Question

21.43 • in a rectangular coordinate system a positive point charge $q = 6.00\times10^{-9} c$ 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) $x = 0$, $y = 0.200 m$.

Explanation:

Step1: Recall electric - field formula

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

Step2: Analyze case (a) at the origin

The two charges $q_1 = 6.00\times10^{-9}\ C$ at $x = 0.150\ m,y = 0$ and $q_2 = 6.00\times10^{-9}\ C$ at $x=-0.150\ m,y = 0$. The electric - field due to $q_1$ at the origin is $\vec{E}_1=\frac{kq_1}{r_1^{2}}\hat{r}_1$, with $r_1 = 0.150\ m$ and $\hat{r}_1=-\hat{i}$. The electric - field due to $q_2$ at the origin is $\vec{E}_2=\frac{kq_2}{r_2^{2}}\hat{r}_2$, with $r_2 = 0.150\ m$ and $\hat{r}_2=\hat{i}$.
$E_{1x}=-\frac{kq}{r^{2}}$, $E_{2x}=\frac{kq}{r^{2}}$, so $E_x=E_{1x}+E_{2x}=-\frac{kq}{r^{2}}+\frac{kq}{r^{2}} = 0$. $E_y = 0$. The magnitude $E=\sqrt{E_x^{2}+E_y^{2}}=0$.

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

The electric - field due to $q_1$ at this point: $r_1=0.300 - 0.150=0.150\ m$, $\vec{E}_1=\frac{kq}{r_1^{2}}\hat{i}$. The electric - field due to $q_2$ at this point: $r_2=0.300+0.150 = 0.450\ m$, $\vec{E}_2=-\frac{kq}{r_2^{2}}\hat{i}$.
$E_{1x}=\frac{kq}{r_1^{2}}=\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.150)^{2}}=2400\ N/C$.
$E_{2x}=-\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.450)^{2}}=-266.67\ N/C$.
$E_x=E_{1x}+E_{2x}=2400 - 266.67 = 2133.33\ N/C$, $E_y = 0$. The magnitude $E=\sqrt{E_x^{2}+E_y^{2}}=2133.33\ N/C$.

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

The distance from $q_1$ to the point is $r_1 = 0.400\ m$, and the distance from $q_2$ to the point is $r_2=\sqrt{(0.150 + 0.150)^{2}+(0.400)^{2}}=\sqrt{0.09 + 0.16}=0.500\ m$.
The electric - field due to $q_1$: $\vec{E}_1=\frac{kq}{r_1^{2}}\hat{j}$, where $\hat{j}$ is the unit vector in the $y$ - direction. $\vec{E}_1=\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.400)^{2}}\hat{j}=337.5\hat{j}\ N/C$.
The electric - field due to $q_2$: $\vec{E}_2=\frac{kq}{r_2^{2}}\hat{r}_2$, where $\hat{r}_2$ has $x$ - component $\frac{0.300}{0.500}$ and $y$ - component $\frac{-0.400}{0.500}$.
$E_{2x}=\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.500)^{2}}\times\frac{0.300}{0.500}=648\times0.6 = 388.8\ N/C$.
$E_{2y}=\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.500)^{2}}\times\frac{-0.400}{0.500}=648\times(-0.8)=-518.4\ N/C$.
$E_x=E_{2x}=388.8\ N/C$, $E_y=E_{1y}+E_{2y}=337.5-518.4=-180.9\ N/C$.
The magnitude $E=\sqrt{E_x^{2}+E_y^{2}}=\sqrt{(388.8)^{2}+(-180.9)^{2}}\approx428.9\ N/C$.
The direction $\theta=\arctan(\frac{E_y}{E_x})=\arctan(\frac{-180.9}{388.8})\approx - 24.9^{\circ}$ (below the $x$ - axis).

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

The distance from $q_1$ and $q_2$ to the point is $r=\sqrt{(0.150)^{2}+(0.200)^{2}}=0.250\ m$.
The $x$ - components of the electric - fields due to $q_1$ and $q_2$ cancel out because of symmetry.
The $y$ - component of the electric - field due to $q_1$ is $E_{1y}=\frac{kq}{r^{2}}\times\frac{0.200}{0.250}$, and the $y$ - component of the electric - field due to $q_2$ is $E_{2y}=\frac{kq}{r^{2}}\times\frac{0.200}{0.250}$.
$E_y = 2\times\frac{9.0\times10^{9}\times6.00\times10^{-9}}{(0.250)^{2}}\times\frac{0.200}{0.250}=345.6\ N/C$, $E_x = 0$.
The magnitude $E=\sqrt{E_x^{2}+E_y^{2}}=345.6\ N/C$. The direction is along the positive $y$ - axis.

Answer:

(a) $E_x = 0\ N/C$, $E_y = 0\ N/C$, $E = 0\ N/C$, no direction.
(b) $E_x = 2133.33\ N/C$, $E_y = 0\ N/C$, $E = 2133.33\ N/C$, along the positive $x$ - axis.
(c) $E_x = 388.8\ N/C$, $E_y=-180.9\ N/C$, $E\approx428.9\ N/C$, $\theta\approx - 24.9^{\circ}$ (below the $x$ - axis).
(d) $E_x = 0\ N/C$, $E_y = 345.6\ N/C$, $E = 345.6\ N/C$, along the positive $y$ - axis.