Chapter 20, Problem 18P

(a)

To determine

The location of the particle and the charges on them.

Answer to Problem 18P

The location on first particle is $\underset{\_}{7.00\text{cm at 70}\text{.0}°}$ and the charge on the particle is $\underset{\_}{7.00\text{nC}}$.

The location on second particle is $\underset{\_}{3.00\text{cm at 90}\text{.0}°}$ and the charge on the particle is $\underset{\_}{8.00\text{nC}}$.

Explanation of Solution

Write the expression of standard equation for the potential difference.

  $V=\frac{kq}{r}$        (I)

Here, $V$ is the potential difference for distance, $k$ is Coulomb’s constant, $q$ is charge and $r$ is distance.

Write the expression of standard equation for the electric filed.

  $E=\frac{kq}{r^2}$        (II)

Here, $E$ is the electric field.

Write the expression for the given potential difference.

$V_{given}=\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left[\left(\frac{7.00\times {10}^{-9}\text{C}}{\left(0.070\text{m}\right)}\right)-\left(\frac{8.00\times {10}^{-9}\text{C}}{\left(0.030\text{m}\right)}\right)\right]$        (III)

Here, $V_{given}$ is the given potential difference.

Write the expression for the given electric field.

$E_{given}=\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left[\begin{array}{l}-\left(\frac{7.00\times {10}^{-9}\text{C}}{{\left(0.070\text{m}\right)}^2}\cos \left(70.0°\right)\right)\widehat{i}-\\ \left(\frac{7.00\times {10}^{-9}\text{C}}{{\left(0.070\text{m}\right)}^2}\sin \left(70.0°\right)\right)\widehat{j}+\\ \left(\frac{8.00\times {10}^{-9}\text{C}}{\left(0.030\text{m}\right)}\right)\widehat{j}\end{array}\right]$        (IV)

Here, $E_{given}$ is the given electric field.

Conclusion:

Comparing equation (II) and (IV), the first particle is having charge with magnitude

$\begin{array}{c}{Q}_1=7.00\times {10}^{-6}\text{C}\\ =\left(7.00\times {10}^{-6}\text{C}\right)\left(\frac{1\text{μC}}{1\times {10}^{-6}\text{C}}\right)\\ \text{=7}\mu \text{C}\end{array}$

The location of the first particle is

$\begin{array}{c}{r}_1=0.070\text{m}\\ =\left(0.070\text{m}\right)\left(\frac{1\text{cm}}{1\times {10}^{-2}\text{m}}\right)\\ =7\text{cm}\end{array}$

Thus, The location on first particle is $\underset{\_}{7.00\text{cm at 70}\text{.0}°}$ and the charge on the particle is $\underset{\_}{7.00\text{nC}}$.

Comparing equation (II) and (IV), the second particle is having charge with magnitude

$\begin{array}{c}{Q}_2=8.00\times {10}^{-6}\text{C}\\ =\left(8.00\times {10}^{-6}\text{C}\right)\left(\frac{1\text{μC}}{1\times {10}^{-6}\text{C}}\right)\\ \text{=8}\mu \text{C}\end{array}$

The location of the second particle is

$\begin{array}{c}{r}_2=0.030\text{m}\\ =\left(0.030\text{m}\right)\left(\frac{1\text{cm}}{1\times {10}^{-2}\text{m}}\right)\\ =3\text{cm}\end{array}$

Thus, the location on second particle is $\underset{\_}{3.00\text{cm at 90}\text{.0}°}$ and the charge on the particle is $\underset{\_}{8.00\text{nC}}$.

(b)

To determine

The force acting on charge -16.0 nC.

Answer to Problem 18P

The force acting on charge -16.0 nC is $\underset{\_}{\left(7.03\widehat{i}-109\widehat{j}\right)\times {10}^{-5}\text{N}}$.

Explanation of Solution

Rewrite the expression for the given electric field from equation (IV).

${\overrightarrow{E}}_{given}=\left(-4.39\times {10}^3\text{N/C}\right)\widehat{i}+\left(67.8\times {10}^3\text{N/C}\right)\widehat{j}$        (V)

Write the expression for the force action on the charge.

$\overrightarrow{F}=q{\overrightarrow{E}}_{given}$        (VI)

Here, $F$ is the force, $q$ is charge.

Conclusion:

Substitute $\left(-4.39\times {10}^3\text{N/C}\right)\widehat{i}+\left(67.8\times {10}^3\text{N/C}\right)\widehat{j}$ for ${\overrightarrow{E}}_{given}$, $\left(-16\text{nC}\right)$ for $q$ in Equation (VI) to find $\overrightarrow{F}$.

$\begin{array}{c}\overrightarrow{F}=\left[\left(-16\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]\left[\left(-4.39\times {10}^3\text{N/C}\right)\widehat{i}+\left(67.8\times {10}^3\text{N/C}\right)\widehat{j}\right]\\ =\left(-16\times {10}^{-9}\text{C}\right)\left[\left(-4.39\times {10}^3\text{N/C}\right)\widehat{i}+\left(67.8\times {10}^3\text{N/C}\right)\widehat{j}\right]\\ =\left(7.03\widehat{i}-109\widehat{j}\right)\times {10}^{-5}\text{N}\end{array}$

Thus, the force acting on charge -16.0 nC is $\underset{\_}{\left(7.03\widehat{i}-109\widehat{j}\right)\times {10}^{-5}\text{N}}$.

(c)

To determine

The work done required to move the charge.

Answer to Problem 18P

The work done required to move the charge is $\underset{\_}{2.40\times {10}^{-5}\text{J}}$.

Explanation of Solution

Rewrite the expression for the given potential difference by using equation (III).

$\begin{array}{c}{V}_{given}=\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left[\left(\frac{7.00\times {10}^{-9}\text{C}}{\left(0.070\text{m}\right)}\right)-\left(\frac{8.00\times {10}^{-9}\text{C}}{\left(0.030\text{m}\right)}\right)\right]\\ =-1.50\times {10}^3\text{V}\end{array}$        (VII)

Write the expression for the required work done.

$W=qV$        (VIII)

Here, $W$ is the work done and $q$ is the charge.

Conclusion:

Substitute $-\left(1.50\times {10}^3\right)\text{V}$ for $V_{given}$, $\left(-16\text{nC}\right)$ for $q$ in Equation (VIII) to find $W$.

  $\begin{array}{c}W=\left[\left(-16\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]\left(-1.50\times {10}^3\right)\text{V}\\ =2.40\times {10}^{-5}\text{J}\end{array}$

Thus, the work done required to move the charge is $\underset{\_}{2.40\times {10}^{-5}\text{J}}$.