Question

electric field due to multiple point charges
two point charges are placed on the x axis. (figure 1)the
first charge, ( q_1 = 8.00 , \text{nc} ), is placed a distance 16.0 m
from the origin along the positive x axis, the second
charge, ( q_2 = 6.00 , \text{nc} ), is placed a distance 9.00 m from
the origin along the negative x axis.
figure
(shows coordinate system: ( q_2 ) at (-9 m, 0 m), ( q_1 ) at (+16 m, 0 m), point a at (0 m, 12 m), point b at (0 m, 15 m))
part a
calculate the electric field at point a, located at coordinates (0 m, 12.0 m).
give the x and y components of the electric field as an ordered pair. express your answer in newtons per coulomb to three significant
figures.
view available hint(s)
( e_{ax}, e_{ay} = ) n/c
submit
part b complete previous part(s)

Explanation:

Step1: Recall the electric field formula

The electric field due to a point charge \( q \) at a distance \( r \) is given by \( \vec{E} = k\frac{q}{r^2}\hat{r} \), where \( k = 8.988\times 10^{9}\ N\cdot m^2/C^2 \), and \( \hat{r} \) is the unit vector from the charge to the point.

Step2: Find the distance from \( q_1 \) to point A

For \( q_1 = 8.00\ nC = 8.00\times 10^{-9}\ C \) at \( (16.0\ m, 0) \), point A is at \( (0, 12.0\ m) \). The distance \( r_1 \) is \( \sqrt{(16.0 - 0)^2+(0 - 12.0)^2}=\sqrt{256 + 144}=\sqrt{400}=20.0\ m \). The unit vector \( \hat{r}_{1A} \) has components \( \frac{-16.0}{20.0}\hat{i}+\frac{12.0}{20.0}\hat{j}=-0.8\hat{i}+ 0.6\hat{j} \). The electric field due to \( q_1 \) at A is \( E_{1A}=k\frac{q_1}{r_1^2}\hat{r}_{1A} \). Substituting values: \( E_{1A}=8.988\times 10^{9}\times\frac{8.00\times 10^{-9}}{20.0^2}(-0.8\hat{i}+ 0.6\hat{j}) \). Calculate the magnitude factor: \( \frac{8.988\times 8.00\times 10^{0}}{400}= \frac{71.904}{400}=0.17976\ N/C \). So \( E_{1A}=0.17976(-0.8\hat{i}+ 0.6\hat{j})=(-0.1438\hat{i}+ 0.1079\hat{j})\ N/C \).

Step3: Find the distance from \( q_2 \) to point A

For \( q_2 = 6.00\ nC = 6.00\times 10^{-9}\ C \) at \( (-9.00\ m, 0) \), the distance \( r_2 \) is \( \sqrt{( - 9.00-0)^2+(0 - 12.0)^2}=\sqrt{81 + 144}=\sqrt{225}=15.0\ m \). The unit vector \( \hat{r}_{2A} \) has components \( \frac{9.00}{15.0}\hat{i}+\frac{12.0}{15.0}\hat{j}=0.6\hat{i}+ 0.8\hat{j} \). The electric field due to \( q_2 \) at A is \( E_{2A}=k\frac{q_2}{r_2^2}\hat{r}_{2A} \). Substituting values: \( E_{2A}=8.988\times 10^{9}\times\frac{6.00\times 10^{-9}}{15.0^2}(0.6\hat{i}+ 0.8\hat{j}) \). Calculate the magnitude factor: \( \frac{8.988\times 6.00\times 10^{0}}{225}=\frac{53.928}{225}=0.23968\ N/C \). So \( E_{2A}=0.23968(0.6\hat{i}+ 0.8\hat{j})=(0.1438\hat{i}+ 0.1917\hat{j})\ N/C \).

Step4: Find the total electric field components

The total \( x \)-component \( E_{Ax}=E_{1Ax}+E_{2Ax}=- 0.1438 + 0.1438 = 0\ N/C \). The total \( y \)-component \( E_{Ay}=E_{1Ay}+E_{2Ay}=0.1079+0.1917 = 0.2996\ N/C\approx0.300\ N/C \).

Answer:

\( 0.000, 0.300 \) (or in ordered pair format \( (0.000, 0.300) \) N/C)