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

What he said /\

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Step-by-step explanation:

Answer 2

The electric field is
`\frac{10.08*10^(-18)\hat{i}}{2295.2}-\frac{7.2*10^(-18)\hat{j}}{2295.2}-\frac{14.4*10^(-18)\hat{k}}{2295.2}`

Step-by-step explanation:

Given that,

Location of charge
`r_(1) = <4.00*10^(-9),-3.00*10^(-9),-6.00*10^(-9)>/ m`

Location of electric field
`r_(12) = <-3.00*10^(-9),2.00*10^(-9),4.00*10^(-9)>/ 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})*10^(-9)`

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

We need to calculate the electric field

Using formula of electric field

`E=(1)/(4\pi\epsilon_(0))(q)/(r^3)*\vec{r}`

Put the value into the formula

`E=\frac{9*10^(9)*(1.6*10^(-19))*((-7\hat{i}+5\hat{j}+10\hat{k})*10^(-9))}{(√((-7)^2+(5)^2+(10)^2))^3}`

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

`E=\frac{10.08*10^(-18)\hat{i}}{2295.2}-\frac{7.2*10^(-18)\hat{j}}{2295.2}-\frac{14.4*10^(-18)\hat{k}}{2295.2}`

Hence, The electric field is
`\frac{10.08*10^(-18)\hat{i}}{2295.2}-\frac{7.2*10^(-18)\hat{j}}{2295.2}-\frac{14.4*10^(-18)\hat{k}}{2295.2}`