Chapter 22, Problem 35AP

Three charged particles are aligned along the x axis as shown in Figure P22.35. Find the electric field at (a) the position (2.00 m, 0) and (b) the position (0, 2.00 m).

Figure P22.35

To determine

The electric field at the position $\left(2.00\text{ m, 0}\right)$.

Answer to Problem 35AP

The electric field at the position $\left(2.00\text{ m, 0}\right)$. is $24.24\widehat{\text{i}}\text{ N}/\text{C}$

Explanation of Solution

Given Info: Three charges are acting on the same line along the $x$-axis.

According to Coulomb’s law, the electric field created by a charge $q$ is,

$\overrightarrow{E}=k_e\frac{q}{r^2}\widehat{r}$

Here,

$E$ is the electric field

$k_e$ is the Coulomb’s constant

$q$ is the charge

$r$ is the distance between two charges.

The electric field at a point due to number of charges is resultant of the electric field due to the individual charges.

The figure for the position of point charges is shown below.

Figure (1)

The electric field due to charge $q_1$ at $P$ can be given as,

$\overrightarrow{E_1}=k_e\frac{q_1}{r_1{}^2}{\widehat{r}}_1$

Here,

$E_1$ is the electric field due to charge $q_1$.

$r_1$ is the distance between the charge $q_1$ and point $P$.

Substitute $9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ for $k_e$, $-4.00\text{ nC}$ for $q_1$ and $2.50\text{ m}$ for $r_1$ in above expression.

$\begin{array}{c}\overrightarrow{E_1}=\frac{\left(9\times {10}^9{\text{Nm}}^2/{\text{C}}^2\right)\left(-4.00\text{ nC}\right)\left(\frac{1\text{C}}{{10}^9\text{ nC}}\right)}{{\left(2.50\text{ m}\right)}^2}\stackrel{\wedge }{i}\\ =-5.76\widehat{\text{i}}\text{N}/\text{C}\end{array}$

The electric field due to charge $q_2$ at $P$ can be given as,

$\overrightarrow{E_2}=k_e\frac{q_2}{r_2{}^2}{\widehat{r}}_2$

Here,

$E_2$ is the electric field due to charge $q_2$.

$r_2$ is the distance between the charge $q_2$ and point $P$.

Substitute $9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ for $k_e$, $5.00\text{ nC}$ for $q_2$ and $2.00\text{ m}$ for $r_2$ in above expression.

$\begin{array}{c}\overrightarrow{E_2}=\frac{\left(9\times {10}^9{\text{Nm}}^2/{\text{C}}^2\right)\left(5.00\text{ nC}\right)\left(\frac{1\text{C}}{{10}^9\text{ nC}}\right)}{{\left(2.00\text{ m}\right)}^2}\stackrel{\wedge }{i}\\ =11.25\widehat{\text{i}}\text{N}/\text{C}\end{array}$

The electric field due to charge $q_3$ at $P$ can be given as,

$\overrightarrow{E_3}=k_e\frac{q_3}{r_3{}^2}{\widehat{r}}_3$

Here,

$E_3$ is the electric field due to charge $q_3$.

$r_3$ is the distance between the charge $q_3$ and point $P$.

Substitute $9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ for $k_e$, $3.00\text{ nC}$ for $q_3$ and $1.20\text{ m}$ for $r_3$ in above expression.

$\begin{array}{c}\overrightarrow{E_3}=\frac{\left(9\times {10}^9{\text{Nm}}^2/{\text{C}}^2\right)\left(3.00\text{ nC}\right)\left(\frac{1\text{C}}{{10}^9\text{ nC}}\right)}{{\left(1.20\text{ m}\right)}^2}\widehat{\text{i}}\\ =18.75\widehat{\text{i}}\text{N}/\text{C}\end{array}$

The resultant electric field can be given as,

$\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E_3}$

Here,

$E$ is the resultant electric field

Substitute $-5.76\widehat{\text{i}}\text{N}/\text{C}$ for $\overrightarrow{E_1}$, $11.25\widehat{\text{i}}\text{N}/\text{C}$ for $\overrightarrow{E_2}$ and $18.75\widehat{\text{i}}\text{N}/\text{C}$ for $\overrightarrow{E_3}$.

