Question

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 field at location < -3.00 10-9, 2.00 10-9, 4.00 10-9 > m, due to the π_ particle?

Answer 1

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 charge$r_{1} = / m$

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