Chapter 16, Problem 110P

(a)

To determine

The net electric force acting on the dipole.

Answer to Problem 110P

The net force is $0\text{N}$.

Explanation of Solution

The charges of dipole are $\pm 3.0\text{μC}$, separation between charges is $7.0\text{cm}$, electric field is $2.0\times {10}^4\text{N}/\text{C}$, mass of charge is $5.0\text{g}$, and the mass of rod is $20.0\text{g}$.

Write the equation to find the x-component of net force.

$\sum F_x=q{E}_x-q{E}_x$

Here, $\sum F_x$ is the x-component of net force, $q$ is the magnitude of charge, and $E_x$ is the x-component of electric field.

The first terms on the right hand side of above equation denotes x-component of force due to positive charge and the second terms denotes the x-component of force due to negative charge on the dipole.

Rewrite the above relation.

$\sum F_x=0\text{N}$

Write the equation to find the y-component of net force.

$\sum F_y=q{E}_y-q{E}_y$

Here, $\sum F_y$ is the y-component of net force and $E_y$ is the y-component of electric field.

The first terms on the right hand side of above equation denotes y-component of force due to positive charge and the second terms denotes the y-component of force due to negative charge on the dipole.

Rewrite the above relation.

$\sum F_y=0\text{N}$

Write the equation for net electric force.

$F_{net}=\sqrt{{\left(\sum F_x\right)}^2+{\left(\sum F_y\right)}^2}$

Here, $F_{net}$ is the net electric force.

Conclusion:

Substitute $0\text{N}$ for $\sum F_x$ and $\sum F_y$ in the above equation to find $F_{net}$.

$\begin{array}{c}{F}_{net}=\sqrt{{\left(0\text{N}\right)}^2+{\left(0\text{N}\right)}^2}\\ =0\text{N}\end{array}$

Therefore, the net force is $0\text{N}$.

(b)

To determine

Prove that the magnitude of torque on the dipole is $\tau =qEd\sin \theta$.

Answer to Problem 110P

It is proved that the magnitude of torque on the dipole is $\tau =qEd\sin \theta$.

Explanation of Solution

The charges of dipole are $\pm 3.0\text{μC}$, separation between charges is $7.0\text{cm}$, electric field is $2.0\times {10}^4\text{N}/\text{C}$, mass of charge is $5.0\text{g}$, and the mass of rod is $20.0\text{g}$.

Write the equation for net torque.

$\sum \tau =F_{-q}{r}_{-q}\sin \theta +F_q{r}_q\sin \theta$

Here, $\sum \tau$ is the net torque, $F_{-q}$ is the force due to negative charge, $r_{-q}$ is the moment arm of negative charge (half of the separation between the charges), $F_q$ is the force due to positive charge, $r_q$ is the moment arm of negative charge (half of the separation between the charges), and $\theta$ is the angle made by dipole with the electric field.

Write the equation for $F_{-q}$.

$F_{-q}=qE$

Write the equation for $F_q$.

$F_q=qE$

Write the equation for $r_q$.

$r_q=\frac{d}{2}$

Here, $d$ is the separation between charges.

Write the equation for $r_{-q}$.

$r_{-q}=\frac{d}{2}$

Rewrite the equation for $\sum \tau$ by substituting the above relations for $r_{-q}$.

$\begin{array}{c}\sum \tau =qE\left(\frac{d}{2}\right)\sin \theta +qE\left(\frac{d}{2}\right)\sin \theta \\ =qEd\sin \theta \end{array}$ (I)

Conclusion:

Therefore, it is proved that the magnitude of torque on the dipole is $\tau =qEd\sin \theta$.

(c)

To determine

Torque acting on the dipoles at angles $\theta =0°,36.9°,90.0°$

Answer to Problem 110P

Torques at angles $\theta =0°,36.9°,90.0°$ are $0\text{N}\cdot \text{m, }0.0025\text{N}\cdot \text{m},\text{and}0.0042\text{N}\cdot \text{m}$ respectively.

Explanation of Solution

The charges of dipole are $\pm 3.0\text{μC}$, separation between charges is $7.0\text{cm}$, electric field is $2.0\times {10}^4\text{N}/\text{C}$, mass of charge is $5.0\text{g}$, and the mass of rod is $20.0\text{g}$.

Conclusion:

Substitute $3.0\text{μC}$ for $q$, $2.0\times {10}^4\text{N}/\text{C}$ for $E$, $7.0\text{cm}$ for $d$, and $0°$ for $\theta$ in equation (I) to find $\sum \tau$.

$\begin{array}{c}\sum \tau =\left(3.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(2.0\times {10}^4\text{N}/\text{C}\right)\left(7.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\left(\sin 0°\right)\\ =\left(42\times {10}^{-4}\text{N}\cdot \text{m}\right)\left(0\right)\\ =0\text{N}\cdot \text{m}\end{array}$

Substitute $3.0\text{μC}$ for $q$, $2.0\times {10}^4\text{N}/\text{C}$ for $E$, $7.0\text{cm}$ for $d$, and $90.0°$ for $\theta$ in equation (I) to find $\sum \tau$.

Substitute $3.0\text{μC}$ for $q$, $2.0\times {10}^4\text{N}/\text{C}$ for $E$, $7.0\text{cm}$ for $d$, and $90.0°$ for $\theta$ in equation (I) to find $\sum \tau$.

$\begin{array}{c}\sum \tau =\left(3.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(2.0\times {10}^4\text{N}/\text{C}\right)\left(7.0\text{cm}\left(\frac{{10}^{-2}\text{cm}}{1\text{cm}}\right)\right)\left(\sin 90.0°\right)\\ =\left(42\times {10}^{-4}\text{N}\cdot \text{m}\right)\left(1.0\right)\\ =0.0042\text{N}\cdot \text{m}\end{array}$

Therefore, torques at angles $\theta =0°,36.9°,90.0°$ are $0\text{N}\cdot \text{m, }0.0025\text{N}\cdot \text{m},\text{and}0.0042\text{N}\cdot \text{m}$ respectively.