Chapter 26, Problem 26.50P

A small rigid object carries positive and negative 3 .50-nC charges. It is oriented so that the positive charge has coordinates (–1.20 mm, 1.10 mm) and the negative charge is at the point (1.40 mm. –1.30 nun), (a) Find the electric dipole moment of the object. The object is placed in an electric field $\stackrel{\to }{E}$ = (7.80 × 10³ $\hat{\text{i}}$  – 4.90 × 10³ $\hat{\text{j}}$ ) N/C. (b) Find the torque acting on the object, (c) Find the potential energy of the object-field system when the object is in this orientation, (d) Assuming the orientation of the object can change. find the difference between the maximum and minimum potential energies of the system.

To determine

The electric dipole moments of the objects.

Answer to Problem 26.50P

The electric dipole moments of the objects is $\left(-9.1\widehat{\text{i}}+8.4\widehat{\text{j}}\right)\times {10}^{-12}\text{C-m}$.

Explanation of Solution

Given info: The electric field is $\left(7.80\times {10}^3\widehat{\text{i}}-4.90\times {10}^3\widehat{\text{j}}\right)\text{N}/\text{C}$ and the magnitude of the charge is $3.50\text{nC}$. The co ordinate of the positive charge is $\left(-1.2\text{mm},1.10\text{mm}\right)$ and the coordinate of the negative charge is $\left(1.40\text{mm},-1.30\text{mm}\right)$.

Write the expression for the distance between the negative and positive charge particle.

$\overrightarrow{r}=\left(x_2-x_1\right)\widehat{\text{i}}+\left(y_2-y_1\right)\widehat{\text{j}}$

Here,

$x_1$ is the initial co ordinate on the $x$ axis.

$x_2$ is the final co ordinate on the $x$ axis.

$y_1$ is the initial co ordinate on the $y$ axis.

$y_2$ is the final co ordinate on the $y$ axis.

Substitute $-1.2\text{mm}$ for $x_2$, $1.10\text{mm}$ for $y_2$, $1.40\text{mm}$ for $x_2$ and $-1.30\text{mm}$ for $y_1$ in the above expression for the distance between the point.

$\begin{array}{c}\overrightarrow{r}=\left(-1.2\text{mm}-1.40\text{mm}\right)\widehat{\text{i}}+\left(1.10\text{mm}-\left(-1.30\text{mm}\right)\right)\widehat{\text{j}}\\ =\left(-2.6\text{mm}\right)\widehat{\text{i}}+\left(2.4\text{mm}\right)\widehat{\text{j}}\end{array}$

Thus, the distance between the point is $\left(-2.6\text{mm}\right)\widehat{\text{i}}+\left(2.4\text{mm}\right)\widehat{\text{j}}$.

Write the expression for the electric dipole moments.

$\overrightarrow{P}=q\cdot \overrightarrow{r}$

Here,

$q$ is the charge on the each plates.

$\overrightarrow{r}$ is the displacement between the positive and negative charge.

Substitute $3.50\text{nC}$ for $q$ and $\left(-2.6\text{mm}\right)\widehat{\text{i}}+\left(2.4\text{mm}\right)\widehat{\text{j}}$ for $\overrightarrow{r}$ in the above expression for the values of the dipole moments.

$\begin{array}{c}\overrightarrow{P}=\left(3.50\text{nC}\right)\cdot \left(\left(-2.6\text{mm}\right)\widehat{\text{i}}+\left(2.4\text{mm}\right)\widehat{\text{j}}\right)\\ =\left(3.50\text{nC}\times \frac{1\text{C}}{{10}^{-9}\text{nC}}\right)\cdot \left(\left(-2.6\right)\widehat{\text{i}}+\left(2.4\right)\widehat{\text{j}}\right)\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\\ =\left(\left(-9.1\right)\widehat{\text{i}}+\left(8.4\right)\widehat{\text{j}}\right)\times {10}^{-12}\text{C-m}\end{array}$

Conclusion:

Therefore, the electric dipole moments of the objects is $\left(\left(-9.1\right)\widehat{\text{i}}+\left(8.4\right)\widehat{\text{j}}\right)\times {10}^{-12}\text{C-m}$.

To determine

The amount of torque acting on the objects.

Answer to Problem 26.50P

The amount of torque acting on the objects is $-2.09\times {10}^{-8}\widehat{\text{k}}\text{N-m}$.

Explanation of Solution

Given info: The electric field is $\left(7.80\times {10}^3\widehat{\text{i}}-4.90\times {10}^3\widehat{\text{j}}\right)\text{N}/\text{C}$ and the magnitude of the charge is $3.50\text{nC}$. The co ordinate of the positive charge is $\left(-1.2\text{mm},1.10\text{mm}\right)$ and the coordinate of the negative charge is $\left(1.40\text{mm},-1.30\text{mm}\right)$.

