Question

Two protons are released from rest when they are 0.750 ${\rm nm}$ apart.
a) What is the maximum speed they will reach?
b) What is the max acceleration they will reach?

Answer 1

Concepts and reason

The concept required to solve the given question is the electrostatic potential energy.

Initially, calculate the electrostatic potential energy at the initial point using the electrostatic potential energy formula. The protons of same mass will attain the same speed. Later, use the conservation of energy and assume finally, there is only kinetic energy to solve for the maximum speed of the protons. Substitute the values to calculate the maximum speed of the protons. Use the Coulomb force equation to calculate the magnitude of maximum force and Newton’s second law to solve for the maximum acceleration.

Fundamentals

The electrostatic potential energy is,

$U = \frac{{k{q_1}{q_2}}}{r}$

Here, $k$ is the electrostatic constant, ${q_1}$ and ${q_2}$ are the magnitudes of the charges, and $r$ is the distance between them.

The kinetic energy is given as,

$K = \frac{1}{2}m{v^2}$

Here, $m$ is the mass, and $v$ is the speed.

The conservation of energy principle states that total energy of an isolated system is conserved that is,

${U_{\rm{i}}} + {K_{\rm{i}}} = {U_{\rm{f}}} + {K_{\rm{f}}}$

Here, the subscripts ${\rm{i}}$ stands for initial, and ${\rm{f}}$ stands for final.

The Coulomb’s force equation is,

$F = \frac{{k{q_1}{q_2}}}{{{r^2}}}$

Here, $k$ is the electrostatic constant, ${q_1}$ and ${q_2}$ are the magnitudes of the charges, and $r$ is the distance between them.

The Newton’s second law gives,

$\sum {F = ma}$

Here, $\sum F$ is the magnitude of net force on the body, $m$ is mass, and $a$ is the acceleration.

(a)

Substitute $e$ for ${q_1}$ , and $e$ for ${q_2}$ in the electrostatic potential energy equation $U = \frac{{k{q_1}{q_2}}}{r}$ to solve for the initial electrostatic potential energy of the protons ${U_{\rm{i}}}$.

${U_{\rm{i}}} = \frac{{k{e^2}}}{r}$ …… (1)

Here, $e$ is the magnitude of charge of the proton.

The initial kinetic energy is zero as the initial speed of protons is zero that is ${K_{\rm{i}}} = 0$.

The speed is maximum, when all the potential energy is converted to kinetic energy of the protons. Therefore, the final potential energy is zero that is ${U_{\rm{f}}} = 0$.

The final kinetic energy is sum of the kinetic energy of the protons with same mass and speed that is,

$\begin{array}{c}\\{K_{\rm{f}}} = \frac{1}{2}m{v^2} + \frac{1}{2}m{v^2}\\\\ = m{v^2}\\\end{array}$

…… (2)

Use the energy conservation equation.

${U_{\rm{i}}} + {K_{\rm{i}}} = {U_{\rm{f}}} + {K_{\rm{f}}}$

Substitute $\frac{{k{e^2}}}{r}$ for ${U_{\rm{i}}}$, $0$ for ${K_{\rm{i}}}$, $0$ for ${U_{\rm{f}}}$, and $m{v^2}$ for ${K_{\rm{f}}}$ from the equation (1) and (2) in the above equation ${U_{\rm{i}}} + {K_{\rm{i}}} = {U_{\rm{f}}} + {K_{\rm{f}}}$ to solve for maximum speed of the protons $v$.

$\begin{array}{c}\\\frac{{k{e^2}}}{r} + 0 = 0 + m{v^2}\\\\v = \sqrt {\frac{{k{e^2}}}{{mr}}} \\\end{array}$

Substitute $8.99 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}$ for $k$, $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $e$, $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for $m$, and $0.750{\rm{ nm}}$ for $r$ in the above equation $v = \sqrt {\frac{{k{e^2}}}{{mr}}}$ and calculate $v$.

$v = \sqrt {\frac{{\left( {8.99 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right){{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}^2}}}{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right)\left( {0.750{\rm{ nm}}} \right)}}}$

Convert nm to m by the multiplying with $\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1{\rm{ nm}}}}} \right)$.

$\begin{array}{c}\\v = \sqrt {\frac{{\left( {8.99 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right){{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}^2}}}{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right)\left( {0.750{\rm{ nm}}\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1{\rm{ nm}}}}} \right)} \right)}}} \\\\ = 1.36 \times {10^4}{\rm{ m/s}}\\\end{array}$

(b)

Use the Coulomb force equation.

$F = \frac{{k{q_1}{q_2}}}{{{r^2}}}$

Substitute $e$ for ${q_1}$ , and $e$ for ${q_2}$ in the above equation $F = \frac{{k{q_1}{q_2}}}{{{r^2}}}$ and solve for magnitude of force.

$F = \frac{{k{e^2}}}{{{r^2}}}$

Use the Newton’s second law to solve the acceleration.

$\sum F = ma$

Substitute $\frac{{k{e^2}}}{{{r^2}}}$ for $\sum F$ in the above equation and rearrange the equation to solve acceleration.

$\begin{array}{c}\\\frac{{k{e^2}}}{{{r^2}}} = ma\\\\a = \frac{{k{e^2}}}{{m{r^2}}}\\\end{array}$

…… (3)

Use the equation (3) to calculate acceleration.

$a = \frac{{k{e^2}}}{{m{r^2}}}$

Substitute $8.99 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}$ for $k$, $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $e$, $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for $m$, and $0.750{\rm{ nm}}$ for $r$ in the above equation $a = \frac{{k{e^2}}}{{m{r^2}}}$ and calculate the maximum acceleration.

$a = \frac{{\left( {8.99 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right){{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}^2}}}{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right){{\left( {0.750{\rm{ nm}}} \right)}^2}}}$

Convert nm to m by multiplying with $\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1{\rm{ nm}}}}} \right)$.

$\begin{array}{c}\\a = \frac{{\left( {8.99 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right){{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}^2}}}{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right){{\left( {0.750{\rm{ nm}}\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1{\rm{ nm}}}}} \right)} \right)}^2}}}\\\\ = 2.45 \times {10^{17}}{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Ans: Part a

The maximum speed of the protons is $1.36 \times {10^4}{\rm{ m/s}}$ .

Part b

The maximum acceleration of the protons is $2.45 \times {10^{17}}{\rm{ m/s}}$ .