Electric Field Calculation Using Gauss's Law

The problem asks for the electric field ($E$) outside a solid ball of radius $R$ where the charge density varies with the radial distance $r$ according to $\rho = \rho_0(1 - \frac{r}{R})$ for $0 < r < R$. We need to find the electric field for $r > R$. Due to the spherical symmetry of the charge distribution, the electric field will be radial and its magnitude will depend only on the distance $r$ from the center.

Applying Gauss's Law

We use Gauss's Law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space ($\epsilon_0$).

$ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} $

Choose a spherical Gaussian surface with radius $r > R$. The electric field $\vec{E}$ is perpendicular to the surface element $d\vec{A}$ and has a constant magnitude on this surface.

The left side of Gauss's Law becomes:

$ \oint \vec{E} \cdot d\vec{A} = E \oint dA = E (4\pi r^2) $

Calculating Enclosed Charge ($Q_{enc}$)

Since the Gaussian surface has a radius $r > R$, it encloses the entire charge of the solid ball. We need to calculate the total charge ($Q_{total}$) within the ball.

The total charge is found by integrating the charge density $\rho$ over the volume ($V$) of the ball:

$ Q_{total} = \int_V \rho dV $

Using spherical coordinates, $dV = 4\pi r^2 dr$. The integration limits are from $0$ to $R$.

$ Q_{total} = \int_0^R \rho_0 \left(1 - \frac{r}{R}\right) (4\pi r^2 dr) $

Factor out constants:

$ Q_{total} = 4\pi \rho_0 \int_0^R \left(r^2 - \frac{r^3}{R}\right) dr $

Perform the integration:

$ Q_{total} = 4\pi \rho_0 \left[ \frac{r^3}{3} - \frac{r^4}{4R} \right]_0^R $

Evaluate the limits:

$ Q_{total} = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^4}{4R} \right) $

$ Q_{total} = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^3}{4} \right) $

$ Q_{total} = 4\pi \rho_0 \left( \frac{4R^3 - 3R^3}{12} \right) $

$ Q_{total} = 4\pi \rho_0 \left( \frac{R^3}{12} \right) $

$ Q_{total} = \frac{\pi \rho_0 R^3}{3} $

So, the enclosed charge for $r > R$ is $Q_{enc} = Q_{total} = \frac{\pi \rho_0 R^3}{3}$.

Determining the Electric Field

Now, substitute the results back into Gauss's Law:

$ E (4\pi r^2) = \frac{Q_{enc}}{\epsilon_0} $

$ E (4\pi r^2) = \frac{1}{\epsilon_0} \left( \frac{\pi \rho_0 R^3}{3} \right) $

Solve for $E$:

$ E = \frac{1}{4\pi r^2} \frac{\pi \rho_0 R^3}{3 \epsilon_0} $

Simplify the expression:

$ E = \frac{\rho_0 R^3}{12 \epsilon_0 r^2} $

This is the electric field outside the ball ($r > R$).