From the part (a) the dipole moment for the charge carries rigid body.

$\overrightarrow{P}=\left(\left(-9.1\right)\widehat{\text{i}}+\left(8.4\right)\widehat{\text{j}}\right)\times {10}^{-12}\text{C-m}$

Thus, the dipole moment of the charge carries rigid body is $\left(\left(-9.1\right)\widehat{\text{i}}+\left(8.4\right)\widehat{\text{j}}\right)\times {10}^{-12}\text{C-m}$.

Write the expression for the torque acting on the object.

$\overrightarrow{\tau }=\overrightarrow{P}\times \overrightarrow{E}$

Here,

$\overrightarrow{P}$ is the dipole moment.

$\overrightarrow{E}$ is the electric field.

Substitute $\left(\left(-9.1\right)\widehat{\text{i}}+\left(8.4\right)\widehat{\text{j}}\right)\times {10}^{-12}\text{C-m}$ for $\overrightarrow{P}$ and $\left(7.80\times {10}^3\widehat{\text{i}}-4.90\times {10}^3\widehat{\text{j}}\right)\text{N}/\text{C}$ for $\overrightarrow{E}$ in the above expression for the torque on the body.

$\begin{array}{c}\overrightarrow{\tau }=\left(\left(-9.1\right)\widehat{\text{i}}+\left(8.4\right)\widehat{\text{j}}\right)\times {10}^{-12}\text{C-m}\times \left(7.80\times {10}^3\widehat{\text{i}}-4.90\times {10}^3\widehat{\text{j}}\right)\text{N}/\text{C}\\ =\left(\left(-9.1\right)\widehat{\text{i}}+\left(8.4\right)\widehat{\text{j}}\right)\text{C-m}\times \left(7.80\widehat{\text{i}}-4.90\widehat{\text{j}}\right)\text{N}/\text{C}\times {10}^{-9}\\ =\left(-20.9\times {10}^{-9}\widehat{\text{k}}\right)\text{N-m}\\ \text{=}-2.09\times {10}^{-8}\widehat{\text{k}}\text{N-m}\end{array}$

Conclusion:

Therefore, the amount of torque acting on the objects is $-2.09\times {10}^{-8}\widehat{\text{k}}\text{N-m}$.

To determine

The potential energy of the object field system.

Answer to Problem 26.50P

The potential energy of the object field system is $0.112\text{μJ}$.

Explanation of Solution

Given info: The electric field is $\left(7.80\times {10}^3\widehat{\text{i}}-4.90\times {10}^3\widehat{\text{j}}\right)\text{N}/\text{C}$ and the magnitude of the charge is $3.50\text{nC}$. The co ordinate of the positive charge is $\left(-1.2\text{mm},1.10\text{mm}\right)$ and the coordinate of the negative charge is $\left(1.40\text{mm},-1.30\text{mm}\right)$.

Write the expression for the potential energy of the object field system.

$U=-\overrightarrow{P}\cdot \overrightarrow{E}$

Substitute $\left(\left(-9.1\right)\widehat{\text{i}}+\left(8.4\right)\widehat{\text{j}}\right)\times {10}^{-12}\text{C-m}$ for $\overrightarrow{P}$ and $\left(7.80\times {10}^3\widehat{\text{i}}-4.90\times {10}^3\widehat{\text{j}}\right)\text{N}/\text{C}$ for $\overrightarrow{E}$ in the above expression for the potential energy of the object field system.

$\begin{array}{c}U=-\left(\left(-9.1\right)\widehat{\text{i}}+\left(8.4\right)\widehat{\text{j}}\right)\times {10}^{-12}\text{C-m}\cdot \left(7.80\times {10}^3\widehat{\text{i}}-4.90\times {10}^3\widehat{\text{j}}\right)\text{N}/\text{C}\\ =-\left[\left(-9.1\times 7.80\times {10}^{-12}\times {10}^3\right)-\left(8.4\times {10}^{-12}\times 4.90\times {10}^3\right)\right]\text{J}\\ =0.112\times {10}^{-6}\text{J}\times \frac{{10}^6\text{μJ}}{1\text{J}}\\ =\text{0}\text{.112}\text{μJ}\end{array}$

Conclusion:

Therefore, the potential energy of the object field system is $0.112\text{μJ}$.

To determine

The difference between the maximum and minimum potential energy of the system.

Answer to Problem 26.50P

The difference between the maximum and minimum potential energy of the system is $228\text{nJ}$.

