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77julia77 [94]
3 years ago
10

A π_ ("pi-minus") particle, which has charge _e, is at location ‹ 4.00 10-9, -3.00 10-9, -6.00 10-9 › m. What is the electric fi

eld at location < -3.00 10-9, 2.00 10-9, 4.00 10-9 > m, due to the π_ particle?
Physics
1 answer:
Law Incorporation [45]3 years ago
8 0

Answer:

The electric field is \dfrac{10.08\times10^{-18}\hat{i}}{2295.2}-\dfrac{7.2\times10^{-18}\hat{j}}{2295.2}-\dfrac{14.4\times10^{-18}\hat{k}}{2295.2}

Explanation:

Given that,

Location of charger_{1} = / m

Location of electric field r_{12} = / m

We nee to calculate the distance

Using relation of distance

\vec{r}=((-3-4)\hat{i}+(2-(-3))\hat{j}+(4-(-6))\hat{k})\times10^{-9}

\vec{r}=(-7\hat{i}+5\hat{j}+10\hat{k})\times10^{-9}

We need to calculate the electric field

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^3}\times\vec{r}

Put the value into the formula

E=\dfrac{9\times10^{9}\times(1.6\times10^{-19})\times((-7\hat{i}+5\hat{j}+10\hat{k})\times10^{-9})}{(\sqrt{(-7)^2+(5)^2+(10)^2})^3}

E= \dfrac{-1.44\times10^{-9}((-7\hat{i}+5\hat{j}+10\hat{k})\times10^{-9}))}{2295.2}

E=\dfrac{10.08\times10^{-18}\hat{i}}{2295.2}-\dfrac{7.2\times10^{-18}\hat{j}}{2295.2}-\dfrac{14.4\times10^{-18}\hat{k}}{2295.2}

Hence, The electric field is \dfrac{10.08\times10^{-18}\hat{i}}{2295.2}-\dfrac{7.2\times10^{-18}\hat{j}}{2295.2}-\dfrac{14.4\times10^{-18}\hat{k}}{2295.2}

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