Chapter 25, Problem 17P

(a)

To determine

The electric potential of the configuration of the three fixed charges.

Answer to Problem 17P

The potential energy of the configuration is $-45.0\mu \text{J}$.

Explanation of Solution

The following figure shows the placement of charges $q_1$, $q_2$ and $q_3$. A fourth charge $q_4$ is placed at point $P$.

Figure-(1)

Write the expression for the configuration of the three fixed point’s electric potential energy.

    $U=k_{\text{e}}\left[\frac{q_1\times q_2}{r_{12}}+\frac{q_1\times q_3}{r_{13}}+\frac{q_2\times q_3}{r_{23}}\right]$                                                                             (I)

Here, $U$ is the electric potential of three charges, $k_{\text{e}}$ is the Coulomb constant, $q_1$, $q_2$ and $q_3$ are charges and $r_{12}$, $r_{13}$ and $r_{23}$ are the distance between the charges.

Conclusion:

Substitute $9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $20.0\text{nC}$ for $q_1$, $-20.0\text{nC}$ for $q_2$, $10\text{nC}$ for $q_3$, $4.00\text{cm}$ for $r_{13}$ and $r_{23}$, $8.00\text{cm}$ for $r_{12}$ in Equation (I) to calculate $U$.

    $\begin{array}{c}U=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left[\frac{\left(-20.0\times 20.0\right){\text{nC}}^2}{8.00\text{cm}}+\frac{\left(10.0\times 20.0\right){\text{nC}}^2}{4.00\text{cm}}+\frac{\left(-20.0\times 10.0\right){\text{nC}}^2}{4.00\text{cm}}\right]\\ =9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[\frac{-400{\text{nC}}^2}{8.00\text{cm}}+\frac{200{\text{nC}}^2}{4.00\text{cm}}+\frac{-200{\text{nC}}^2}{4.00\text{cm}}\right]\\ =9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[\frac{\left(-400{\text{nC}}^2\right){\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)}^2}{\left(8.00\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right]\\ =9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[\frac{-4\times {10}^{-16}{\text{C}}^2}{0.08\text{m}}\right]\end{array}$

Further solve the above equation.

    $\begin{array}{c}U=-\left(45\times {10}^{-6}\text{N}\cdot \text{m}\right)\left(\frac{1\text{J}}{1\text{N}\cdot \text{m}}\right)\\ =-45\times {10}^{-6}\text{J}\left(\frac{1\mu \text{J}}{{10}^{-6}\text{J}}\right)\\ =-45\mu \text{J}\end{array}$

Therefore, the potential energy of the configuration is $-45\mu \text{J}$.

(b)

To determine

The speed of the fourth particle, after it has moved freely to a very large distance away from its point.

Answer to Problem 17P

The particle’s speed is $34.6\text{km}/\text{s}$.

Explanation of Solution

Calculate the distance between the charges, $q_1$ and $q_4$ and the distance between the charges, $q_2$ and $q_4$ in Figure-(1).

Write the expression for Pythagoras theorem.

    $r=\sqrt{r_1{}^2+r_2{}^2}$                                                                                                            (II)

Here, $r$ is the hypotenuse side length, $r_1$ is the adjacent side length and $r_2$ is the opposite side length.

Write the expression for net potential due to the three charges at point $P$.

    $V=k_{\text{e}}\left[\frac{q_1}{r_{14}}+\frac{q_2}{r_{24}}+\frac{q_3}{r_{34}}\right]$                                                                                            (III)

Here, $V$ is the net potential due to the three charges at point $P$.

Write expression when kinetic energy is converted into potential energy in the equilibrium position.

    $\frac{1}{2}\times m\times v^2=q_4\times V_{\text{P}}$

Here, $v$ is the particle’s speed, $m$ is the particle’s mass, $q_4$ is the particle’s charge and $V_{\text{P}}$ is the particle’s potential at the point $P$.

Rewrite the above equation for $v$.

    $\begin{array}{c}\frac{1}{2}\times m\times v^2=q_4\times V_{\text{P}}\\ v^2=\frac{q_4\times V_{\text{P}}}{\frac{1}{2}\times m}\\ v=\sqrt{\frac{q_4\times V_{\text{P}}}{\frac{1}{2}\times m}}\\ v=\sqrt{\frac{2\times q_4\times V_{\text{P}}}{m}}\end{array}$                                                                                         (IV)

Conclusion:

Consider triangle $q_1$$q_3$$q_4$ and triangle $q_2$$q_3$$q_4$. in Figure (1).

Substitute $4.00\text{cm}$ for $r_2$ and $3.00\text{cm}$ for $r_1$ in Equation (II) to calculate $r_{14}$ and $r_{24}$.

    $\begin{array}{c}{r}_{14}=r_{24}\\ =\sqrt{{\left(4.00\text{cm}\right)}^2+{\left(3.00\text{cm}\right)}^2}\\ =\sqrt{\left(16+9\right){\text{cm}}^2}\\ =\sqrt{25{\text{cm}}^2}\end{array}$

Further solve the above equation.

  $\begin{array}{c}{r}_{14}=r_{24}\\ =5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.05\text{m}\end{array}$

Substitute $9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $20.0\text{nC}$ for $q_1$, $-20.0\text{nC}$ for $q_2$, $10.0\text{nC}$ for $q_3$, $0.05\text{m}$ for $r_{14}$ and $r_{24}$, $3.0\text{cm}$ for $r_{34}$ in Equation (III) to calculate $V_{\text{P}}$.

    $\begin{array}{c}{V}_{\text{P}}=9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(\frac{20.0\text{nC}}{0.05\text{m}}+\frac{\left(-20.0\text{nC}\right)}{0.05\text{m}}+\frac{10.0\text{nC}}{3\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\right)\\ =9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(\frac{20}{0.05\text{m}}+\frac{\left(-20\right)}{0.05\text{m}}+\frac{10}{0.03\text{m}}\right)\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\\ =9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(333.33\right)\times {10}^{-9}\text{C}/\text{m}\\ =3\times {10}^3\text{V}\end{array}$

Substitute $40.0\times {10}^{-9}\text{C}$ for $q_4$, $2.00\times {10}^{-13}\text{kg}$ for $m$ and $3\times {10}^3\text{V}$ for $V_{\text{P}}$ in equation (IV) to calculate $v$.

    $\begin{array}{c}v=\sqrt{\frac{2\left(40.0\times {10}^{-9}\text{C}\right)\times \left(3\times {10}^3\text{V}\right)}{2.00\times {10}^{-13}\text{kg}}}\\ =\sqrt{\frac{240\times {10}^7\text{CV}}{2\text{kg}}}\\ =\sqrt{120\times {10}^7\text{CV}/\text{kg}}\\ =34.6\times {10}^3\text{m}/\text{s}\left(\frac{1\text{km}/\text{s}}{{10}^3\text{m}/\text{s}}\right)\end{array}$

Further solve the above equation.

    $v=34.6\text{km}/\text{s}$

Therefore, the particle’ speed is $34.6\text{km}/\text{s}$.