Question

How much work would it take to push two protons very slowly from a separation of 2e-10 m. (a typical atomic distance) to 3e-15 m. (a typical nuclear distance)?

If the protons are both released from rest at the closer distance, how fast are they moving when they reach their original separation?

Answer 1

Concepts and reason

The concepts used here are electric potential energy, and work energy theorem.

First, calculate the potential energy using the electric potential energy equation.

Later, use work energy theorem to calculate the work done.

Finally, calculate the speed of the electron by using the work done in terms of kinetic energy.

Fundamentals

Work is a type of force which when acts on the object, there is a displacement of the point of application in the direction of force. It is measured in Joules and is denoted by W. The work done transfers energy from one place to another.

Mathematically, the work done is equal to the force multiplied by the distance or displacement.

Also, the work done is equal to the change in the potential energy of the system.

According to the work energy theorem, the work done is id equal to the change in the kinetic energy of the system.

The energy possessed by a body by virtue of its position relative to other objects is the potential energy of the object. It is measured in joules and is denoted by U.

The expression for the kinetic energy is as follows:

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

Here, m is the mass of the object and v is the speed of the object.

The expression for the potential energy is as follows:

$U = \frac{{{q_{\rm{p}}}{q_{\rm{P}}}}}{{4\pi {\varepsilon _0}r}}$

Here, ${q_{\rm{P}}}$ is the charge on a proton, ${r_1}$ is the separation between the charges, and ${\varepsilon _0}$ is the permittivity of the free space.

The expression for the work done in terms of the potential energy is as follows:

$W = {U_{\rm{f}}} - {U_{\rm{i}}}$ ……. (1)

Here, ${U_{\rm{f}}}$ is the final potential energy and ${U_{\rm{i}}}$ is the initial potential energy.

The expression for the initial potential energy is as follows:

${U_{\rm{i}}} = \frac{{{q_{\rm{p}}}{q_{\rm{P}}}}}{{4\pi {\varepsilon _0}{r_{\rm{i}}}}}$ ……. (2)

The expression for the final potential energy is as follows:

${U_{\rm{f}}} = \frac{{{q_{\rm{p}}}{q_{\rm{P}}}}}{{4\pi {\varepsilon _0}{r_{\rm{f}}}}}$ ……. (3)

And, the expression for the work energy theorem is as follows:

$W = {K_{\rm{f}}} - {K_{\rm{i}}}$ ……. (4)

Here, ${K_{\rm{f}}}$ is the final kinetic energy and ${K_{\rm{i}}}$ is the initial kinetic energy.

The expression for the initial kinetic energy is as follows:

${K_{\rm{i}}} = \frac{1}{2}mv_{\rm{i}}^2$

Here, ${v_{\rm{i}}}$ is the initial speed.

And, the expression for the final kinetic energy is as follows:

${K_{\rm{f}}} = \frac{1}{2}mv_{\rm{f}}^2$

Here, ${v_{\rm{f}}}$ is the final speed.

(a)

Calculate the initial potential energy by substituting $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{{\rm{ - 2}}}}$ for $\frac{1}{{4\pi {\varepsilon _0}}}$ , $1.6 \times {10^{ - 19}}{\rm{ C}}$ for ${q_{\rm{P}}}$ , and $2 \times {10^{ - 10}}{\rm{ m}}$ for ${r_{\rm{i}}}$ in the equation (2).

$\begin{array}{c}\\{U_{\rm{i}}} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{{\rm{ - 2}}}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}{{\left( {2 \times {{10}^{ - 10}}{\rm{ m}}} \right)}}\\\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{{\rm{ - 2}}}}} \right){{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}^2}}}{{\left( {2 \times {{10}^{ - 10}}{\rm{ m}}} \right)}}\\\\ = 11.52 \times {10^{ - 19}}{\rm{ J}}\\\end{array}$

Calculate the final potential energy by substituting $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{{\rm{ - 2}}}}$ for $\frac{1}{{4\pi {\varepsilon _0}}}$ , $1.6 \times {10^{ - 19}}{\rm{ C}}$ for ${q_{\rm{P}}}$ , and $3 \times {10^{ - 15}}{\rm{ m}}$ for ${r_{\rm{f}}}$ in the equation (3).

$\begin{array}{c}\\{U_{\rm{f}}} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{{\rm{ - 2}}}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}{{3 \times {{10}^{ - 15}}{\rm{ m}}}}\\\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{{\rm{ - 2}}}}} \right){{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}^2}}}{{3 \times {{10}^{ - 15}}{\rm{ m}}}}\\\\ = 7.68 \times {10^{ - 14}}{\rm{ J}}\\\end{array}$

Calculate the work done by substituting $7.68 \times {10^{ - 14}}{\rm{ J}}$ for ${U_{\rm{f}}}$ and $11.52 \times {10^{ - 19}}{\rm{ J}}$ for ${U_{\rm{i}}}$ in the equation (1).

$\begin{array}{c}\\W = 7.68 \times {10^{ - 14}}{\rm{ J}} - 11.52 \times {10^{ - 19}}{\rm{ J}}\\\\ = 7.67 \times {10^{ - 14}}{\rm{ J}}\\\end{array}$

(b)

Calculate the velocity by substituting $\frac{1}{2}mv_{\rm{f}}^2$ for ${K_{\rm{f}}}$ and $\frac{1}{2}mv_{\rm{i}}^2$ for ${K_{\rm{i}}}$ in the equation (4).

$W = \frac{1}{2}mv_{\rm{f}}^2 - \frac{1}{2}mv_{\rm{i}}^2$

Solve for v.

$\begin{array}{l}\\W = 2\left( {\frac{1}{2}m{v^2}} \right)\\\\v = \sqrt {\frac{W}{m}} \\\end{array}$

Therefore, the velocity is equal to,

$v = \sqrt {\frac{W}{m}}$ …… (5)

Calculate the speed of each proton by substituting $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for m and $7.67 \times {10^{ - 14}}{\rm{ J}}$ for W in the equation (5) $v = \sqrt {\frac{W}{m}}$ .

$\begin{array}{c}\\v = \sqrt {\frac{{7.67 \times {{10}^{ - 14}}{\rm{ J}}}}{{1.67 \times {{10}^{ - 27}}{\rm{ kg}}}}} \\\\ = 6.77 \times {10^6}{\rm{ m/s}}\\\end{array}$

Ans: Part a

The work done to push two protons is equal to $7.67 \times {10^{ - 14}}{\rm{ J}}$ .

Part b

The speed of each proton is equal to $6.77 \times {10^6}{\rm{ m/s}}$ .