Question

A point charge is at the origin. With this point charge as the source point, what is the unit vector r^ in the direction of (a) the field point at x = 0 , y = -1.35m; (b) the field point at x = 12.0cm, y = 12.0cm; (c) the field point at x = - 1.10m, y = 2.60m? Express your results in terms of the unit vectors i^ and j^. You may want to review (Pages 695 - 699) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Electric-field vector for a point charge.

Answer 1

Answer:

• a. $\hat{r} =- \hat{j}$
• b. $\hat{r} = \ \frac{1}{\sqrt{2}} \ \hat{i} + \ \frac{1}{\sqrt{2}} \ \hat{j}$
• c. $\hat{r} = \ -0.3871 \ \hat{i} + \ 0.91501\ \hat{j}$

Explanation:

Using Coulomb's Law we know that the electric field E at point $\vec{r}$ is:

$\vec{E(\vec{r})} = k_e \frac{q}{d^2} \frac{\vec{r}-\vec{r'}}{d}$

where  $k_e$ is the Coulomb's Constant, q is the source charge, d is the distance between point and position of the source point charge, and $\vec{r}'$ is the position of the source point charge.

Taking all this in consideration, the unit vector clearly is:

$\hat{r} =\frac{\vec{r}-\vec{r'}}{d}$

For our problem, $\vec{r'} = (0,0)$, as the charge is located at the origin.

So

$\hat{r} =\frac{\vec{r}}{d}$

and d will be the magnitude of $\vec{r}$

Now, we can take the values for each point.

a.

$\vec{r}= (0,-1.35 \ m)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(0 \ m)^2 + (-1.35 \ m )^2}$

$|\vec{r}| =1.35 \ m$

So, the unit vector is:

$\hat{r} =\frac{(0,-1.35 \ m)}{1.35 \ m}$

$\hat{r} =(0,-1,0)$

$\hat{r} =- \hat{j}$

b.

$\vec{r}= (12 \ cm,12 \ cm)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(12 \ cm)^2 + (12 \ cm )^2}$

$|\vec{r}| = \sqrt{2} \ 12 \ cm$

So, the unit vector is:

$\hat{r} =\frac{(12 \ cm,12 \ cm)}{\sqrt{2} \ 12 \ cm}$

$\hat{r} =(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$

$\hat{r} = \ \frac{1}{\sqrt{2}} \ \hat{i} + \ \frac{1}{\sqrt{2}} \ \hat{j}$

c.

$\vec{r}= (-1.10 \ m, 2.60 \ m)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(-1.10 \ m)^2 + (2.60 \ m)^2}$

$|\vec{r}| = 2.8415 \ m$

So, the unit vector is:

$\hat{r} =\frac{ (-1.10 \ m, 2.60 \ m)}{2.8415 \ m}$

$\hat{r} =(-0.3871 ,0.91501)$

$\hat{r} = \ -0.3871 \ \hat{i} + \ 0.91501\ \hat{j}$