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3 years ago

A point charge q = -6.0nC is located at the origin. The electric field (in N/C) vector at the point x = -8.0m, y= +1.5m is?

Physics

1 answer:

Luba_88 [7]3 years ago

4 0

Answer:

Explanation:

The position between point charge <em>q</em> to x = -8.0 m and y = 1.5 m is:

$r=\sqrt{x^{2}+y^{2}}=\sqrt{(-8)^{2}+(1.5)^{2}}=\sqrt{66.25}\approx 8.13$ meter

Then the magnitude of the electric field E is:

$E=\frac{1}{4\pi\epsilon_{o}} \frac{q}{r^{2}}=(9\times 10^{9}) \frac{6.0\times 10^{-9}}{66.25}\approx 0.81 N/C$

For the vector of E:

$\tan\theta=\frac{1.5}{-8}=-0.1875 \rightarrow \theta = -10.61^{0}$

$E_{x}=E\cos (-10.16)=(0.81)(0.984)=0.79704 N/C$

$E_{y}=E\sin\theta = -0.81(0.184)=-0.14904 N/C$

$\vec{E}=0.79704\hat{i}-0.14904\hat{j}$ in N/C

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