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iragen [17]
3 years ago
13

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.
Physics
1 answer:
KiRa [710]3 years ago
3 0

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.

<h3>a.</h3>

\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}

<h3>b.</h3>

\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}

<h3>c.</h3>

\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}

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