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Step-by-Step Solution
Step 1: State the Problem Clearly
We have a spherically symmetric charge distribution whose charge density is given by
$ \rho(r) = \rho_0 \left(\frac{5}{4} - \frac{r}{R}\right)$ for $r \le R$, and $\rho(r) = 0$ for $r > R$. We want to find the electric field $E(r)$ at a distance $r < R$ from the center.
Step 2: Enclose the Charge Using Gauss's Law
According to Gauss's law, for a spherically symmetric charge distribution, if we take a Gaussian surface of radius $r < R$, the electric flux through the surface is
$ \Phi = E(r) \cdot 4\pi r^2 $,
and this must equal the total charge enclosed divided by $ \epsilon_0 $:
$ E(r) \cdot 4\pi r^2 = \frac{Q_{\mathrm{enc}}(r)}{\epsilon_0}.$
Hence,
$ E(r) = \frac{Q_{\mathrm{enc}}(r)}{4 \pi \epsilon_0 \, r^2}. $
Step 3: Calculate the Enclosed Charge $Q_{\mathrm{enc}}(r)$
The enclosed charge up to a radius $r$ is found by integrating the volume charge density over the spherical volume of radius $r$:
$$
Q_{\mathrm{enc}}(r)
= \int_{0}^{r} \rho_0 \left(\frac{5}{4} - \frac{r'}{R}\right) \, 4\pi {r'}^2 \, dr'.
$$
Let us break this integration into parts:
Inside the integral:
$ \rho_0 \left(\frac{5}{4} - \frac{r'}{R}\right) \, 4\pi {r'}^2 = 4\pi \rho_0 \left(\frac{5}{4}{r'}^2 - \frac{{r'}^3}{R}\right). $
Integrate with respect to $r'$:
$
Q_{\mathrm{enc}}(r)
= 4\pi \rho_0 \int_{0}^{r} \left(\frac{5}{4}{r'}^2 - \frac{{r'}^3}{R}\right) dr'.
$
This splits into two separate integrals:
$$
Q_{\mathrm{enc}}(r)
= 4\pi \rho_0 \left[\int_{0}^{r} \frac{5}{4}{r'}^2 \, dr' \;-\; \int_{0}^{r} \frac{{r'}^3}{R} \, dr' \right].
$$
Compute each integral:
$ \displaystyle \int_{0}^{r} {r'}^2 \, dr' = \frac{r^3}{3}, \quad
\int_{0}^{r} {r'}^3 \, dr' = \frac{r^4}{4}. $
Hence,
$
\int_{0}^{r} \frac{5}{4}{r'}^2 \, dr' = \frac{5}{4} \cdot \frac{r^3}{3} = \frac{5r^3}{12},
$
and
$
\int_{0}^{r} \frac{{r'}^3}{R} \, dr' = \frac{1}{R} \cdot \frac{r^4}{4} = \frac{r^4}{4R}.
$
Combine the results:
$$
Q_{\mathrm{enc}}(r)
= 4\pi \rho_0 \left[\frac{5r^3}{12} - \frac{r^4}{4R}\right].
$$
Simplify further by factoring out $r^3$:
$$
Q_{\mathrm{enc}}(r)
= 4\pi \rho_0 \, r^3 \left[\frac{5}{12} - \frac{r}{4R}\right].
$$
Step 4: Substitute $Q_{\mathrm{enc}}(r)$ into Gauss's Law
Recall:
$$
E(r) = \frac{Q_{\mathrm{enc}}(r)}{4 \pi \epsilon_0 \, r^2}.
$$
Substitute the expression for $Q_{\mathrm{enc}}(r)$:
$$
E(r)
= \frac{4\pi \rho_0 \, r^3 \left[\frac{5}{12} - \frac{r}{4R}\right]}{4 \pi \epsilon_0 \, r^2}
= \frac{\rho_0 \, r \left[\frac{5}{12} - \frac{r}{4R}\right]}{\epsilon_0}.
$$
Next, factor out common denominators and simplify carefully:
$$
E(r)
= \frac{\rho_0 \, r}{\epsilon_0} \left[\frac{5}{12} - \frac{r}{4R}\right].
$$
Combine the fractions:
$$
\frac{5}{12} = \frac{5}{12} = \frac{5}{3} \cdot \frac{1}{4};
\quad
\frac{r}{4R} = \frac{1}{4} \cdot \frac{r}{R}.
$$
It can be shown that the final form rearranges to:
$$
E(r)
= \frac{\rho_0 \, r}{4\,\epsilon_0} \left(\frac{5}{3} - \frac{r}{R}\right),
$$
which matches the given correct answer.
Step 5: State the Final Result
Thus, for $r < R$, the electric field inside the sphere is:
$$
E(r)
= \frac{\rho_0 \, r}{4\,\epsilon_0} \left(\frac{5}{3} - \frac{r}{R}\right).
$$
Reference Image
