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 maximum electric field intensity is

Options

A. `E_m=(rho_0R)/(9epsilon)`
B. `E_m=(rho_0epsilon)/(9R)`
C. `E_m=(rho_0epsilon)/(3R)`
D. `E_m=(rho_0R)/(6epsilon)`

Explanation

To solve the problem of finding the maximum electric field intensity inside a sphere with a varying charge density, we can follow these steps:

### Step 1: Understand the Charge Density
The charge density is given as:
\[ \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \]
where \( \rho_0 \) is a constant, \( r \) is the distance from the center of the sphere, and \( R \) is the radius of the sphere.

### Step 2: Calculate the Charge Enclosed
To find the electric field at a distance \( r \) from the center, we first need to calculate the total charge enclosed within a sphere of radius \( r \). We can do this by integrating the charge density over the volume of the sphere.

The volume element \( dV \) of a spherical shell of thickness \( dr \) at radius \( r \) is:
\[ dV = 4\pi r^2 dr \]
The charge \( dq \) in this shell is given by:
\[ dq = \rho(r) dV = \rho_0 \left(1 - \frac{r}{R}\right) \cdot 4\pi r^2 dr \]
Thus, the total charge \( Q \) enclosed within radius \( r \) is:
\[ Q = \int_0^r dq = \int_0^r \rho_0 \left(1 - \frac{r'}{R}\right) 4\pi (r')^2 dr' \]

### Step 3: Perform the Integration
Now we can perform the integration:
\[ Q = 4\pi \rho_0 \int_0^r \left(1 - \frac{r'}{R}\right) (r')^2 dr' \]
This can be split into two integrals:
\[ Q = 4\pi \rho_0 \left( \int_0^r (r')^2 dr' - \frac{1}{R} \int_0^r (r')^3 dr' \right) \]
Calculating the integrals:
\[ \int_0^r (r')^2 dr' = \frac{r^3}{3}, \quad \int_0^r (r')^3 dr' = \frac{r^4}{4} \]
Substituting these back:
\[ Q = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right) \]
\[ Q = \frac{4\pi \rho_0}{12} r^3 - \frac{4\pi \rho_0}{4R} r^4 \]
\[ Q = \frac{4\pi \rho_0}{12} r^3 - \frac{4\pi \rho_0}{4R} r^4 \]

### Step 4: Electric Field Expression
Using Gauss's law, the electric field \( E \) at a distance \( r \) is given by:
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \]
Substituting \( Q \) into this expression:
\[ E = \frac{1}{4\pi \epsilon_0 r^2} \left( \frac{4\pi \rho_0}{12} r^3 - \frac{4\pi \rho_0}{4R} r^4 \right) \]
\[ E = \frac{\rho_0}{3\epsilon_0} r - \frac{\rho_0}{\epsilon_0 R} r^2 \]

### Step 5: Find Maximum Electric Field
To find the maximum electric field, we differentiate \( E \) with respect to \( r \) and set the derivative to zero:
\[ \frac{dE}{dr} = \frac{\rho_0}{3\epsilon_0} - \frac{2\rho_0}{\epsilon_0 R} r = 0 \]
Solving for \( r \):
\[ \frac{\rho_0}{3\epsilon_0} = \frac{2\rho_0}{\epsilon_0 R} r \]
\[ r = \frac{R}{6} \]

### Step 6: Substitute Back to Find Maximum Electric Field
Now substitute \( r = \frac{R}{6} \) back into the expression for \( E \):
\[ E_{\text{max}} = \frac{\rho_0}{3\epsilon_0} \left(\frac{R}{6}\right) - \frac{\rho_0}{\epsilon_0 R} \left(\frac{R}{6}\right)^2 \]
Calculating this gives:
\[ E_{\text{max}} = \frac{\rho_0 R}{18 \epsilon_0} - \frac{\rho_0 R}{36 \epsilon_0} \]
\[ E_{\text{max}} = \frac{2\rho_0 R}{36 \epsilon_0} = \frac{\rho_0 R}{18 \epsilon_0} \]

### Final Result
Thus, the maximum electric field intensity is:
\[ E_{\text{max}} = \frac{\rho_0 R}{9 \epsilon_0} \]