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
1 of 1
(shows coordinates: b (0 m, 15 m), a (0 m, 12 m), o (0,0), ( q_2 ) at (-9 m, 0), ( q_1 ) at (16 m, 0))
part b
an unknown additional charge ( q_3 ) is now placed at point b, located at coordinates (0 m, 15.0 m ). find the magnitude and sign of ( q_3 ) needed to make the total electric field at point a equal to zero.
express your answer in nanocoulombs to three significant figures.
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( q_3 = ) box nc
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Explanation:

Step1: Find distance from \( q_1 \) to \( A \)

Using Pythagoras: \( r_{1A} = \sqrt{(16.0)^2 + (12.0)^2} = 20.0 \, \text{m} \)

Step2: Find distance from \( q_2 \) to \( A \)

Using Pythagoras: \( r_{2A} = \sqrt{(9.00)^2 + (12.0)^2} = 15.0 \, \text{m} \)

Step3: Find distance from \( q_3 \) to \( A \)

\( r_{3A} = 15.0 - 12.0 = 3.00 \, \text{m} \)

Step4: Electric field due to \( q_1 \) at \( A \)

\( E_1 = k\frac{|q_1|}{r_{1A}^2} \), direction: angle \( \theta_1 \) with y - axis, \( \cos\theta_1=\frac{12}{20}=0.6 \)
\( E_{1y}=E_1\cos\theta_1 = k\frac{8.00\times10^{-9}}{20^2}\times0.6 \)

Step5: Electric field due to \( q_2 \) at \( A \)

\( E_2 = k\frac{|q_2|}{r_{2A}^2} \), direction: angle \( \theta_2 \) with y - axis, \( \cos\theta_2=\frac{12}{15}=0.8 \)
\( E_{2y}=E_2\cos\theta_2 = k\frac{6.00\times10^{-9}}{15^2}\times0.8 \)

Step6: Total electric field from \( q_1 \) and \( q_2 \) at \( A \) (y - component)

\( E_{total - y}=E_{1y}+E_{2y} \)
\( E_{1y}=9\times10^{9}\times\frac{8\times10^{-9}}{400}\times0.6 = 9\times8\times0.6\times10^{-9 + 9}/400=43.2/400 = 0.108 \, \text{N/C} \)
\( E_{2y}=9\times10^{9}\times\frac{6\times10^{-9}}{225}\times0.8 = 9\times6\times0.8\times10^{-9+9}/225 = 43.2/225 = 0.192 \, \text{N/C} \)
\( E_{total - y}=0.108 + 0.192=0.3 \, \text{N/C} \) (upward)

Step7: Electric field due to \( q_3 \) at \( A \)

To cancel \( E_{total - y} \), \( E_3 = E_{total - y} \) (downward), so \( q_3 \) must be negative.
\( E_3 = k\frac{|q_3|}{r_{3A}^2}\Rightarrow|q_3|=\frac{E_3r_{3A}^2}{k} \)
\( |q_3|=\frac{0.3\times(3)^2}{9\times10^{9}}=\frac{0.3\times9}{9\times10^{9}} = 3\times10^{-10}\, \text{C}=0.300\times10^{-9}\, \text{C} = 0.300 \, \text{nC} \) (negative)

Answer:

\( q_3=\boxed{- 0.300} \) nC