Question

Three particles, charge ${\mathbf{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{10}\mathbf{}\mathbf{\mu C}$, ${\mathbf{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{20}\mathbf{}\mathbf{\mu C}$ , and ${\mathbf{q}}_{\mathbf{3}}\mathbf{=}\mathbf{+}\mathbf{}\mathbf{\mu C}$ , are positioned at the vertices of an isosceles triangle as shown in Fig. 24-62. If $\mathbf{a}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{cm}$ and $\mathbf{b}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ , how much work must an external agent do to exchange the positions of (a) ${\mathit{q}}_{\mathbf{1}}$ and ${\mathit{q}}_{\mathbf{3}}$ and, instead, (b) ${\mathit{q}}_{\mathbf{1}}$ and${\mathit{q}}_{\mathbf{2}}$?

Short answer

(a) The work is $W=-24\text{J}$.

(b) The work is $W=0\text{J}$.

Step-by-step solution

Given data:

Draw the free body diagram as below.

From the above figure:

The charge, $q_1=+10\mu C=+10\times {10}^{-6}C$

The charge, $q_2=-20\mu C=-20\times {10}^{-6}C$

The charge, $q_3=+30\mu C=+30\times {10}^{-6}C$

The distance,$a=0.10\text{m}$$b=0.06\text{m}$

The distance,

Understanding the concept:

Conservation of work energy principle state that,the total energy of a particle in a conservative force field is constant.

(a) Calculate how much work must an external agent do to exchange the positions of  q1 and q3 :

The net energy of the system in initial set-up of charges:

$$\begin{gathered} {\left(U_{net}\right)}_i=U_{12}+U_{23}+U_{13} \\ =\frac{1}{4\pi {\epsilon }_o}\left[\frac{q_1{q}_2}{b}+\frac{q_2{q}_3}{a}+\frac{q_1{q}_3}{a}\right] \end{gathered}$$

$$\begin{gathered} {\left(U_{net}\right)}_i=9.0\times {10}^{9}N.m^2/C^2\left[\frac{\left(+10\times {10}^{-6}\text{C}\right)\left(-20\times {10}^{-6}\text{C}\right)}{0.06\text{m}}+\frac{\left(-20\times {10}^{-6}\text{C}\right)\left(+30\times {10}^{-6}\text{C}\right)}{0.10\text{m}} \\ +\frac{\left(+10\times {10}^{-6}\text{C}\right)\left(+30\times {10}^{-6}\text{C}\right)}{0.10\text{m}} \\ \right] \end{gathered}$$

$=9.0\times {10}^{9}N.m^2/C^2\left[\frac{\left(-200\times {10}^{-12}{\text{C}}^2\right)}{0.06\text{m}}+\frac{\left(-600\times {10}^{-12}{\text{C}}^2\right)}{0.10\text{m}}+\frac{\left(+300\times {10}^{-12}{\text{C}}^2\right)}{0.10\text{m}}\right]$

$=9.0\times {10}^{9}N.m^2/C^2\left[\begin{array}{c}\left(-3333.33\times {10}^{-12}{C}^2/m\right)-\left(6000\times {10}^{-12}{C}^2/m\right)\\ =\left(3000\times {10}^{-12}{C}^2/m\right)\\ \end{array}\right]$

$=9.0\times {10}^{9}N.m^2/C^2\left(6333.33\times {10}^{-12}{C}^2/m\right)$

$$\begin{gathered} {\left(U_{net}\right)}_i=-56999.97\times {10}^{-3}\text{N}·\text{m} \\ =-57\text{J} \end{gathered}$$

Now, after exchanging $q_1$ and $q_3$:

The net energy of the system with final set-up:

$$\begin{gathered} {\left(U_{net}\right)}_f=U_{12}+U_{23}+U_{12} \\ =\frac{1}{4\pi {\epsilon }_o}\left[\frac{q_1{q}_2}{a}+\frac{q_2{q}_3}{b}+\frac{q_1{q}_3}{b}\right] \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} {\left(U_{net}\right)}_f=9.0\times {10}^9\left[\frac{-200}{0.10}-\frac{600}{0.06}+\frac{300}{0.10}\right]\times {10}^{-12} \\ =9.0\times {10}^9\left[-9000\times {10}^{-12}\right] \\ =-81\text{J} \end{gathered}$$

Thus, the net work done is,

$$\begin{gathered} W={\left(U_{net}\right)}_f-{\left(U_{net}\right)}_i \\ =\left(-81+57\right)\text{J} \\ =-24\text{J} \end{gathered}$$

Hence, the work is $-24J$.

(b) Calculate how much work must an external agent do to exchange the positions of  q1 and q2 :

The net energy of the system in initial set-up of charges:

${\left(U_{net}\right)}_i=-57J$

Now, after exchanging$q_1$ and $q_2$.

The net energy of the system with final set-up:

$$\begin{gathered} {\left(U_{net}\right)}_f=U_{12}+U_{23}+U_{12} \\ =\frac{1}{4\pi {\epsilon }_o}\left[\frac{q_1{q}_2}{b}+\frac{q_2{q}_3}{a}+\frac{q_1{q}_3}{a}\right] \end{gathered}$$

$$\begin{gathered} {\left(U_{net}\right)}_f=9.0\times {10}^9\left[\frac{-200}{0.06}+\frac{300}{0.10}-\frac{600}{0.10}\right]\times {10}^{-12} \\ =-57\text{J} \end{gathered}$$

Therefore, the net work done is,

$$\begin{gathered} W={\left(U_{net}\right)}_f-{\left(U_{net}\right)}_i \\ =\left(-57+57\right)\text{J} \\ =0\text{J} \end{gathered}$$

Hence, the work is $0J$.