Chapter 23, Problem 31P

(a)

To determine

The electric field at point $P$.

Answer to Problem 31P

The electric field at point $P$ is $1.8\times {10}^4\text{N}/\text{C}$ towards the right along x-axis.

Explanation of Solution

The net electric field at point $P$ due to other three charges is as shown below.

Figure-(1)

Write the expression for the electric field.

    $E=\frac{k_{\text{e}}q}{r^2}$                                                                                                                      (I)

Here, $E$ is the electric field, $k_{\text{e}}$ is the electric constant, $q$ is the charge and $r$ is the distance between the charge and location.

Write the expression for the electric field due to the bottom $3.00\text{nC}$ charge which is at a distance of $4.00\text{cm}$ from point $P$.

    ${\overrightarrow{E}}_{\text{1}}=\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\cos 30°\widehat{i}+\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\sin 30°\widehat{j}$                                                   (II)

Here, ${\overrightarrow{E}}_{\text{1}}$ is the electric field due to the bottom $3.00\text{nC}$ charge at point $P$, $k_{\text{e}}$ is a coulombs constant.

Write the expression for the electric field due to upper $3.00\text{nC}$ charge which is at a distance of $4.00\text{cm}$ from point $P$.

    ${\overrightarrow{E}}_2=\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\cos 30°\widehat{i}-\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\sin 30°\widehat{j}$                                                 (III)

Here, ${\overrightarrow{E}}_2$ is the electric field due to the bottom $3.00\text{nC}$ charge at point $P$, $k_{\text{e}}$ is a coulombs constant.

Write the expression for the electric field due to $-2.00\text{nC}$ charge which is at a distance of $4.00\text{cm}$ from point $P$.

    ${\overrightarrow{E}}_3=-\frac{k_{\text{e}}\left(-2.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\widehat{i}$                                                                                            (IV)

Here, ${\overrightarrow{E}}_3$ is the electric field due to the charge of $-2.00\text{nC}$ at point $P$, $k_{\text{e}}$ is a coulomb’s constant.

Write an expression to obtain the net electric field at point $P$.

    $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_3$                                                                                                      (V)

Here, $\overrightarrow{E}$ is the net electric field at point $P$.

Conclusion:

Substitute $\left[\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\cos 30°\widehat{i}+\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\sin 30°\widehat{j}\right]$ for ${\overrightarrow{E}}_1$, $\left[\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\cos 30°\widehat{i}-\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\sin 30°\widehat{j}\right]$ for ${\overrightarrow{E}}_2$ and $\left[-\frac{k_{\text{e}}\left(-2.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\widehat{i}\right]$ for ${\overrightarrow{E}}_3$ in equation (V) to calculate $\overrightarrow{E}$.

    $\begin{array}{c}\overrightarrow{E}=\left[\begin{array}{l}\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\cos 30°\widehat{i}+\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\sin 30°\widehat{j}\\ +\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\cos 30°\widehat{i}-\frac{k_{\text{e}}\left(3.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\sin 30°\widehat{j}-\frac{k_{\text{e}}\left(-2.00\text{nC}\right)}{{\left(4.00\text{cm}\right)}^2}\widehat{i}\end{array}\right]\\ =\left[\begin{array}{l}\frac{k_{\text{e}}\left(3.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\frac{\sqrt{3}}{2}\widehat{i}+\frac{k_{\text{e}}\left(3.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\frac{1}{2}\widehat{j}\\ +\frac{k_{\text{e}}\left(3.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\frac{\sqrt{3}}{2}\widehat{i}-\frac{k_{\text{e}}\left(3.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\frac{1}{2}\widehat{j}\\ -\frac{k_{\text{e}}\left(-2.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\widehat{i}\end{array}\right]\end{array}$

Substitute $8.99\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^2$ for $k_{\text{e}}$ in the above equation.

    $\begin{array}{c}\overrightarrow{E}=\left[\begin{array}{l}\frac{8.99\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^2\left(3.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\frac{\sqrt{3}}{2}\widehat{i}\\ +\frac{8.99\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^2\left(3.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\frac{1}{2}\widehat{j}\\ +\frac{8.99\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^2\left(3.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\frac{\sqrt{3}}{2}\widehat{i}\\ -\frac{8.99\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^2\left(3.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\frac{1}{2}\widehat{j}\\ -\frac{8.99\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^2\left(-2.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{{\left(4.00\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}^2}\widehat{i}\end{array}\right]\\ =\left[\begin{array}{c}14594\text{N}/\text{C}\widehat{i}+8425.8\text{N}/\text{C}\widehat{j}+14594\text{N}/\text{C}\widehat{i}-8425.8\text{N}/\text{C}\widehat{j}\\ -11234.5\text{N}/\text{C}\widehat{i}\end{array}\right]\\ =17953.5\text{N}/\text{C}\widehat{i}-0\text{N}/\text{C}\widehat{j}\\ \approx 1.8\times {10}^5\text{N}/\text{C}\widehat{i}\end{array}$

Therefore, the electric field at point $P$ is $1.8\times {10}^4\text{N}/\text{C}$ towards the right along x-axis.

(b)

To determine

The force at point $P$.

Answer to Problem 31P

The force at point $P$ is $8.98\times {10}^{-5}\text{N}$ in the left direction along x-axis.

Explanation of Solution

Write the expression to obtain the force at point $P$.

    $\overrightarrow{F}=q\overrightarrow{E}$

Here, $\overrightarrow{F}$ is the force at point $P$, $q$ is the charge at that point and $\overrightarrow{E}$ is the net electric field at point $P$.

Conclusion:

Substitute, $17953.5\text{N}/\text{C}\widehat{i}$ for $\overrightarrow{E}$ and $-5.00\text{nC}$ for $q$ in the above equation to calculate $\overrightarrow{F}$.

    $\begin{array}{c}\overrightarrow{F}=\left(-5.00\text{nC}\right)\times 17953.5\text{N}/\text{C}\widehat{i}\\ =\left(-5.00\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)\times 17953.5\text{N}/\text{C}\widehat{i}\\ =-8.98\times {10}^{-5}\text{N}\widehat{i}\end{array}$

Therefore, the force at point $P$ is $8.98\times {10}^{-5}\text{N}$ in the left direction along x-axis.