21.43 • in a rectangular coordinate system a positive point charge q = 6.00×10^(-9) c is placed at the point…

21.43 • in a rectangular coordinate system a positive point charge q = 6.00×10^(-9) c is placed at the point x = +0.150 m, y = 0, and an identical point charge is placed at x = -0.150 m, y = 0. find the x - and y - components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) x = 0.300 m, y = 0; (c) x = 0.150 m, y = -0.400 m; (d) x = 0, y = 0.200 m.

21.43 • in a rectangular coordinate system a positive point charge q = 6.00×10^(-9) c is placed at the point x = +0.150 m, y = 0, and an identical point charge is placed at x = -0.150 m, y = 0. find the x - and y - components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) x = 0.300 m, y = 0; (c) x = 0.150 m, y = -0.400 m; (d) x = 0, y = 0.200 m.

Answer

Explanation:

Step1: Recall electric - field formula

The electric - field due to a point charge $q$ at a distance $r$ is given by $\vec{E}=\frac{kq}{r^{2}}\hat{r}$, where $k = 9.0\times10^{9}\ N\cdot m^{2}/C^{2}$.

Step2: Analyze case (a) at the origin

The two charges $q_1 = 6.00\times10^{-9}\ C$ at $x = 0.150\ m$ and $q_2 = 6.00\times10^{-9}\ C$ at $x=-0.150\ m$ create electric - fields. The electric - field due to $q_1$ at the origin is $\vec{E}1=\frac{kq_1}{r_1^{2}}\hat{i}$, and due to $q_2$ is $\vec{E}2=\frac{kq_2}{r_2^{2}}(-\hat{i})$. Since $r_1 = r_2=0.150\ m$ and $q_1 = q_2$, the $x$ - components of the electric - fields due to the two charges cancel out. $E{x}=0$, $E{y}=0$. Magnitude $E = 0$, direction is undefined.

Step3: Analyze case (b) at $x = 0.300\ m,y = 0$

The electric - field due to $q_1$ (at $x = 0.150\ m$) is $\vec{E}1=\frac{kq_1}{(0.300 - 0.150)^{2}}\hat{i}=\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.150)^{2}}\hat{i}=2400\ N/C\hat{i}$. The electric - field due to $q_2$ (at $x=-0.150\ m$) is $\vec{E}2=\frac{kq_2}{(0.300+0.150)^{2}}\hat{i}=\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.450)^{2}}\hat{i}\approx266.7\ N/C\hat{i}$. $E{x}=E_1 + E_2=(2400 + 266.7)\ N/C=2666.7\ N/C$, $E{y}=0$. Magnitude $E = 2666.7\ N/C$, direction is along the positive $x$ - axis.

Step4: Analyze case (c) at $x = 0.150\ m,y=-0.400\ m$

The distance from $q_1$ to the point is $r_1 = 0.400\ m$. The electric - field due to $q_1$ is $\vec{E}1=\frac{kq_1}{r_1^{2}}\hat{j}=\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.400)^{2}}\hat{j}=337.5\ N/C\hat{j}$. The distance from $q_2$ to the point is $r_2=\sqrt{(0.150 + 0.150)^{2}+(0.400)^{2}}=\sqrt{(0.300)^{2}+(0.400)^{2}} = 0.500\ m$. The $x$ - component of the electric - field due to $q_2$ is $E{2x}=-\frac{kq_2}{r_2^{2}}\cos\theta\hat{i}$, where $\cos\theta=\frac{0.300}{0.500}$. $E_{2x}=-\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.500)^{2}}\times\frac{0.300}{0.500}\hat{i}=-129.6\ N/C\hat{i}$. The $y$ - component of the electric - field due to $q_2$ is $E_{2y}=-\frac{kq_2}{r_2^{2}}\sin\theta\hat{j}$, where $\sin\theta=\frac{0.400}{0.500}$. $E_{2y}=-\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.500)^{2}}\times\frac{0.400}{0.500}\hat{j}=-172.8\ N/C\hat{j}$. $E_{x}=-129.6\ N/C$, $E_{y}=337.5-172.8 = 164.7\ N/C$. Magnitude $E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-129.6)^{2}+(164.7)^{2}}\approx210.7\ N/C$. $\tan\theta=\frac{E_{y}}{E_{x}}=\frac{164.7}{-129.6}$, $\theta\approx128.7^{\circ}$ counter - clockwise from the negative $x$ - axis.

Step5: Analyze case (d) at $x = 0,y = 0.200\ m$

The distance from $q_1$ to the point is $r_1=\sqrt{(0.150)^{2}+(0.200)^{2}} = 0.250\ m$. The $x$ - component of the electric - field due to $q_1$ is $E_{1x}=-\frac{kq_1}{r_1^{2}}\cos\theta_1\hat{i}$, where $\cos\theta_1=\frac{0.150}{0.250}$. $E_{1x}=-\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.250)^{2}}\times\frac{0.150}{0.250}\hat{i}=-1296\ N/C\hat{i}$. The $y$ - component of the electric - field due to $q_1$ is $E_{1y}=\frac{kq_1}{r_1^{2}}\sin\theta_1\hat{j}$, where $\sin\theta_1=\frac{0.200}{0.250}$. $E_{1y}=\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.250)^{2}}\times\frac{0.200}{0.250}\hat{j}=1728\ N/C\hat{j}$. The distance from $q_2$ to the point is also $r_2 = 0.250\ m$. The $x$ - component of the electric - field due to $q_2$ is $E_{2x}=\frac{kq_2}{r_2^{2}}\cos\theta_2\hat{i}$, where $\cos\theta_2=\frac{0.150}{0.250}$. $E_{2x}=\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.250)^{2}}\times\frac{0.150}{0.250}\hat{i}=1296\ N/C\hat{i}$. The $y$ - component of the electric - field due to $q_2$ is $E_{2y}=\frac{kq_2}{r_2^{2}}\sin\theta_2\hat{j}$, where $\sin\theta_2=\frac{0.200}{0.250}$. $E_{2y}=1728\ N/C\hat{j}$. $E_{x}=E_{2x}+E_{1x}=0$, $E_{y}=E_{1y}+E_{2y}=3456\ N/C$. Magnitude $E = 3456\ N/C$, direction is along the positive $y$ - axis.

Answer:

(a) $E_{x}=0$, $E_{y}=0$, $E = 0$, direction undefined. (b) $E_{x}=2666.7\ N/C$, $E_{y}=0$, $E = 2666.7\ N/C$, direction along positive $x$ - axis. (c) $E_{x}=-129.6\ N/C$, $E_{y}=164.7\ N/C$, $E\approx210.7\ N/C$, $\theta\approx128.7^{\circ}$ counter - clockwise from negative $x$ - axis. (d) $E_{x}=0$, $E_{y}=3456\ N/C$, $E = 3456\ N/C$, direction along positive $y$ - axis.