A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r)
 given as follows:
ρ(r)=ρ0(1−r/R)
for r≤R
ρ(r)=0
for r≥R
where ρ0=3Q/πR3
 is a positive constant.
find electric field for r<=R

We are asked to find the electric field \( \mathbf{E}(r) \) for \( r \leq R \) given the charge density distribution:

\[
\rho(r) = \rho_0 \left( 1 - \frac{r}{R} \right), \quad \text{for } r \leq R
\]
\[
\rho(r) = 0, \quad \text{for } r \geq R
\]

### Step 1: Find the total charge enclosed within a radius \( r \leq R \)
We will use Gauss's law in integral form to compute the electric field. First, we need to find the charge enclosed within a sphere of radius \( r \leq R \).

The charge enclosed \( Q_{\text{enc}}(r) \) is obtained by integrating the charge density over the volume inside a radius \( r \):

\[
Q_{\text{enc}}(r) = \int_0^r \rho(r') \, 4\pi r'^2 \, dr'
\]

Substitute the expression for \( \rho(r') \):

\[
Q_{\text{enc}}(r) = \int_0^r \rho_0 \left( 1 - \frac{r'}{R} \right) 4\pi r'^2 \, dr'
\]

### Step 2: Perform the integration
We will split the integral into two terms:

\[
Q_{\text{enc}}(r) = 4\pi \rho_0 \int_0^r r'^2 \, dr' - 4\pi \rho_0 \frac{1}{R} \int_0^r r'^3 \, dr'
\]

The integrals are:

\[
\int_0^r r'^2 \, dr' = \frac{r^3}{3}, \quad \int_0^r r'^3 \, dr' = \frac{r^4}{4}
\]

Thus, we have:

\[
Q_{\text{enc}}(r) = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)
\]

### Step 3: Substitute \( \rho_0 \)
We know that \( \rho_0 = \frac{3Q}{\pi R^3} \), so substitute this into the expression for \( Q_{\text{enc}}(r) \):

\[
Q_{\text{enc}}(r) = 4\pi \frac{3Q}{\pi R^3} \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)
\]

Simplifying:

\[
Q_{\text{enc}}(r) = \frac{12Q}{R^3} \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)
\]

\[
Q_{\text{enc}}(r) = Q \left( \frac{r^3}{R^3} - \frac{r^4}{R^4} \right)
\]

### Step 4: Apply Gauss's law
Now we apply Gauss's law in integral form:

\[
\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}(r)}{\epsilon_0}
\]

For a spherically symmetric distribution, the electric field \( \mathbf{E}(r) \) is radial and constant over the surface of a sphere of radius \( r \). The surface area of the sphere is \( 4\pi r^2 \), so Gauss's law becomes:

\[
E(r) \cdot 4\pi r^2 = \frac{Q_{\text{enc}}(r)}{\epsilon_0}
\]

Substitute \( Q_{\text{enc}}(r) \) from Step 3:

\[
E(r) \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \left( \frac{r^3}{R^3} - \frac{r^4}{R^4} \right)
\]

Solve for \( E(r) \):

\[
E(r) = \frac{Q}{4\pi \epsilon_0 r^2} \left( \frac{r^3}{R^3} - \frac{r^4}{R^4} \right)
\]

Simplifying:

\[
E(r) = \frac{Q}{4\pi \epsilon_0 R^4} \left( r - \frac{r^2}{R} \right)
\]

Thus, the electric field for \( r \leq R \) is:

\[
E(r) = \frac{Q}{4\pi \epsilon_0 R^4} \left( r - \frac{r^2}{R} \right)
\]

### Final Answer:
For \( r \leq R \), the electric field is:

\[
E(r) = \frac{Q}{4\pi \epsilon_0 R^4} \left( r - \frac{r^2}{R} \right)
\]

Do you need further details or have any questions?

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### Related Questions:
1. What is the electric field outside the charge distribution (\( r > R \))?
2. How would the electric potential be derived for \( r \leq R \)?
3. What happens to the electric field if the charge density was uniform instead of nonuniform?
4. How does this solution compare to the electric field of a uniformly charged solid sphere?
5. What is the total charge of the distribution when integrated over the entire volume?

### Tip:
Remember that Gauss's law is most useful when the symmetry of the problem allows easy integration over surfaces like spheres.

A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows: ρ(r) = ρ0(1−r/R) for r ≤ R, ρ(r) = 0 for r ≥ R, where ρ0 = 3Q/πR³ is a positive constant. Find the electric field for r ≤ R.

Find the electric field for a spherically symmetric, nonuniform charge distribution using Gauss's Law. This solution covers key physics concepts like charge density, spherical symmetry, and integration. Ideal for undergraduate physics students.

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