Respuesta :
Answer:
The electric field due to charge at distance r is [tex](-0.8169\times10^{6},0.8169\times10^{6}, 2.800\times10^{6})\ N/C[/tex]
Explanation:
Given that,
Initial position of particle [tex]r_{1}=4.00\times10^{-9}i-5.00\times10^{-9}j-2.00\times10^{-9}k\ m[/tex]
Final position of particle [tex]r_{2}=-2.00\times10^{-9}i+4.00\times10^{-9}j+ 3.00\times10^{-9}k\ m[/tex]
We need to calculate the magnitude of the position vector
Using formula of position vector
[tex]\vec{r}=\vec{r_{2}}-\vec{r_{1}}[/tex]
Put the value into the formula
[tex]\vec{r}=-2.00\times10^{-9}i+4.00\times10^{-9}j+ 3.00\times10^{-9}k-(4.00\times10^{-9}i-5.00\times10^{-9}j-2.00\times10^{-9}k)[/tex]
[tex]\vec{r}=-9\times10^{-9}i+9\times10^{-9}j+5\times10^{-9}k[/tex]
[tex]\vec{|r|}=\sqrt{(9^2+9^2+5^2)\times10^{-18}}[/tex]
[tex]\vec{|r|}=13.6\times10^{-9}\ m[/tex]
We need to calculate the unit vector along electric field direction
Using formula of unit vector
[tex]\hat{r}=\dfrac{\vec{r}}{|\vec{r}|}[/tex]
Put the value into the formula
[tex]\hat{r}=\dfrac{-9\times10^{-9}i+9\times10^{-9}j+5\times10^{-9}k}{13.6\times10^{-9}}[/tex]
[tex]\hat{r}=-0.66i+0.66j+0.36k[/tex]
We need to calculate the electric field due to charge at distance r
Using formula of electric field
[tex]E=\dfrac{kq}{r^2}\hat{r}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times1.6\times10^{-19}}{(13.6\times10^{-9})^2}\times(-0.66i+0.66j+0.36k)[/tex]
[tex]E=7.78\times10^{6}\times(-0.66i+0.66j+0.36k)[/tex]
[tex]E=-0.8169\times10^{6}i+0.8169\times10^{6}j+2.800\times10^{6}k\ N/C[/tex]
[tex]E=(-0.8169\times10^{6},0.8169\times10^{6}, 2.800\times10^{6})\ N/C[/tex]
Hence, The electric field due to charge at distance r is [tex](-0.8169\times10^{6},0.8169\times10^{6}, 2.800\times10^{6})\ N/C[/tex]
