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Step 1: Understand the Problem
We are given a spherical charge distribution of radius $R$ where the charge density varies as:
$$\rho(r) = \rho_0 \Bigl(1 - \frac{r}{R}\Bigr)\quad\text{for}\quad 0 \le r \le R.$$
We need to find the electric field at a point outside the sphere, i.e., at a distance $r > R$ from the center.
Step 2: Total Charge Within the Sphere
To find the electric field outside the sphere, we first need the total charge $q$ contained in the sphere. Consider a thin spherical shell of radius $r$ and thickness $dr$. The volume element of this shell is
$$dV = 4\pi r^2 \, dr.$$
Thus, a small element of charge in this shell is
$$dq = \rho(r)\, dV = \rho_0\Bigl(1 - \frac{r}{R}\Bigr) \times 4\pi r^2 \, dr.$$
To get the total charge, integrate from $r=0$ to $r=R$:
$$
q = \int_0^R \rho_0\Bigl(1 - \frac{r}{R}\Bigr)4\pi r^2 \, dr.
$$
Step 3: Evaluate the Integral
Expand and integrate term by term:
$$
q = 4\pi \rho_0 \int_0^R \Bigl(r^2 - \frac{r^3}{R}\Bigr)\,dr
= 4\pi \rho_0 \Bigl[\int_0^R r^2\,dr - \int_0^R \frac{r^3}{R}\,dr\Bigr].
$$
Now compute these integrals separately:
$$
\int_0^R r^2\,dr = \frac{R^3}{3}, \quad
\int_0^R \frac{r^3}{R}\,dr = \frac{1}{R}\cdot \frac{R^4}{4} = \frac{R^3}{4}.
$$
Substituting these results back:
$$
q = 4\pi \rho_0 \left(\frac{R^3}{3} - \frac{R^3}{4}\right)
= 4\pi \rho_0 \cdot \frac{R^3}{12}
= \frac{4\pi \rho_0 R^3}{12}.
$$
We can simplify this:
$$
q = \frac{\pi \rho_0 R^3}{3}.
$$
However, for the final form of the electric field, we keep it factored as
$$
q = \left(4\pi \rho_0\right)\frac{R^3}{12}
$$
since it will align neatly with the standard electrostatic form.
Step 4: Use Coulomb's Law for Outside the Sphere
For a spherically symmetric charge distribution, the electric field at a distance $r > R$ (outside the sphere) is given by treating the entire charge $q$ as if it were concentrated at the center. Thus,
$$
E = \frac{1}{4\pi \epsilon_0}\,\frac{q}{r^2}.
$$
Substitute $q = \left(4\pi \rho_0\right)\frac{R^3}{12}$:
$$
E = \frac{1}{4\pi \epsilon_0} \times \frac{\left(4\pi \rho_0\right)\frac{R^3}{12}}{r^2}.
$$
Simplify:
$$
E = \frac{\rho_0 R^3}{12\,\epsilon_0\,r^2}.
$$
Final Answer
The electric field at a point outside the ball (for $r > R$) is
$$
\boxed{ E = \frac{\rho_0 R^3}{12 \,\epsilon_0 \,r^2}.}
$$
