Question

Given a spherically symmetric charge density $\rho(r)=\begin{cases}kr^2, & r<R \\ 0, & r>R\end{cases}$ (k being a constant), the electric field for $r<R$ is (take the total charge as $Q$) 
 

• A. $\dfrac{Qr^3}{4\pi \epsilon_0 R^5}\,\hat{r}$
• B. $\dfrac{3Qr^2}{4\pi \epsilon_0 R^3}\,\hat{r}$
• C. $\dfrac{5Qr^3}{8\pi \epsilon_0 R^5}\,\hat{r}$
• D. $\dfrac{Q}{4\pi \epsilon_0 r^2}\,\hat{r}$

Hint:

For spherically symmetric charge distributions, Gauss's law simplifies the calculation of electric fields dramatically.

Answer 1

The Correct Option is A

Step 1: Compute enclosed charge for $r<R$.
$\rho(r)=kr^2$. Total charge inside radius $r$ is $\displaystyle Q_{\text{enc}}=\int_0^r \rho(r')\,4\pi r'^2\,dr' =4\pi k\int_0^r r'^4\,dr' = \dfrac{4\pi k r^5}{5}.$

Step 2: Compute total charge $Q$ of the sphere.
$Q = \dfrac{4\pi k R^5}{5}.$ Thus $k = \dfrac{5Q}{4\pi R^5}.$

Step 3: Substitute $k$ into $Q_{\text{enc}}$.
$Q_{\text{enc}} = \dfrac{5Q}{4\pi R^5}\cdot \dfrac{4\pi r^5}{5} = Q\left(\dfrac{r}{R}\right)^5.$

Step 4: Apply Gauss's law.
$E(4\pi r^2)=\dfrac{Q_{\text{enc}}}{\epsilon_0}$.
Thus $\displaystyle E = \dfrac{Q}{4\pi\epsilon_0 R^5}\,r^3.$

Step 5: Conclusion.
Electric field for $r<R$ is $\dfrac{Qr^3}{4\pi\epsilon_0 R^5}\hat{r}.$