Chapter 17, Problem 35P

(a)

To determine

The electric force on the particle moving from $b$ to $a$ in a straight line.

Answer to Problem 35P

The electric force on the particle is $1.0\text{μN}$ directed to the right.

Explanation of Solution

The magnitude of the electric field is $240{\text{NC}}^{-1}$ directed to the right and the charge of the particle is $+4.2\text{nC}$.

Write the expression for electric force

    $\overrightarrow{F}=q\overrightarrow{E}$                                                                                                        (I)

Here, $\overrightarrow{F}$ is the electric force, $q$ is the charge of the particle and $\overrightarrow{E}$ is the electric field. The direction of electric force is same as the direction of the electric field.

Substitute $240{\text{NC}}^{-1}$ for $\overrightarrow{E}$ and $+4.2\text{ nC}$ for $q$ in (I) to find $\overrightarrow{F}$

  $\begin{array}{c}\overrightarrow{F}=\left(+4.2\text{ nC}\right)\left(240{\text{NC}}^{-1}\right)\\ =\left(+4.2\times {10}^{-9}\text{ C}\right)\left(240{\text{NC}}^{-1}\right)\\ =1.0\times {10}^{-6}\text{N}\\ =1.0\text{μN}\end{array}$

Thus, the electric force is $1.0\text{μN}$ directed to the right.

(b)

To determine

The potential difference along a straight line from $b$ to $a$.

Answer to Problem 35P

The potential difference is $60\text{V}$.

Explanation of Solution

Electric field is the negative gradient of potential and hence points in the direction of decreasing potential.

The potential at point $b$ is lesser than the potential at point $a$.

Write the expression for potential difference if the particle moves in a straight line from $b$ to $a$

    $\Delta V=V_a-V_b=Ed$                                                                                            (II)

Here,$\Delta V$ is the potential difference, $V_a$ is the potential at point $a$, $V_b$ is the potential at point $b$, $E$ is the magnitude of electric field and $d$ is the distance between point $a$ and $b$.

Substitute $240{\text{NC}}^{-1}$ for $E$ and $0.25\text{m}$ for $d$ in (II) to find $\Delta V$.

  $\begin{array}{c}\Delta V=\left(240{\text{NC}}^{-1}\right)\left(0.25\text{m}\right)\\ =60\text{V}\end{array}$

Thus, the potential difference is $60\text{V}$.

(c)

To determine

The change in electric potential energy of the particle moving from $b$ to $a$ in a straight line.

Answer to Problem 35P

The change in electric potential energy of the particle is $0.25\text{μJ}$.

Explanation of Solution

The charge being moved is $+4.2\text{nC}$.

Write the expression for change in electric potential energy moving from $b$ to $a$ in a straight line

  $\Delta U=q\Delta V$                                                                                                  (III)

Here, $\Delta U$ is the change in electric potential energy, $q$ is the charge being moved from the point $a$ to $b$.

Substitute  $+4.2\text{nC}$ for $q$ and $60\text{V}$ for $\Delta V$ in (III) to find $\Delta U$

    $\begin{array}{c}\Delta U=+4.2\text{nC}\left[60\text{V}\right]\\ =+4.2\times {10}^{-9}\text{C}\left(60\text{V}\right)\\ =0.25\times {10}^{-6}\text{J}\\ =0.25\text{μJ}\end{array}$

Thus, the change in electric potential energy is $0.25\text{μJ}$.

(d)

To determine

The work done on the particle by the electric field.

Answer to Problem 35P

The work done by the electric field is $-0.25\text{μJ}$.

Explanation of Solution

Write the expression for work done by the electric field in terms of force and displacement

  $W=Fd\cos \theta$                                                                                              (IV)

Here, $W$ is the work done by the electric field, $F$ is the magnitude of the electric force, $d$ is the distance between the points $a$ and $b$ whereas $\theta$ is the angle between the force and the displacement vector.

Since force and the displacement vector is along the opposite direction in the given path, $\theta =180°$

Substitute $1.0\text{μN}$ for $F$ and $0.25\text{m}$ for $d$ in (IV) to find $W$

  $\begin{array}{c}W=1.0\text{μN}\left(0.25\text{m}\right)\cos 180°\\ =0.25\times {10}^{-6}{\text{Nm}}^{-1}\times \left(-1\right)\\ =-0.25\text{μJ}\end{array}$

Write the expression for work done by the electric field in moving a charge

  $W=-\Delta U$                                                                                                       (V)

Here, $W$ is the work done by the electric field and $\Delta U$ is the change in electric potential in moving the charge from $a$ to $b$.

Substitute $0.25\text{μJ}$ for $\Delta U$ in (V) to find $W$

    $\begin{array}{c}W=-\left(0.25\text{μJ}\right)\\ =-0.25\text{μJ}\end{array}$

The result using (IV) and (V) are the same.

Thus, the work done by the electric field is $-0.25\text{μJ}$.