$\begin{array}{c}\overrightarrow{E}=\left(-5.76+11.25+18.75\right)\widehat{\text{i}}\text{ N}/\text{C}\\ =24.24\widehat{\text{i}}\text{ N}/\text{C}\end{array}$

Thus, the electric field at position $\left(2.00\text{ m, 0}\right)$ is $24.24\widehat{\text{i}}\text{ N}/\text{C}$.

Conclusion:

Therefore, the electric field at the position $\left(2.00\text{ m, 0}\right)$. is $24.24\widehat{\text{i}}\text{ N}/\text{C}$

To determine

The electric field at the position $\left(0,2.00\text{ m}\right)$.

Answer to Problem 35AP

The electric field at the position $\left(0,2.00\text{ m}\right)$ is $\left(-4.21\right)\widehat{\text{i}}+\left(8.42\right)\widehat{\text{j}}\text{ N}/\text{C}$

Explanation of Solution

Given Info: Three charges are acting on the same line along the $x$-axis.

The figure for the position of point charges is shown below,

Figure (2)

According to Pythagoras theorem, the distance between the charge $q_1$ and position $Q\left(0,2.00\text{ m}\right)$ is,

$\begin{array}{c}{d}_1=\sqrt{{\left(0.50\text{ m}\right)}^2+{\left(2.00\text{ m}\right)}^2}\\ =2.06155\text{ m}\end{array}$

Here,

$d_1$ is the distance between the charge $q_1$ and point $Q$

According to Pythagoras theorem, the distance between the charge $q_3$ and position $Q\left(0,2.00\text{ m}\right)$ is,

$\begin{array}{c}{d}_3=\sqrt{{\left(0.80\text{ m}\right)}^2+{\left(2.00\text{ m}\right)}^2}\\ =2.154\text{ m}\end{array}$

Here,

$d_3$ is the distance between the charge $q_3$ and point $Q$.

The electric field due to charge $q_1$ at $Q$ can be given as,

$\overrightarrow{E_1}=k_e\frac{q_1}{d_1{}^2}{\widehat{d}}_1$

Here,

$E_1$ is the electric field due to charge $q_1$.

Substitute $9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ for $k_e$, $-4.00\text{ nC}$ for $q_1$ and $2.06155\text{ m}$ for $d_1$,

$\begin{array}{c}\overrightarrow{E_1}=\frac{\left(9\times {10}^9{\text{Nm}}^2/{\text{C}}^2\right)\left(-4.00\text{ nC}\right)\left(\frac{1\text{C}}{{10}^9\text{ nC}}\right)}{{\left(2.06155\text{ m}\right)}^2}\widehat{d}\\ =-8.47\widehat{d}\text{N}/\text{C}\end{array}$

In component form the electric field can be written as,

$\overrightarrow{E_1}=\left(-8.47\cos \theta \widehat{i}-8.47\sin \theta \widehat{\text{j}}\right)\text{N}/\text{C}$

According to right angle triangle property,

$\begin{array}{c}\cos \theta =\frac{0.50}{2.06}\\ =0.2427\end{array}$

And,

$\begin{array}{c}\sin \theta =\frac{2.00}{2.06}\\ =0.971\end{array}$

Substitute $0.2427$ for $\cos \theta$ and $0.971$ for $\sin \theta$ according to right angle triangle property to get $\overrightarrow{E_1}$,

$\begin{array}{c}\overrightarrow{E_1}=\left(-8.47\times 0.2427\right)\widehat{\text{i}}-\left(8.47\times 0.971\right)\widehat{\text{j}}\text{N}/\text{C}\\ =-2.05\widehat{\text{i}}-8.23\widehat{\text{j}}\text{N}/\text{C}\end{array}$

The electric field due to charge $q_2$ at $Q$ can be given as,

$\overrightarrow{E_2}=k_e\frac{q_2}{d_2{}^2}{\widehat{d}}_2$

Here,

$E_2$ is the electric field due to charge $q_2$.

