship between the charge density \( \rho \) and the radius \( r \) is given by:
\[ \rho(r) \propto \frac{1}{r} \]
Thus, the correct option is (C).