For spherical charge distribution gives as .
`{{:(,rho = rho_(0)(1-(r)/(3)),"when" r le 3m),(,rho = 0 ,"when" r gt 3m):}`
( where x is the distance form the centre of spherical charge distribution)
The electric field intensity is maximum for the value of r=_______m.

Explanation

To find the value of \( r \) at which the electric field intensity is maximum for the given spherical charge distribution, we will follow these steps:

### Step 1: Understand the Charge Distribution
The charge density \( \rho \) is given as:
\[ \rho = \rho_0 \left(1 - \frac{r}{3}\right) \quad \text{for} \quad r \leq 3 \, \text{m} \]
\[ \rho = 0 \quad \text{for} \quad r > 3 \, \text{m} \]
This indicates that the charge density decreases linearly from \( \rho_0 \) to 0 as \( r \) approaches 3 m.

### Step 2: Apply Gauss's Law
To find the electric field \( E \) at a distance \( r \) from the center, we will use Gauss's Law:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
For a spherical Gaussian surface of radius \( r \), the area \( dA \) is \( 4\pi r^2 \) and the electric field \( E \) is constant over this surface.

### Step 3: Calculate the Enclosed Charge \( Q_{\text{enc}} \)
The charge enclosed within the Gaussian surface of radius \( r \) is given by:
\[ Q_{\text{enc}} = \int_0^r \rho \, dV = \int_0^r \rho_0 \left(1 - \frac{r'}{3}\right) \, dV \]
Where \( dV = 4\pi (r')^2 \, dr' \). Thus,
\[ Q_{\text{enc}} = \int_0^r \rho_0 \left(1 - \frac{r'}{3}\right) 4\pi (r')^2 \, dr' \]

### Step 4: Evaluate the Integral
Calculating the integral:
\[ Q_{\text{enc}} = 4\pi \rho_0 \int_0^r \left( (r')^2 - \frac{(r')^3}{3} \right) \, dr' \]
Evaluating this gives:
\[ = 4\pi \rho_0 \left[ \frac{(r')^3}{3} - \frac{(r')^4}{12} \right]_0^r = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{12} \right) \]

### Step 5: Substitute into Gauss's Law
Now substituting back into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{12} \right)}{\epsilon_0} \]
Thus,
\[ E = \frac{\rho_0}{\epsilon_0} \left( \frac{r}{3} - \frac{r^2}{12} \right) \]

### Step 6: Find Maximum Electric Field
To find the maximum electric field, we take the derivative of \( E \) with respect to \( r \) and set it to zero:
\[ \frac{dE}{dr} = \frac{\rho_0}{\epsilon_0} \left( \frac{1}{3} - \frac{r}{6} \right) = 0 \]
Solving for \( r \):
\[ \frac{1}{3} - \frac{r}{6} = 0 \implies r = 2 \, \text{m} \]

### Conclusion
The electric field intensity is maximum at \( r = 2 \, \text{m} \).

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