Explanation of Solution

Given info: The electric field is $\left(7.80\times {10}^3\widehat{\text{i}}-4.90\times {10}^3\widehat{\text{j}}\right)\text{N}/\text{C}$ and the magnitude of the charge is $3.50\text{nC}$. The co ordinate of the positive charge is $\left(-1.2\text{mm},1.10\text{mm}\right)$ and the coordinate of the negative charge is $\left(1.40\text{mm},-1.30\text{mm}\right)$.

Write the expression for the magnitude of the electric field.

$\left|\overrightarrow{E}\right|=\sqrt{E_{\text{x}}^2+E_{\text{y}}^2}$

Here,

$E_{\text{x}}$ is the component of the electric field along the $x$ axis.

$E_{\text{y}}$ is the, magnitude of the electric field along the $y$ axis.

Substitute $7.80\times {10}^3$ for $E_{\text{x}}$ and $-4.90\times {10}^3$ for $E_{\text{y}}$ in the above expression for the magnitude of the electric field.

$\begin{array}{c}\left|\overrightarrow{E}\right|=\sqrt{{\left(7.80\times {10}^3\right)}^2+{\left(-4.90\times {10}^3\right)}^2}\text{N}/\text{C}\\ =9.21\times {10}^3\text{N}/\text{C}\end{array}$

Write the expression for the magnitude of the electric dipole.

$\left|\overrightarrow{P}\right|=\sqrt{P_{\text{x}}^2+P_{\text{y}}^2}$

Here,

$P_{\text{x}}$ is the component of the electric dipole along the $x$ axis.

$P_{\text{y}}$ is the, magnitude of the electric dipole along the $y$ axis.

Substitute $\left(-9.1\right)\times {10}^{-12}$ for $P_{\text{x}}$ and $\left(8.4\right)\times {10}^{-12}$ for $P_{\text{y}}$ in the above expression for the magnitude of the electric field.

$\begin{array}{c}\left|\overrightarrow{P}\right|=\sqrt{{\left(\left(-9.1\right)\times {10}^{-12}\right)}^2+{\left(\left(8.4\right)\times {10}^{-12}\right)}^2}\text{C-m}\\ =12.4\times {10}^{-12}\text{C-m}\end{array}$

Write the expression for the maximum potential energy of the object field system.

${\left|U\right|}_{\max }=\left|\overrightarrow{P}\right|\cdot \left|\overrightarrow{E}\right|$

Substitute $12.4\times {10}^{-12}\text{C-m}$ for $\overrightarrow{\left|P\right|}$ and $9.21\times {10}^3\text{N}/\text{C}$ for $\overrightarrow{\left|E\right|}$ in the above expression for the maximum potential energy of the object field system.

$\begin{array}{c}{\left|U\right|}_{\max }=12.4\times {10}^{-12}\text{C-m}\times 9.21\times {10}^3\text{N}/\text{C}\\ =114\times {10}^{-9}\text{J}\times \frac{{10}^9\text{nJ}}{1\text{J}}\\ =114\text{nJ}\end{array}$

Thus, the maximum potential energy is $114\text{nJ}$.

Write the expression for the minimum potential energy of the object field system.

${\left|U\right|}_{\min }=-\left|\overrightarrow{P}\right|\cdot \left|\overrightarrow{E}\right|$

Substitute $12.4\times {10}^{-12}\text{C-m}$ for $\overrightarrow{\left|P\right|}$ and $9.21\times {10}^3\text{N}/\text{C}$ for $\overrightarrow{\left|E\right|}$ in the above expression for the maximum potential energy of the object field system.

$\begin{array}{c}{\left|U\right|}_{\min }=-12.4\times {10}^{-12}\text{C-m}\times 9.21\times {10}^3\text{N}/\text{C}\\ =-114\times {10}^{-9}\text{J}\times \frac{{10}^9\text{nJ}}{1\text{J}}\\ =-114\text{nJ}\end{array}$

Thus, the minimum potential energy is $-114\text{nJ}$.

Write the expression for the change in the maximum and minimum potential energy of the electric field.

$\Delta U={\left|U\right|}_{\max }-{\left|U\right|}_{\min }$

Substitute $-114\text{nJ}$ for ${\left|U\right|}_{\min }$ and $114\text{nJ}$ for ${\left|U\right|}_{\max }$ in the above expression for the change in the potential energy.

$\begin{array}{c}\Delta U=114\text{nJ}-\left(-114\text{nJ}\right)\\ =228\text{nJ}\end{array}$

Conclusion:

Therefore, the difference between the maximum and minimum potential energy of the system is $228\text{nJ}$.