Chapter 7.5, Problem 41E

a.

To determine

To state: The work that will be needed to move the second electron along the x- axis from the point $\left(-1,0\right)$ to the origin if one electron is held fixed at the point $\left(1,0\right)$ on the x- axis (units in meters). Two electrons r meters apart repel each other with a force of $F=\frac{23\times {10}^{-29}}{r^2}\text{ newton}$ .

Answer to Problem 41E

The resultant answer is $1.15\times {10}^{-28}\text{J}$ .

Explanation of Solution

Given information:

It is given that two electrons r meters apart repel each other with a force of $F=\frac{23\times {10}^{-29}}{r^2}$ .

Two electrons r meters apart repel each other with a force of $F=\frac{23\times {10}^{-29}}{r^2}$ . If the force $F\left(x\right)$ is not constant, then the work done in moving an object from $x=a\text{ to }x=b$ is the definite integral $W={\displaystyle {\int }_a^{b}F\left(x\right)dx}$ .

Consider the given expression:

  $F\left(r\right)=\frac{23\times {10}^{-29}}{r^2}\text{ and }a=1,b=2$

The distance goes from $2m\text{ to }1m$ . The work done by an external force equals the work done by repulsion in moving the electrons from a distance of $1m\text{ to 2}m$ .

The work done to move a second electron along x-axis from the point $\left(-1,0\right)$

to the origin is given by:

  $\begin{array}{c}W={\displaystyle {\int }_a^{b}F\left(r\right)dr}\\ ={\displaystyle {\int }_1^2\frac{23\times {10}^{-29}}{r^2}dr}\\ =23\times {10}^{-29}{\left[-\frac{1}{r}\right]}_1^2\\ =1.15\times {10}^{-28}\text{J}\end{array}$

Therefore, $1.15\times {10}^{-28}\text{J}$ work will be needed to move the second electron along the x- axis.

b.

To determine

To state: The work that will be needed to move the third electron along the x- axis from the point $\left(5,0\right)\text{ to }\left(3,0\right)$ if an electron is held fixed at the point $\left(-1,0\right)\text{ and }\left(1,0\right)$ on the x- axis (units in meters). Two electrons r meters apart repel each other with a force of $F=\frac{23\times {10}^{-29}}{r^2}\text{ newton}$ .

Answer to Problem 41E

The resultant answer is $7.6667\times {10}^{-29}\text{J}$ .

Explanation of Solution

Given information:

It is given that an electron is held fixed at the point $\left(-1,0\right)\text{ and }\left(1,0\right)$ on the x- axis (units in meters).

Find the work done by the fixed electrons in pushing the third one way. The total work is the sum of the work done by each fixed electron. The changes in distance are $4m\text{ and }6m$ and $2m\text{ and 4}m$ respectively.

  $\begin{array}{c}W={\displaystyle {\int }_4^6\frac{23\times {10}^{-29}}{r^2}dr}+{\displaystyle {\int }_2^4\frac{23\times {10}^{-29}}{r^2}dr}\\ =23\times {10}^{-29}\left({\left[-\frac{1}{r}\right]}_4^6+{\left[-\frac{1}{r}\right]}_2^4\right)\\ \approx 7.6667\times {10}^{-29}\text{J}\end{array}$

Therefore, $7.6667\times {10}^{-29}\text{J}$ work will be needed to move the third electron along the x- axis.