Chapter 17, Problem 15P

To determine

The work done in bringing a third charge from infinite to the point $a$ given in the figure.

Answer to Problem 15P

The work done by the field is $-1.80\text{ μJ}$.

Explanation of Solution

The charges are $+8.00\text{nC, }-8.00\text{nC}\text{and +2}.00\text{nC}$  and the distances between the charges are given in the figure. Initially the third charge was at $b$ and it is moved to the point $c$.

Write the expression for work done by the electric field

    $W_{field}=-\Delta U$

Here, $W_{field}$ is the work done by the electric field,  $\Delta U$ is the change in potential energy

Substitute for $\Delta U$ in the above expression

  $W_{field}=U_i-U_f$                                                                                        (I)

Here, $U_f$ is the final electric potential energy and $U_i$ is the initial electric potential energy

Write the expression for initial electric potential energy for three charges

    $U_i=k\left[\frac{q_1{q}_2}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_3{q}_1}{r_{31}}\right]$                                                                                (II)

Here, $q_1$, $q_2$,$q_3$ are the three charges,  $r_{12}$ is the initial distance between $q_1$ and $q_2$, $r_{23}$ is the  initial distance between $q_2$ and $q_3$, $r_{31}$ is the initial distance between $q_3$ and $q_1$ , $U_i$ is the initial electric potential energy and $k$ is a constant.

Write the expression for final electric potential energy for three charges

    $U_f=k\left[\frac{q_1{q}_2}{{r^{\prime }}_{12}}+\frac{q_2{q}_3}{{r^{\prime }}_{23}}+\frac{q_3{q}_1}{{r^{\prime }}_{31}}\right]$                                                                                (III)

Here, $q_1$, $q_2$,$q_3$ are the three charges,  ${r^{\prime }}_{12}$ is the final distance between $q_1$ and $q_2$, ${r^{\prime }}_{23}$ is the  final distance between $q_2$ and $q_3$, ${r^{\prime }}_{31}$ is the final distance between $q_3$ and $q_1$ , $U_f$ is the final electric potential energy and $k$ is a constant.

Substitute (II) and (III) in (I) to find $W_{field}$

  $W_{field}=k\left[\frac{q_1{q}_2}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_3{q}_1}{r_{31}}\right]-k\left[\frac{q_1{q}_2}{{r^{\prime }}_{12}}+\frac{q_2{q}_3}{{r^{\prime }}_{23}}+\frac{q_3{q}_1}{{r^{\prime }}_{31}}\right]$

Only the position of the third charge changes and hence the distance between the first two charges remains the same.

Substitute $r_{12}$ for ${r^{\prime }}_{12}$ in the above equation and solve further

  $W_{field}=k\left[q_2{q}_3\left(\frac{1}{r_{23}}-\frac{1}{{r^{\prime }}_{23}}\right)+q_3{q}_1\left(\frac{1}{r_{31}}-\frac{1}{{r^{\prime }}_{31}}\right)\right]$                                                           (IV)

Substitute $+8.00\text{nC}$ for $q_1$,  $-8.00\text{nC}$ for $q_2$,  $\text{+2}.00\text{nC}$ for $q_3$,  $8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}$ for $k$, $4.00\text{cm}$ for $r_{23}$, $8.00\text{cm}$ for $r_{31}$, $12.0\text{cm}$ for ${r^{\prime }}_{23}$ and $12.0\text{cm}$ for ${r^{\prime }}_{31}$ in (IV) to find $W_{field}$

    $\begin{array}{c}{W}_{field}=8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left[\begin{array}{l}\left(-8.00\text{nC}\right)\left(\text{2}.00\text{nC}\right)\left(\frac{1}{4.00\text{cm}}-\frac{1}{8.00\text{cm}}\right)+\\ \left(\text{2}.00\text{nC}\right)\left(8.00\text{nC}\right)\left(\frac{1}{12.0\text{cm}}-\frac{1}{12.0\text{cm}}\right)\end{array}\right]\\ =8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(8.00\text{C}\right)\left(\text{2}.00\text{C}\right)\left[-\left(\frac{1}{4.00\text{m}}-\frac{1}{8.00\text{m}}\right)+\left(\frac{1}{12.0\text{m}}-\frac{1}{12.0\text{m}}\right)\right]\times \frac{{10}^{-18}}{{10}^{-2}}\\ =-1.80\times {10}^{-6}\text{ Nm}\\ =-1.80\text{ μJ}\end{array}$

Thus, the work done by the electrical field is $-1.80\text{ μJ}$.