$d_2$ is the distance between the charge $q_2$ and point $Q$

Substitute $9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ for $k_e$, $5.00\text{ nC}$ for $q_2$ and $2.00\text{ m}$ for $d_2$,

$\begin{array}{c}\overrightarrow{E_2}=\frac{\left(9\times {10}^9{\text{Nm}}^2/{\text{C}}^2\right)\left(5.00\text{ nC}\right)\left(\frac{1\text{C}}{{10}^9\text{ nC}}\right)}{{\left(2.00\text{ m}\right)}^2}\widehat{d}\\ =11.25\widehat{\text{j}}\text{N}/\text{C}\end{array}$

The electric field due to charge $q_3$ at $Q$ can be given as,

$\overrightarrow{E_3}=k_e\frac{q_3}{d_3{}^2}{\widehat{d}}_3$

Here,

$E_3$ is the electric field due to charge $q_3$.

Substitute $9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ for $k_e$, $3.00\text{ nC}$ for $q_3$ and $2.154\text{ m}$ for $d_3$,

$\begin{array}{c}\overrightarrow{E_3}=\frac{\left(9\times {10}^9{\text{Nm}}^2/{\text{C}}^2\right)\left(3.00\text{ nC}\right)\left(\frac{1\text{C}}{{10}^9\text{ nC}}\right)}{{\left(2.154\text{ m}\right)}^2}\widehat{d}\\ =5.819\widehat{d}\text{N}/\text{C}\end{array}$

In component form the electric field can be written as,

$\overrightarrow{E_3}=\left(-5.819\cos \varphi \right)\widehat{\text{i}}+\left(5.819\sin \varphi \right)\widehat{\text{j}}\text{N}/\text{C}$

According to right angle triangle property,

$\begin{array}{c}\cos \varphi =\frac{0.80}{2.154}\\ =0.371\end{array}$

And,

$\begin{array}{c}\sin \varphi =\frac{2.00}{2.154}\\ =0.928\end{array}$

Substitute $0.371$ for $\cos \varphi$ and $0.928$ for $\sin \varphi$ according to right angle triangle property to get $\overrightarrow{E_3}$,

$\begin{array}{c}\overrightarrow{E_3}=\left(-5.819\times 0.371\right)\widehat{\text{i}}+\left(5.819\times 0.928\right)\widehat{\text{j}}\text{N}/\text{C}\\ =-2.16\widehat{\text{i}}+5.40\widehat{\text{j}}\text{N}/\text{C}\end{array}$

The resultant electric field can be given as,

$\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E_3}$

Here,

$E$ is the resultant electric field

Substitute $-2.05\widehat{\text{i}}-8.23\widehat{\text{j}}\text{N}/\text{C}$ for $\overrightarrow{E_1}$, $11.25\widehat{\text{j}}\text{N}/\text{C}$ for $\overrightarrow{E_2}$ and $-2.16\widehat{\text{i}}+5.40\widehat{\text{j}}\text{N}/\text{C}$ for $\overrightarrow{E_3}$,

$\begin{array}{c}\overrightarrow{E}=\left(-2.05\widehat{\text{i}}-8.23\widehat{\text{j}}+11.25\widehat{\text{j}}-2.16\widehat{\text{i}}+5.40\widehat{\text{j}}\right)\text{ N}/\text{C}\\ =\left(-4.21\right)\widehat{\text{i}}+\left(8.42\right)\widehat{\text{j}}\text{ N}/\text{C}\end{array}$

Thus, the electric field at position $\left(0,\text{ 2}\text{.00 m}\right)$ is $\left(-4.21\right)\widehat{\text{i}}+\left(8.42\right)\widehat{\text{j}}\text{ N}/\text{C}$.

Conclusion:

Therefore, the electric field at position $\left(0,\text{ 2}\text{.00 m}\right)$ is $\left(-4.21\right)\widehat{\text{i}}+\left(8.42\right)\widehat{\text{j}}\text{ N}/\text{C}$