Question

in newtons per coulomb to three significant figures Two point charges are placed on the x axis. (Figure 1) The first charge, q 8.00 nC, is placed a distance 16.0 Im from the origin along the positive x axis; the second charge, g2 -6.00 nC, is placed a distance 9.00 m from the origin along the negative x axis Hints EA,EA0,0.300 N/C Submit My Answers Give Up Correct Figure 1 of 1 Part B An unknown additional charge q3 is now placed at point B, located at coordinates (0 m, 15.0 m) Find the magnitude and sign of q3 needed to make the total electric field at point A equal to zero B (0 m, 15 m) A (0 m, 12 m) Express your answer in nanocoulombs to three significant figures Hints 1 O (0 m, 0 m) 41 (-9m, 0 m) 16 m, 0m) Submit My Answers Give Up Provide Feedhack ContinuR

Answer 1

According to the figure let $\small \angle$Aq₁O = $\small \theta$₁ and $\small \angle$Aq₂O = $\small \theta$₂. Thus electric field along x direction is,

  $\small E_{x}=-E_{q_{1}}cos\theta _{1}+E_{q_{2}}cos\theta _{2}$

$\small \Rightarrow E_{x}=E_{q_{_{2}}}\left ( \frac{9}{\sqrt{12^{2}+9^{2}}} \right )-E_{q_{_{1}}}\left ( \frac{16}{\sqrt{12^{2}+16^{2}}} \right )$

$\small \Rightarrow E_{x}=\left ( \frac{k\times 6}{\left ( \sqrt{12^{2}+9^{2}} \right )^{2}} \right )\left ( \frac{9}{\sqrt{12^{2}+9^{2}}} \right )-\left ( \frac{k\times 8}{\left ( \sqrt{12^{2}+16^{2}} \right )^{2}} \right )\left ( \frac{16}{\sqrt{12^{2}+16^{2}}} \right )$

$\small \Rightarrow E_{x}=\left ( \frac{k\times 6}{15^{2}} \right )\left ( \frac{9}{15} \right )-\left ( \frac{k\times 8}{20^{2}} \right )\left ( \frac{16}{20} \right )$

$\small \Rightarrow E_{x}=k(0.016-0.016)=0\: N/C$

Hence the electric field at point A along x - axis is zero. Now along y - axis, the electric field is,

  $\small E_{y}=E_{q_{1}}sin\theta _{1}+E_{q_{2}}sin\theta _{2}-E_{q_{3}}$

As the total electric field at point A should be zero so the above equation will become,

$\small \Rightarrow 0=E_{q_{1}}sin\theta _{1}+E_{q_{2}}sin\theta _{2}-E_{q_{3}}$

Now the charge q₃ can be calculated as,

$\small \Rightarrow 0=\left ( \frac{k\times 8}{\left ( \sqrt{12^{2}+16^{2}} \right )^{2}} \right )\left ( \frac{12}{\sqrt{12^{2}+16^{2}}} \right )+\left ( \frac{k\times 6}{\left ( \sqrt{12^{2}+9^{2}} \right )^{2}} \right )\left ( \frac{12}{\sqrt{12^{2}+9^{2}}} \right )-\left ( \frac{k\times q_{3}}{\left ( 15-12 \right )^{2}} \right )$

$\small \Rightarrow \left ( \frac{k\times 8}{\left ( 20 \right )^{2}} \right )\left ( \frac{12}{20} \right )+\left ( \frac{k\times 6}{\left ( 15 \right )^{2}} \right )\left ( \frac{12}{15} \right )=\left ( \frac{k\times q_{3}}{\left ( 3 \right )^{2}} \right )$

$\small \Rightarrow 0.012+0.0213=\left ( \frac{ q_{3}}{9} \right )$

$\small \Rightarrow q_{3}=0.3\: nC$