Question

A lead nucleus is spherical with a radius of about $\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{\text{m}}$ . The nucleus contains $\mathbf{82}$protons (and typically$\mathbf{126}$ neutrons). Because of their motions the protons can be considered on average to be uniformly distributed throughout the nucleus. Base on the net flux at the surface of the nucleus, calculate the divergence of the electric field as electric flux per unit volume. Repeat the calculation at a radius of$\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{\text{m}}$ . (You can use Gauss’s law to determine the magnitude of the electric field at this radius.) Also calculate the quantity$\mathit{\rho }\mathbf{/}{\mathit{\epsilon }}_{\mathbf{0}}$ inside the nucleus.

Short answer

The divergence of electric field of the sphere with radius$7\times {10}^{-15}\text{m}$ is $1.03\times {10}^{36}N.m/C$.

Step-by-step solution

Given Information.

A lead nucleus is spherical with a radius of about$7\times {10}^{-15}\text{m}$ . The nucleus contains$82$ protons. Because of their motions the protons can be considered on average to be uniformly distributed throughout the nucleus.

Definition/ Concept.

The charge on the each proton is equal to $\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{\text{C}}$ .

Find the divergence of the electric field.

The lead nucleus contains protons. So, the change in the lead nucleus is:

$$\begin{gathered} q=\left(82\right)\left(1.60\times {10}^{-19}\text{C}\right) \\ q=1.31\times {10}^{-17}\text{C} \end{gathered}$$

The volume two of the spherical nucleus of radius $r$can be expressed as:

$V=\frac{4}{3}\pi r^3$

Now substitute $7\times {10}^{-15}\text{m}$for radius of the nucleus in the above equation.

Then:

$$\begin{gathered} V=\frac{4}{3}\pi {\left(7\times {10}^{-15}\right)}^3 \\ V=1.44\times {10}^{-42}\text{m} \end{gathered}$$

Now the change density in the nucleus is$\rho =\frac{q}{V}$ .

Substitute the numerical values in the above equation.

$\begin{array}{rcl}\rho & =& \frac{q}{V}\\ & =& \frac{1.31\times {10}^{-17}}{1.44\times {10}^{-42}}\\ \rho & =& 9.12\times {10}^{24}C/m^3\end{array}$

The divergence of electric field related to charge density as $div\left(\overrightarrow{E}\right)=\frac{\rho }{{\epsilon }_0}$.

Here, the value of constant$\left({\epsilon }_0\right)$ is$8.85\times {10}^{-12}{C}^2/N.m^2$ .

Now substitute the numerical values in the above equation as:

$\begin{array}{rcl}div\left(\overrightarrow{E}\right)& =& \frac{\rho }{{\epsilon }_0}\\ div\left(\overrightarrow{E}\right)& =& \frac{9.12\times {10}^{24}C/m^3}{8.85\times {10}^{-12}{C}^2/N.m^2}\\ div\left(\overrightarrow{E}\right)& =& 1.03\times {10}^{36}N.m/C\end{array}$

Therefore, the divergence of electric field of the sphere with radius$7\times {10}^{-15}\text{m}$ is$1.03\times {10}^{36}N.m/C$ .