Question

23.47: A point charge${\mathbf{q}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{nC}$is placed at the origin, and a second point charge${\mathbf{q}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{nC}$is placed on the x-axis at$\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{+}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$. A third point charge${\mathbf{q}}_{\mathbf{3}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{nC}$is to be placed on the x-axis between${\mathbf{q}}_{\mathbf{1}}$and${\mathbf{q}}_{\mathbf{2}}$. (Take as zero the potential energy of the three charges when they are infinitely far apart.) (a) What is the potential energy of the system of the three charges if ${\mathbf{q}}_{\mathbf{3}}$is placed at x = +10.0 cm? (b) Where should ${\mathbf{q}}_{\mathbf{3}}$be placed to make the potential energy of the system equal to zero?

Short answer

(a) The potential energy of a system of charges is expressed as $U=-3.60*{10}^{-7}J$

(b) The charge $7.40cm$is $q_3$needs to be placed at a distance of $7.40cm$, to make the potential energy of the system equal to zero.

Step-by-step solution

Step-1:The definition of potential energy

The potential energy of a system of charges is given as $\mathit{U}\mathbf{=}\mathit{k}\underset{\mathbf{i}\mathbf{<}\mathbf{j}}{\overset{\mathbf{o}}{\mathbf{a}}}\frac{{\mathbf{q}}_{\mathbf{i}}{\mathbf{q}}_{\mathbf{j}}}{{\mathbf{r}}_{\mathbf{i}\mathbf{j}}}$where $\mathit{k}$is the constant$\mathit{k}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{p}{\mathbf{e}}_{\mathbf{0}}}\mathbf{=}\mathbf{9}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}\mathit{N}{\mathit{m}}^{\mathbf{2}}\mathbf{/}{\mathit{C}}^{\mathbf{2}}\mathbf{,}{\mathit{q}}_{\mathbf{i}}\mathbf{,}{\mathit{q}}_{\mathbf{j}}$are the pair of charges under a separation of ${\mathit{r}}_{\mathbf{i}\mathbf{j}}$.For a system of three charges, the formula can be arranged as $\mathit{U}\mathbf{=}\mathit{k}\left[\frac{q_1{q}_2}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_1{q}_3}{r_{13}}\right]$.

Step-2: Calculating the value of potential energy

(a) The arrangement is as per the diagram shown. The distance between $q_1$and $q_2$is denoted by $r_{12}$, the distance between $q_2$and $q_3$is $r_{23}$, and the distance between $q_1$and $q_3$is $r_{13}$.

Substituting the values of

$$\begin{gathered} k=\frac{1}{4p{e}_0}=9*{10}^{9}N{m}^2/C^2,q_1=4.00nC=4.00*{10}^{-9}C, \\ {q}_2=-3.00nC=-3*{10}^{-9}C,q_3=2nc=2*{10}^{-9}C,r_{12}=20cm=0.20m, \\ {r}_{23}=0.10m,r_{13}=0.10m\mathrm{in}U=k\left[\frac{q_1{q}_2}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_1{q}_3}{r_{13}}\right], \end{gathered}$$

$$\begin{gathered} U=\left(9*{10}^9\right)\left[\frac{\left(4*{10}^{-9}\right)\left(-3*{10}^{-9}\right)}{0.20}+\frac{\left(-3*{10}^{-9}\right)\left(2*{10}^{-9}\right)}{0.10}+\frac{\left(4*{10}^{-9}\right)\left(2*{10}^{-9}\right)}{0.10}\right] \\ U=\left(9*{10}^9\right)\left[\frac{-12*{10}^{-18}}{0.20}+\frac{-6*{10}^{-18}}{0.10}+\frac{8*{10}^{-18}}{0.10}\right] \\ U=\left(9*{10}^9\right)\left(-40\right) \\ U=-3.6*{10}^{-7}J \end{gathered}$$

Thus, the total potential energy possessed by the system of three charges is $U=-3.6*{10}^{-7}J$.

Step-3: Calculating the distance of placing q3.

(b) The charge $q_3$is assumed to be placed at a distance of $dcm$from $q_1$, then the rest distance to $q_2$can be $0.20-d$. ie, ${\mathrm{r}}_{13}=\mathrm{d}\mathrm{cm}$, ${\mathrm{r}}_{32}=0.20-\mathrm{d}$, ${\mathrm{r}}_{12}=0.20$,and this new configuration is of zero potential energy.

The equation $U=k\left[\frac{q_1{q}_2}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_1{q}_3}{r_{13}}\right]$can be modified to, $0=k\left[\frac{q_1{q}_2}{0.20}+\frac{q_2{q}_3}{0.20-d}+\frac{q_1{q}_3}{d}\right]$with the values of $q_1=4.00nC=4.00*{10}^{-9}C$it becomes,

$\begin{array}{r}0=k\left[\frac{\left(\left(4\ast {10}^{-9}\right)\left(-3^{-3}{10}^{-9}\right)\right)}{0.20}+\frac{\left(-3^{\ast }{10}^{-9}\right)\left(2\ast {10}^{-9}\right)}{0.20-d}+\frac{\left(4\ast {10}^{-9}\right)\left(2\ast {10}^{-9}\right)}{d}\right]\\ 0=k\left[\frac{\left(\left(-{12}^{\ast }{10}^{-18}\right)\right.}{0.20}+\frac{\left(-6\ast {10}^{-18}\right)}{0.20-d}+\frac{\left(8\ast {10}^{-18}\right)}{d}\right]\\ 0=k\left[-60+\frac{(-6)}{0.20-d}+\frac{\left(8\right)}{d}\right]\\ \\ 60d^2-26d+1.60=0\end{array}$

Solving, the values of d as are obtained as solutions, among which, that value coming within the limit of the total distance 20cm is

$0.074\mathrm{m}=7.40\mathrm{cm}$

Thus, the charge $q_3$needs to be placed at$7.40cm$ from the origin.