Question

Electric Fields in an Atom. The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately $\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{m}$. (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{m}$? (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

Short answer

(a) The electric field this nucleus produces just outside its surface is $2.42\times {10}^{21}\mathrm{N}/\mathrm{C}$.

(b) The magnitude of electric field is $1.3\times {10}^{13}\mathrm{N}/\mathrm{C}$.

(c) The net electric field is zero.

Step-by-step solution

Definition of Electric field

The term electric field may be defined as the area around the charge where a positive test charge experienced a force.

Determine the electric field

The electric field can be calculated by the relation $\mathrm{E}=\frac{{\mathrm{q}}_{\mathrm{enq}}}{4{\mathrm{\tau \tau \epsilon }}_0{\mathrm{r}}^2}$

(a) For ${\mathrm{r}}_{\mathrm{n}}$the electric field is

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{q}}_{\mathrm{enq}}}{4{\mathrm{\tau \tau \epsilon }}_0{\mathrm{r}}^2} \\ \mathrm{E}=\frac{9\times {10}^9\left(92\times 1.6\times {10}^{-19}\right)}{{\left(7.4\times {10}^{-15}\right)}^2} \\ \mathrm{E}=2.42\times {10}^{21}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence,the electric field this nucleus produces just outside its surface is $2.42\times {10}^{21}\mathrm{N}/\mathrm{C}$.

For${\mathrm{r}}_1$ the electric field is

$\mathrm{E}=\frac{{\mathrm{q}}_{\mathrm{enq}}}{4{\mathrm{\tau \tau \epsilon }}_0{\mathrm{r}}_1^2}$

$$\begin{gathered} \mathrm{E}=\frac{9\times {10}^9\left(92\times 1.6\times {10}^{-19}\right)}{{\left(1.0\times {10}^{-10}\right)}^2} \\ \mathrm{E}=1.3\times {10}^{13}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence,the magnitude of electric field is $1.3\times {10}^{13}\mathrm{N}/\mathrm{C}$.

(c) The total enclosed charged at the center of the shell is zero that’s why the net electric field is zero.