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1 year ago

The charge per unit length on a long, straight filament is -90.0μC/m . Find the electric field (b) 20.0cm

Physics

1 answer:

Anna71 [15]1 year ago

5 0

The electric field for 20 cm is 8.09 MN/C

Length of the straight filament = (20cm) (1m/100cm) = 0.2

Electric field for the straight conductor = $E = \frac{2k_{e} \lambda}{r}$

E = [2(8.99 × 10^9 N·m^2/C^2)(90.0 × 10^-6 C/m)] / 0.2 m

E = 8.09 MN/C

The electric field is directly radially inward, toward the filament.

Electric field is a force produced by a charge near its surroundings. This force is exerted on other charges when brought in the vicinity of this field. SI unit of electric field is N/C (Force/Charge).

Learn more about electric field here:

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During a lab, a student tapes a ruler to a lab table and sets the ruler in motion. A laser detector pointed at the ruler records

zimovet [89]

Answer:

Four; Two; One; Three

Explanation:

During a lab, a student tapes a ruler to a lab table and sets the ruler in motion. A laser detector pointed at the ruler records how many times the ruler vibrates back and forth in a period of time.

Pitch or frequency is defined as the number of cycles per unit time. We can calculate the pitch for each trials i.e.

Trial one, $pitch=\dfrac{10}{0.25}=40\ s^{-1}$

Trial two, $pitch=\dfrac{11}{0.30}=36.67\ s^{-1}$

Trial three, $pitch=\dfrac{13}{0.31}=41.93\ s^{-1}$

Trial four, $pitch=\dfrac{12}{0.33}=36.36\ s^{-1}$

Hence, the order of the trials from least to greatest pitch is :

Trial four, Trial two, Trial one and Trial three i.e. option (d) is correct.

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3 years ago

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)

KiRa [710]

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.

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

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

Now, we can take the values for each point.

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

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:

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

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

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

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

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:

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

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

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

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

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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