Answer:
The electric field is [tex]\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}[/tex]
Explanation:
Given that,
Location of charge[tex]r_{1} = <4.00\times10^{-9},-3.00\times10^{-9},-6.00\times10^{-9}>/ m[/tex]
Location of electric field [tex]r_{12} = <-3.00\times10^{-9},2.00\times10^{-9},4.00\times10^{-9}>/ m[/tex]
We nee to calculate the distance
Using relation of distance
[tex]\vec{r}=((-3-4)\hat{i}+(2-(-3))\hat{j}+(4-(-6))\hat{k})\times10^{-9}[/tex]
[tex]\vec{r}=(-7\hat{i}+5\hat{j}+10\hat{k})\times10^{-9}[/tex]
We need to calculate the electric field
Using formula of electric field
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^3}\times\vec{r}[/tex]
Put the value into the formula
[tex]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}[/tex]
[tex]E= \dfrac{-1.44\times10^{-9}((-7\hat{i}+5\hat{j}+10\hat{k})\times10^{-9}))}{2295.2}[/tex]
[tex]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}[/tex]
Hence, The electric field is [tex]\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}[/tex]
