Answer:
Explanation:
Using Coulomb's Law we know that the electric field E at point [tex]\vec{r}[/tex] is:
[tex]\vec{E(\vec{r})} = k_e \frac{q}{d^2} \frac{\vec{r}-\vec{r'}}{d}[/tex]
where [tex]k_e[/tex] is the Coulomb's Constant, q is the source charge, d is the distance between point and position of the source point charge, and [tex]\vec{r}'[/tex] is the position of the source point charge.
Taking all this in consideration, the unit vector clearly is:
[tex]\hat{r} =\frac{\vec{r}-\vec{r'}}{d}[/tex]
For our problem, [tex]\vec{r'} = (0,0)[/tex], as the charge is located at the origin.
So
[tex]\hat{r} =\frac{\vec{r}}{d}[/tex]
and d will be the magnitude of [tex]\vec{r}[/tex]
Now, we can take the values for each point.
[tex]\vec{r}= (0,-1.35 \ m)[/tex]
and, the magnitude of the vector is
[tex]|\vec{r}| = \sqrt{r_x^2 + r_y^2}[/tex]
[tex]|\vec{r}| = \sqrt{(0 \ m)^2 + (-1.35 \ m )^2}[/tex]
[tex]|\vec{r}| =1.35 \ m[/tex]
So, the unit vector is:
[tex]\hat{r} =\frac{(0,-1.35 \ m)}{1.35 \ m}[/tex]
[tex]\hat{r} =(0,-1,0) [/tex]
[tex]\hat{r} =- \hat{j} [/tex]
[tex]\vec{r}= (12 \ cm,12 \ cm)[/tex]
and, the magnitude of the vector is
[tex]|\vec{r}| = \sqrt{r_x^2 + r_y^2}[/tex]
[tex]|\vec{r}| = \sqrt{(12 \ cm)^2 + (12 \ cm )^2}[/tex]
[tex]|\vec{r}| = \sqrt{2} \ 12 \ cm[/tex]
So, the unit vector is:
[tex]\hat{r} =\frac{(12 \ cm,12 \ cm)}{\sqrt{2} \ 12 \ cm}[/tex]
[tex]\hat{r} =(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0) [/tex]
[tex]\hat{r} = \ \frac{1}{\sqrt{2}} \ \hat{i} + \ \frac{1}{\sqrt{2}} \ \hat{j} [/tex]
[tex]\vec{r}= (-1.10 \ m, 2.60 \ m)[/tex]
and, the magnitude of the vector is
[tex]|\vec{r}| = \sqrt{r_x^2 + r_y^2}[/tex]
[tex]|\vec{r}| = \sqrt{(-1.10 \ m)^2 + (2.60 \ m)^2}[/tex]
[tex]|\vec{r}| = 2.8415 \ m[/tex]
So, the unit vector is:
[tex]\hat{r} =\frac{ (-1.10 \ m, 2.60 \ m)}{2.8415 \ m}[/tex]
[tex]\hat{r} =(-0.3871 ,0.91501) [/tex]
[tex]\hat{r} = \ -0.3871 \ \hat{i} + \ 0.91501\ \hat{j} [/tex]
