Question

(2) a point charge $q_1 = 6.00 nc$ is at the point $x = 0.600 m$, $y = 0.800 m$, and a second point charge $q_2=-4.00 nc$ is at the point $x = 1.600 m$, $y = 0$. calculate the magnitude and direction of the net electric - field at the origin due to these two point charges.

Explanation:

Step1: Recall electric - field formula

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

Step2: Calculate the electric - field due to $q_1$

For $q_1 = 6.00\ nC=6.00\times 10^{-9}\ C$, $r_1=\sqrt{(0.600)^{2}+(0.800)^{2}}=1.00\ m$.
The $x$ - component of $\vec{E}_1$ is $E_{1x}=\frac{kq_1}{r_1^{2}}\frac{x_1}{r_1}$, and the $y$ - component is $E_{1y}=\frac{kq_1}{r_1^{2}}\frac{y_1}{r_1}$.
$x_1 = 0.600\ m$, $y_1 = 0.800\ m$.
$E_{1x}=\frac{9\times 10^{9}\times6.00\times 10^{-9}}{(1.00)^{2}}\times\frac{0.600}{1.00}=32.4\ N/C$
$E_{1y}=\frac{9\times 10^{9}\times6.00\times 10^{-9}}{(1.00)^{2}}\times\frac{0.800}{1.00}=43.2\ N/C$

Step3: Calculate the electric - field due to $q_2$

For $q_2=-4.00\ nC = - 4.00\times 10^{-9}\ C$, $r_2 = 1.600\ m$.
The electric - field due to $q_2$ is $\vec{E}_2=\frac{kq_2}{r_2^{2}}\hat{r}_2$. Since $q_2$ is on the $x$ - axis, $E_{2x}=\frac{kq_2}{r_2^{2}}=\frac{9\times 10^{9}\times(-4.00\times 10^{-9})}{(1.600)^{2}}=-14.0625\ N/C$, and $E_{2y} = 0$.

Step4: Calculate the net $x$ and $y$ components of the electric - field

$E_x=E_{1x}+E_{2x}=32.4-14.0625 = 18.3375\ N/C$
$E_y=E_{1y}+E_{2y}=43.2\ N/C$

Step5: Calculate the magnitude of the net electric - field

$E=\sqrt{E_x^{2}+E_y^{2}}=\sqrt{(18.3375)^{2}+(43.2)^{2}}\approx46.9\ N/C$

Step6: Calculate the direction of the net electric - field

$\theta=\tan^{-1}(\frac{E_y}{E_x})=\tan^{-1}(\frac{43.2}{18.3375})\approx67.0^{\circ}$ above the $x$ - axis.

Answer:

Magnitude: $46.9\ N/C$, Direction: $67.0^{\circ}$ above the $x$ - axis.