Chapter 30, Problem 28E

(a)

To determine

The magnitude of the electric force.

Answer to Problem 28E

The magnitude of the electric force is $230.4\text{N}$.

Explanation of Solution

Write the expression for the magnitude of the electric force.

    $F_{pp}=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$        (I)

Here, $F_{pp}$ is the magnitude of the force, ${\epsilon }_0$ is the permittivity of free space,  $q$ is the charge of each proton and $r$ is the separation distance between the protons.

Conclusion:

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$, $9\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}$ for $1/4\pi {\epsilon }_0$ and ${10}^{-15}\text{m}$ for $r$ in equation (I).

    $\begin{array}{c}{F}_{pp}=\left(9\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}\right)\frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2}{{\left({10}^{-15}\text{m}\right)}^2}\\ =230.4\text{N}\end{array}$

Thus, the magnitude of the electric force is $230.4\text{N}$.

(b)

To determine

The electric field between a proton and an electron.

Answer to Problem 28E

The electric field between a proton and an electron is $230.4\times {10}^{-6}\text{N}$.

Explanation of Solution

Write the expression for the magnitude of the electric force.

    $F_{pe}=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$        (II)

Here, $F_{pe}$ is the magnitude of the force, ${\epsilon }_0$ is the permittivity of free space,  $q$ is the charge of proton and electron and $r$ is the separation distance between the proton and the electron.

Conclusion:

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$, $9\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}$ for $1/4\pi {\epsilon }_0$ and ${10}^{-10}\text{m}$ for $r$ in expression (II).

  $\begin{array}{c}{F}_{pe}=\left(9\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}\right)\frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2}{{\left({10}^{-10}\text{m}\right)}^2}\\ =230.4\times {10}^{-6}\text{N}\end{array}$

Thus, the electric field between a proton and an electron is $230.4\times {10}^{-6}\text{N}$.

(c)

To determine

The ratio of the forces.

Answer to Problem 28E

The ratio of the forces is ${10}^6$.

Explanation of Solution

Write the expression for the ratio between the two forces.

    $r=\frac{F_{pp}}{F_{pe}}$        (III)

Here, $F_{pp}$ is the magnitude of the field between two protons and $F_{pe}$ is the magnitude of the field between a proton and an electron.

Conclusion:

Substitute $230.4\text{N}$ for $E_{pp}$ and $230.4\times {10}^{-6}\text{N}$ for $E_{pe}$ in equation (III).

    $\begin{array}{c}r=\frac{230.4\text{N}}{230.4\times {10}^{-6}\text{N}}\\ ={10}^6\end{array}$

Thus, the ratio of the forces is ${10}^6$.