Chapter 4, Problem 38P

 (a)

To determine

 The total charge.

Explanation of Solution

 Given:

 The atomic radius is $a$.

 The volume charge density $\left({\rho }_v\right)$ is ${\rho }_o\left(\frac{1-r^2}{a^2}\right)$.

 Calculation:

 Calculate the total charge $\left(Q\right)$ using the relation.

$\begin{array}{l}Q={\displaystyle {\int }_v{\rho }_{v}dv}\\ Q={\displaystyle {\int }_{\varphi =0}^{2\pi }{\displaystyle {\int }_{\theta =0}^{\pi }{\displaystyle {\int }_{r=0}^a{\rho }_v{r}^2\sin \theta drd\theta d\varphi }}}\end{array}$

$\begin{array}{c}Q={\displaystyle {\int }_{\varphi =0}^{2\pi }{\displaystyle {\int }_{\theta =0}^{\pi }{\displaystyle {\int }_{r=0}^a{\rho }_o\left(\frac{1-r^2}{a^2}\right)r^2\sin \theta drd\theta d\varphi }}}\\ =\frac{{\rho }_o}{a^2}{\displaystyle {\int }_{\varphi =0}^{2\pi }{\displaystyle {\int }_{\theta =0}^{\pi }{\displaystyle {\int }_{r=0}^a\left(r^2-r^4\right)\sin \theta drd\theta d\varphi }}}\\ =\frac{{\rho }_o}{a^2}{\left[\frac{r^3}{3}-\frac{r^5}{5}\right]}_0^a{\left[-\cos \theta \right]}_0^{\pi }{\left[\varphi \right]}_0^{2\pi }\\ =\frac{{\rho }_o}{a^2}\left[\frac{a^3}{3}-\frac{a^5}{5}-0\right]\left[1+1\right]\left[2\pi -0\right]\\ =4\pi {\rho }_o\left(\frac{a}{3}-\frac{a^3}{5}\right)\end{array}$

 Thus, the total charge $\left(Q\right)$ is $\underset{\_}{4\pi {\rho }_o\left(\frac{a}{3}-\frac{a^3}{5}\right)}$.

 (b)

To determine

 The field $E$ and potential $V$ outside the nucleus.

Explanation of Solution

 Given:

 The atomic radius is $a$.

 The volume charge density $\left({\rho }_v\right)$ is ${\rho }_o\left(\frac{1-r^2}{a^2}\right)$.

 Calculation:

 Outside the nucleus, the radius is $r>a$.

 Calculate the electric field intensity $\left(E\right)$ using the relation.

$E=\frac{Q}{4\pi {\epsilon }_0{r}^2}{a}_r$

$\begin{array}{c}E=\frac{4\pi {\rho }_o\left(\frac{a}{3}-\frac{a^3}{5}\right)}{4\pi {\epsilon }_0{r}^2}\\ =\frac{{\rho }_o\left(\frac{a}{3}-\frac{a^3}{5}\right)}{{\epsilon }_0{r}^2}\end{array}$

 Calculate the potential $\left(V\right)$ using the relation.

$V={\displaystyle {\int }_{r}Edr}$

$\begin{array}{c}V={\displaystyle \int \left(\frac{{\rho }_o\left(\frac{a}{3}-\frac{a^3}{5}\right)}{{\epsilon }_0{r}^2}\right)dr}\\ =\frac{{\rho }_o\left(\frac{a}{3}-\frac{a^3}{5}\right)}{{\epsilon }_0}{\displaystyle \int \frac{1}{r^2}dr}\\ =\frac{{\rho }_o\left(\frac{a}{3}-\frac{a^3}{5}\right)}{{\epsilon }_0}\left(\frac{-1}{r}\right)\\ =\frac{{\rho }_o\left(\frac{a^3}{5}-\frac{a}{3}\right)}{{\epsilon }_{0}r}\end{array}$

 Thus, the electric field intensity $\left(E\right)$ is $\underset{\_}{\frac{{\rho }_o\left(\frac{a}{3}-\frac{a^3}{5}\right)}{{\epsilon }_0{r}^2}}$ and the potential is $\underset{\_}{\frac{{\rho }_o\left(\frac{a^3}{5}-\frac{a}{3}\right)}{{\epsilon }_{0}r}}$.

 (c)

To determine

 The field $E$ and potential $V$ inside the nucleus.

Explanation of Solution

 Given:

 The atomic radius is $a$.

 The volume charge density $\left({\rho }_v\right)$ is ${\rho }_o\left(\frac{1-r^2}{a^2}\right)$.

 Calculation:

 Inside the nucleus, the radius is $r<a$.

 Calculate the electric field intensity $\left(E\right)$ using the relation.

$\begin{array}{l}E=\frac{Q}{4\pi {\epsilon }_0{r}^2}{a}_r\\ E=\frac{4\pi {\rho }_o\left(\frac{a}{3}-\frac{a^3}{5}\right)}{4\pi {\epsilon }_0{r}^2}\end{array}$

 Here, $r<a$. Thus, $r\to a$ and $a\to r$.

$E=\frac{{\rho }_o}{{\epsilon }_0{a}^2}\left(\frac{r}{3}-\frac{r^3}{5}\right)$

 Calculate the potential $\left(V\right)$ using the relation.

$V={\displaystyle {\int }_{r}Edr}$

$\begin{array}{c}V={\displaystyle \int \left(\frac{{\rho }_o\left(\frac{r}{3}-\frac{r^3}{5}\right)}{{\epsilon }_0{a}^2}\right)dr}\\ =\frac{{\rho }_o}{{\epsilon }_0{a}^2}{\displaystyle {\int }_0^r\left(\frac{r}{3}-\frac{r^3}{5}\right)dr}\\ =\frac{{\rho }_o}{{\epsilon }_0{a}^2}\left(\frac{r^2}{6}-\frac{r^4}{20}\right)\end{array}$

 Thus, the electric field intensity $\left(E\right)$ is $\underset{\_}{\frac{{\rho }_o}{{\epsilon }_0{a}^2}\left(\frac{r}{3}-\frac{r^3}{5}\right)}$ and the potential is $\underset{\_}{\frac{{\rho }_o}{{\epsilon }_0{a}^2}\left(\frac{r^2}{6}-\frac{r^4}{20}\right)}$.

 (d)

To determine

 The prove that $E$ is maximum at $r=0.745a$.

Explanation of Solution

 Given:

 The atomic radius is $a$.

 The volume charge density $\left({\rho }_v\right)$ is ${\rho }_o\left(\frac{1-r^2}{a^2}\right)$.

 Calculation:

 Calculate the radius $\left(r\right)$ for the maximum electric field using the relation.

$\frac{dE}{dr}=0$

$\begin{array}{l}\frac{d}{dr}\left(\frac{{\rho }_o\left(\frac{r}{3}-\frac{r^3}{5}\right)}{{\epsilon }_0{a}^2}\right)=0\\ \frac{{\rho }_o}{{\epsilon }_0{a}^2}\left[\frac{d}{dr}\left(\frac{r}{3}-\frac{r^3}{5}\right)\right]=0\\ \frac{{\rho }_o}{{\epsilon }_0{a}^2}\left(\frac{1}{3}-\frac{3r^2}{5}\right)=0\\ \frac{1}{3}=\frac{3r^2}{5}\end{array}$

$r=0.745$

 Thus, the electric field intensity $\left(E\right)$ is maximum at $r=0.745$ not at $r=0.745a$.