Answer:
Explanation:
The position between point charge q to x = -8.0 m and y = 1.5 m is:
[tex]r=\sqrt{x^{2}+y^{2}}=\sqrt{(-8)^{2}+(1.5)^{2}}=\sqrt{66.25}\approx 8.13[/tex] meter
Then the magnitude of the electric field E is:
[tex]E=\frac{1}{4\pi\epsilon_{o}} \frac{q}{r^{2}}=(9\times 10^{9}) \frac{6.0\times 10^{-9}}{66.25}\approx 0.81 N/C[/tex]
For the vector of E:
[tex]\tan\theta=\frac{1.5}{-8}=-0.1875 \rightarrow \theta = -10.61^{0}[/tex]
[tex]E_{x}=E\cos (-10.16)=(0.81)(0.984)=0.79704 N/C[/tex]
[tex]E_{y}=E\sin\theta = -0.81(0.184)=-0.14904 N/C[/tex]
[tex]\vec{E}=0.79704\hat{i}-0.14904\hat{j}[/tex] in N/C
