Chapter 23, Problem 63PQ

(a)

To determine

The net-electrostatic force on the third particle with charge $+2.0\mu \text{C}$.

Answer to Problem 63PQ

The net-electrostatic force on the third particle with charge $+2.0\mu \text{C}$ is $\text{8}\text{.79}\text{N}$.

Explanation of Solution

The net-electrostatic force on the particle with charge $+2.0\mu \text{C}$ will be the sum of the force by charge 1 on 3 and the force by charge 2 on 3.

Write the expression for Coulomb’s law.

  $\overrightarrow{F}=\frac{k{q}_1{q}_2}{r^2}\widehat{r}$

Here, $F$ is the electrostatic force between the particles, $k$ is the Coulomb’s constant, $q_1$ is the charge of particle 1, $q_2$ is the charge of particle 2, $r$ is the distance between the centers of the particles.

The force due to 1 on 3 is in the $x$ direction repulsive and the force due to 2 on 3 is the $-x$ axis attractive.

Substitute $5.0\text{cm}$ for $r$ and $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $4.0\mu \text{C}$ for $q_1$ and $2.0\mu \text{C}$ for $q_2$ to find the electrostatic force by charge 1 on 3.

  $\begin{array}{c}{F}_{13}=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(4.0\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)\left(2.0\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}{{\left(5.0\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(4.0\times {10}^{-6}\text{C}\right)\left(2.0\times {10}^{-6}\text{C}\right)}{{\left(5.0\times {10}^{-2}\text{m}\right)}^2}\\ =28.77\text{N}\end{array}$

Substitute $5.0\text{cm}-\text{2}\text{.0}\text{cm}$ for $r$ and $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $1.0\mu \text{C}$ for $q_1$ and $2.0\mu \text{C}$ for $q_2$ to find the electrostatic force by charge 2 on 3.

  $\begin{array}{c}{F}_{23}=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(1.0\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)\left(2.0\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}{{\left(\left(5.0\text{cm}-\text{2}\text{.0}\text{cm}\right)\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(1.0\times {10}^{-6}\text{C}\right)\left(2.0\times {10}^{-6}\text{C}\right)}{{\left(3.0\times {10}^{-2}\text{m}\right)}^2}\\ =19.98\text{N}\end{array}$

Write the expression to find the net-electrostatic force on the third particle with charge $+2.0\mu \text{C}$.

  $F=F_{13}-F_{23}$

Here, ${\overrightarrow{F}}_{13}$ is the electrostatic force by charge 1 on 2, ${\overrightarrow{F}}_{23}$ is the electrostatic force by charge 2 on 3.

Conclusion:

Substitute $28.77\text{N}$ for ${\overrightarrow{F}}_{13}$ and $19.98\text{N}$ for ${\overrightarrow{F}}_{23}$ to find the net-electrostatic force on the third particle with charge $+2.0\mu \text{C}$.

  $\begin{array}{c}F=28.77\text{N}-19.98\text{N}\\ \text{=8}\text{.79}\text{N}\end{array}$

Therefore, the net-electrostatic force on the third particle with charge $+2.0\mu \text{C}$ is $\text{8}\text{.79}\text{N}$.

(b)

To determine

Compere the force on $+2.0\mu \text{C}$ by $+3.0\mu \text{C}$ with part (a).

Answer to Problem 63PQ

The force on $+2.0\mu \text{C}$ by $+3.0\mu \text{C}$ is $21.6\widehat{i}\text{N}$.

Explanation of Solution

Write the expression for Coulomb’s law.

  $\overrightarrow{F}=\frac{k{q}_1{q}_2}{r^2}\widehat{r}$

Here, $F$ is the electrostatic force between the particles, $k$ is the Coulomb’s constant, $q_1$ is the charge of particle 1, $q_2$ is the charge of particle 2, $r$ is the distance between the centers of the particles.

Conclusion:

Substitute $5.0\text{cm}$ for $r$ and $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $3.0\mu \text{C}$ for $q_1$ and $2.0\mu \text{C}$ for $q_2$ to find the force on $+2.0\mu \text{C}$ by $+3.0\mu \text{C}$

  $\begin{array}{c}{F}_{13}=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(3.0\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)\left(2.0\mu \text{C}\frac{{10}^{-6}\text{C}}{1.0\mu \text{C}}\right)}{{\left(5.0\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2}\widehat{i}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(3.0\times {10}^{-6}\text{C}\right)\left(2.0\times {10}^{-6}\text{C}\right)}{{\left(5.0\times {10}^{-2}\text{m}\right)}^2}\widehat{i}\\ =21.6\widehat{i}\text{N}\end{array}$

Therefore, the force on $+2.0\mu \text{C}$ by $+3.0\mu \text{C}$ is $21.6\widehat{i}\text{N}$.

In the case, the separation and the distance to the 3^rd charge is comparable. The first two charges cannot be replaced by an effective charge at origin.