Chapter 26, Problem 50P

(a)

To determine

The electric dipole moment of the object.

Answer to Problem 50P

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

Explanation of Solution

Write the expression to obtain the distance between the positive and negative charges.

    $\overrightarrow{d}=\left[\left(x_1-x_2\right)\widehat{i}+\left(y_1-y_2\right)\widehat{j}\right]$                                                                               (I)

Here, $\overrightarrow{d}$ is the between the positive and negative charges, $\left(x_1,y_1\right)$ is the coordinate of positive charge particle and $\left(x_2,y_2\right)$ is the coordinate of negative charge particle.

Write the expression to obtain the electric dipole moment.

    ${\overrightarrow{m}}_{\text{d}}=\left(q\right)\left(\overrightarrow{d}\right)$                                                                                                       (II)

Here, ${\overrightarrow{m}}_{\text{d}}$ is the electric dipole moment, $q$ is the charge of electric dipole and $\overrightarrow{d}$ is the between the positive and negative charges, $\left(x_1,y_1\right)$ is the coordinate of positive

Conclusion:

Substitute $-1.20\text{mm}$ for $x_1$, $1.10\text{mm}$ for $y_1$, $1.40\text{mm}$ for $x_2$ and $-1.30\text{mm}$ for $y_2$ in equation (I) to calculate $\overrightarrow{d}$.

    $\begin{array}{c}\overrightarrow{d}=\left[\left(-1.20\text{mm}-1.40\text{mm}\right)\widehat{i}+\left(1.10\text{mm+}1.30\text{mm}\right)\widehat{j}\right]\\ =\left[\left(-2.60\right)\widehat{i}+\left(2.40\right)\widehat{j}\right]\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\\ =\left[\left(-2.60\right)\widehat{i}+\left(2.40\right)\widehat{j}\right]\times {10}^{-3}\text{m}\end{array}$

Substitute $\left[\left(-2.60\right)\widehat{i}+\left(2.40\right)\widehat{j}\right]\times {10}^{-3}\text{m}$ for $\overrightarrow{d}$ and $3.50\text{nC}$ for $q$ in equation (II) to calculate $\overrightarrow{m}$.

    $\begin{array}{c}{\overrightarrow{m}}_{\text{d}}=\left(3.50\text{nC}\right)\left(\left[\left(-2.60\right)\widehat{i}+\left(2.40\right)\widehat{j}\right]\times {10}^{-3}\text{m}\right)\\ =\left(3.50\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)\left(\left[\left(-2.60\right)\widehat{i}+\left(2.40\right)\widehat{j}\right]\times {10}^{-3}\text{m}\right)\\ =\left(-9.1\widehat{i}+8.4\widehat{j}\right)\times {10}^{-12}\text{C-m}\end{array}$

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

(b)

To determine

The torque acting on the object.

Answer to Problem 50P

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

Explanation of Solution

Write the expression to obtain the torque acting on the object.

    $\overrightarrow{\tau }=\left(\overrightarrow{m}\right)\times \left(\overrightarrow{E}\right)$

Here, $\overrightarrow{\tau }$ is the torque acting on the object, $\overrightarrow{m}$ is the dipole moment acting on the object and $\overrightarrow{E}$ is the electric field.

Conclusion:

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

    $\begin{array}{c}\overrightarrow{\tau }=\left(\left(-9.1\widehat{i}+8.4\widehat{j}\right)\times {10}^{-12}\text{C-m}\right)\times \left(\left(7.80\times {10}^3\widehat{i}-4.90\times {10}^3\widehat{j}\right)\text{N}/\text{C}\right)\\ =\left(44590\widehat{k}-65520\widehat{k}\right)\times {10}^{-12}\text{N-m}\\ =-2.09\times {10}^{-8}\widehat{k}\text{N-m}\end{array}$

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

(c)

To determine

The potential energy of the object field system.

Answer to Problem 50P

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

Explanation of Solution

Write the expression to obtain the potential energy of the object field system.

    $U=-\left(m\right).\left(E\right)$

Here, $U$ is the potential energy of the object field system, $m$ is the dipole moment acting on the object and $E$ is the electric field.

Conclusion:

Substitute $\left(-9.1\widehat{i}+8.4\widehat{j}\right)\times {10}^{-12}\text{C-m}$ for $m$ and $\left(7.80\times {10}^3\widehat{i}-4.90\times {10}^3\widehat{j}\right)\text{N}/\text{C}$ for $E$ in the above equation to calculate $U$.

    $\begin{array}{c}U=\left(\left(-9.1\widehat{i}+8.4\widehat{j}\right)\times {10}^{-12}\text{C-m}\right).\left(\left(7.80\times {10}^3\widehat{i}-4.90\times {10}^3\widehat{j}\right)\text{N}/\text{C}\right)\\ =\left(-9.1\times 7800-8.4\times 4900\right)\times {10}^{-12}\text{J}\\ \text{=1}\text{.1214}\times {\text{10}}^{-7}\text{J}\\ \text{=1}\text{.1214}\times {\text{10}}^{-7}\text{J}\times \frac{{10}^6\mu \text{J}}{1\text{J}}\end{array}$

Further solve the above equation.

    $U=0.112\mu \text{J}$

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

(d)

To determine

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

Answer to Problem 50P

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

Explanation of Solution

Write the expression to obtain the maximum potential energy.

    $U_{\max }=\left|\overrightarrow{m}\right|\left|\overrightarrow{E}\right|$                                                                                                        (III)

Here, $U_{\max }$ is the maximum potential energy of the object field system, $\overrightarrow{m}$ is the dipole moment acting on the object and $\overrightarrow{E}$ is the electric field.

Write the expression to obtain the minimum potential energy.

    $U_{\min }=-\left|\overrightarrow{m}\right|\left|\overrightarrow{E}\right|$                                                                                                       (IV)

Here, $U_{\min }$ is the minimum potential energy of the object field system, $\overrightarrow{m}$ is the dipole moment acting on the object and $\overrightarrow{E}$ is the electric field.

Write the expression to obtain the difference in the maximum and the minimum potential energy of the system.

    $\Delta U=U_{\max }-U_{\min }$                                                                                                     (V)

Here, $\Delta U$ is the difference in the maximum and the minimum potential energy of the system.

Conclusion:

Substitute $12.4\times {10}^{-12}\text{C-m}$ for $\left|\overrightarrow{m}\right|$ and $9.21\times {10}^3\text{N}/\text{C}$ for $\left|\overrightarrow{E}\right|$ in equation (III) to calculate $U_{\max }$

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

Further solve the above equation.

    $U_{\max }=114\text{nJ}$

Substitute $12.4\times {10}^{-12}\text{C-m}$ for $\left|\overrightarrow{m}\right|$ and $9.21\times {10}^3\text{N}/\text{C}$ for $\left|\overrightarrow{E}\right|$ in equation (IV) to calculate $U_{\min }$

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

Further solve the above equation.

    $U_{\min }=-114\text{nJ}$

Substitute $114\text{nJ}$ for $U_{\max }$ and $-114\text{nJ}$ for $U_{\min }$ in equation (V) to calculate $\Delta U$.

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

Therefore, the difference in the maximum and the minimum potential energy of the system is

$228\text{nJ}$.