A sphere of charges of radius R carries a positive charge whose volume chasrge density depends only on the distance r from the ball's centre as `rho=rho_0(1-r/R)`, where `rho_0` is constant. Assume epsilon as theh permittivity of space.
The value of distance `r_m` at which electric field intensity is maximum is given by

Options

A. `r_m=R/3`
B. `r_m=(3R)/2`
C. `r_m=(2R)/3`
D. `r_m=(4R)/3`

Explanation

To find the distance \( r_m \) at which the electric field intensity is maximum for a sphere with a volume charge density given by \( \rho = \rho_0 \left(1 - \frac{r}{R}\right) \), we can follow these steps:

### Step 1: Determine the charge density
The charge density is given as:
\[ \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \]
This indicates that the charge density decreases linearly from \( \rho_0 \) at the center of the sphere to \( 0 \) at the surface.

### Step 2: Calculate the charge enclosed within a radius \( r \)
To find the electric field at a distance \( r \) from the center, we first need to find the total charge enclosed within a sphere of radius \( r \):
\[ Q_{\text{enc}} = \int_0^r \rho(r') \, dV \]
The volume element in spherical coordinates is \( dV = 4\pi r'^2 dr' \). Thus,
\[ Q_{\text{enc}} = \int_0^r \rho_0 \left(1 - \frac{r'}{R}\right) 4\pi r'^2 dr' \]

### Step 3: Evaluate the integral
Calculating the integral:
\[ Q_{\text{enc}} = 4\pi \rho_0 \int_0^r \left(1 - \frac{r'}{R}\right) r'^2 dr' \]
This can be split into two integrals:
\[ = 4\pi \rho_0 \left( \int_0^r r'^2 dr' - \frac{1}{R} \int_0^r r'^3 dr' \right) \]
Calculating these integrals:
\[ \int_0^r r'^2 dr' = \frac{r^3}{3} \quad \text{and} \quad \int_0^r r'^3 dr' = \frac{r^4}{4} \]
Thus,
\[ Q_{\text{enc}} = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right) \]
\[ = \frac{4\pi \rho_0}{12} r^3 - \frac{4\pi \rho_0}{4R} r^4 \]
\[ = \frac{4\pi \rho_0}{12} r^3 - \frac{\pi \rho_0}{R} r^4 \]

### Step 4: Apply Gauss's Law to find the electric field
Using Gauss's Law, the electric field \( E(r) \) at a distance \( r \) is given by:
\[ E(r) \cdot 4\pi r^2 = Q_{\text{enc}} \]
Thus,
\[ E(r) = \frac{Q_{\text{enc}}}{4\pi r^2} \]
Substituting \( Q_{\text{enc}} \):
\[ E(r) = \frac{1}{4\pi \epsilon_0} \left( \frac{4\pi \rho_0}{12} r^3 - \frac{\pi \rho_0}{R} r^4 \right) \cdot \frac{1}{r^2} \]
\[ = \frac{\rho_0}{3\epsilon_0} r - \frac{\rho_0}{4\epsilon_0 R} r^2 \]

### Step 5: Maximize the electric field
To find the maximum electric field, we differentiate \( E(r) \) with respect to \( r \) and set the derivative to zero:
\[ \frac{dE}{dr} = \frac{\rho_0}{3\epsilon_0} - \frac{\rho_0}{2\epsilon_0 R} r = 0 \]
Solving for \( r \):
\[ \frac{\rho_0}{3\epsilon_0} = \frac{\rho_0}{2\epsilon_0 R} r \]
\[ r = \frac{2R}{3} \]

### Conclusion
Thus, the distance \( r_m \) at which the electric field intensity is maximum is:
\[ r_m = \frac{2R}{3} \]