A ball of radius R carries a positive charges whose volume density at a point is given as `rho=rho_(0)(1-r//R)`, Where `rho_(0)` is a constant and r is the distance of the point from the center. Assume the permittivities of the ball and the enviroment to be equal to unity.
(a) Find the magnitude of the electric field strength as a function of the distance r both inside and outside the ball.
(b) Find the maximum intensity `E_("max")` and the corresponding distance `r_(m)`.

Explanation

To solve the problem step-by-step, we will break it down into two parts as outlined in the question: (a) finding the electric field strength both inside and outside the ball, and (b) finding the maximum intensity and the corresponding distance.

### Part (a): Electric Field Strength Inside and Outside the Ball

1. **Understanding Charge Density**:
The volume charge density is given by:
\[ \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \]
where \( \rho_0 \) is a constant, \( r \) is the distance from the center, and \( R \) is the radius of the ball.

2. **Electric Field Outside the Ball**:
For points outside the ball (i.e., \( r > R \)), we can use Gauss's Law:
\[ \Phi_E = \frac{Q_{\text{inside}}}{\epsilon_0} \]
Here, \( \Phi_E = E \cdot A \) where \( A = 4\pi r^2 \) is the surface area of the Gaussian surface.

To find \( Q_{\text{inside}} \), we integrate the charge density over the volume of the ball:
\[ Q_{\text{inside}} = \int_0^R \rho(r) \, dV = \int_0^R \rho_0 \left(1 - \frac{r}{R}\right) 4\pi r^2 \, dr \]
Evaluating this integral:
\[ Q_{\text{inside}} = 4\pi \rho_0 \int_0^R \left( r^2 - \frac{r^3}{R} \right) dr = 4\pi \rho_0 \left[ \frac{r^3}{3} - \frac{r^4}{4R} \right]_0^R \]
\[ = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^4}{4R} \right) = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^3}{4} \right) = 4\pi \rho_0 \left( \frac{4R^3 - 3R^3}{12} \right) = \frac{\pi \rho_0 R^3}{3} \]

Now substituting \( Q_{\text{inside}} \) into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{\frac{\pi \rho_0 R^3}{3}}{\epsilon_0} \]
\[ E = \frac{\rho_0 R^3}{12 \epsilon_0 r^2} \quad \text{for } r > R \]

3. **Electric Field Inside the Ball**:
For points inside the ball (i.e., \( r < R \)), we again use Gauss's Law:
\[ Q_{\text{inside}} = \int_0^r \rho(r) \, dV = \int_0^r \rho_0 \left(1 - \frac{r'}{R}\right) 4\pi (r')^2 \, dr' \]
Evaluating this integral:
\[ Q_{\text{inside}} = 4\pi \rho_0 \int_0^r \left( (r')^2 - \frac{(r')^3}{R} \right) dr' = 4\pi \rho_0 \left[ \frac{(r')^3}{3} - \frac{(r')^4}{4R} \right]_0^r \]
\[ = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right) \]

Now substituting \( Q_{\text{inside}} \) into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)}{\epsilon_0} \]
\[ E = \frac{\rho_0}{3 \epsilon_0} r \left(1 - \frac{3r}{4R}\right) \quad \text{for } r < R \]

### Part (b): Finding Maximum Intensity \( E_{\text{max}} \) and Corresponding Distance \( r_m \)

1. **Finding Maximum Electric Field**:
To find the maximum electric field, we differentiate the expression for \( E \) inside the ball with respect to \( r \) and set it to zero:
\[ \frac{dE}{dr} = 0 \]
From the expression:
\[ E = \frac{\rho_0}{3 \epsilon_0} r \left(1 - \frac{3r}{4R}\right) \]
Differentiate:
\[ \frac{dE}{dr} = \frac{\rho_0}{3 \epsilon_0} \left(1 - \frac{3r}{4R}\right) + \frac{\rho_0}{3 \epsilon_0} r \left(-\frac{3}{4R}\right) = 0 \]
Setting the derivative to zero gives:
\[ 1 - \frac{3r}{4R} - \frac{3r}{4R} = 0 \implies 1 - \frac{6r}{4R} = 0 \implies r_m = \frac{2R}{3} \]

2. **Calculating Maximum Electric Field**:
Substitute \( r_m \) back into the expression for \( E \):
\[ E_{\text{max}} = \frac{\rho_0}{3 \epsilon_0} \left(\frac{2R}{3}\right) \left(1 - \frac{3 \cdot \frac{2R}{3}}{4R}\right) \]
Simplifying:
\[ = \frac{\rho_0}{3 \epsilon_0} \left(\frac{2R}{3}\right) \left(1 - \frac{1.5}{4}\right) = \frac{\rho_0}{3 \epsilon_0} \left(\frac{2R}{3}\right) \left(\frac{5}{8}\right) = \frac{5\rho_0 R}{72 \epsilon_0} \]

### Summary of Results
- Electric Field Strength:
- Inside the ball (\( r < R \)):
\[ E = \frac{\rho_0}{3 \epsilon_0} r \left(1 - \frac{3r}{4R}\right) \]
- Outside the ball (\( r > R \)):
\[ E = \frac{\rho_0 R^3}{12 \epsilon_0 r^2} \]

- Maximum Electric Field \( E_{\text{max}} \):
\[ E_{\text{max}} = \frac{5\rho_0 R}{72 \epsilon_0} \]
- Corresponding distance \( r_m \):
\[ r_m = \frac{2R}{3} \]