91影视

Draw the free body diagram as below.

From the above figure:

The charge, $q_1=+10渭C=+10脳{10}^{-6}C$

The charge, $q_2=-20渭C=-20脳{10}^{-6}C$

The charge, $q_3=+30渭C=+30脳{10}^{-6}C$

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

The distance,

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

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

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

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

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

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

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

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

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

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

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

Substitute known values in the above equation.

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

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 $-24J$.

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:

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

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

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 $0J$.