Mathematics, 06.09.2019 16:30 fperez616

Part a what is the electric field at the position (x1,y1)=(5.0 cm , 0 cm) in component form? express your answer in terms of the unit vectors i^ and j^. use the 'unit vector' button to denote unit vectors in your answer. e⃗ 1 e → 1 = n/c
part b what is the electric field at the position (x2,y2)=(−5.0 cm , 0 cm) in component form? express your answer in terms of the unit vectors i^ and j^. use the 'unit vector' button to denote unit vectors in your answer. e⃗ 2 e → 2 = nothing n/c
part c what is the electric field at the position (x3,y3)=(0 cm , 5.0 cm) in component form? express your answer in terms of the unit vectors i^ and j^. use the 'unit vector' button to denote unit vectors in your answer. e⃗ 3 e → 3 = nothing n/c

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Answer from: keenonwhite6801

The complete Question is

A −12 nC charge is located at (x, y)=(1.0 cm, 0 cm).

Part A) What is the electric field at the position (x1, y1)=(5.0 cm, 0 cm) in component form?

Part B) What is the electric field at the position (x2, y2)=(−5.0 cm, 0 cm) in component form?

Part C) What is the electric field at the position (x3, y3)=(0 cm, 5.0 cm) in component form?

Express your answer in terms of the unit vectors i^ and j^. Use the 'unit vector' button to denote unit vectors in your answer.

A) E = - 67500i + 0j N/C

B) E = 30000i + 0j N/C

C) E = 8143.18i - -40715.66j N/C

Step-by-step explanation:

Part A)

Magnitude of the charge = Q = -12 nC = $-12 \times 10^{-9}$ C

Since, its the negative charge, the direction of Electric Field lines will be directed toward the charge

Location of the charge = (1, 0)

We need to find the electric field at point (5, 0)

The formula for the magnitude of electric field due to a point charge is:

$E=\frac{kQ}{r^{2}}$

Here,

k = Coulomb's Law Constant = $9 \times 10^{9}$

Q = Magnitude of the charge

r = Distance between the charge and the point where we need to find the value of E

We can find r by using the Distance Formula.

So,

$r=\sqrt{(5-1)^{2}+(0-0)^{2}}=4$ cm = 0.04 m

Using these values in the formula, we get:

$E=\frac{9\times 10^{9} \times 12 \times 10^{-9}}{(0.04)^{2}}= 67500$ N/C

Since, two point (5, 0) is to the right of the given charge (as shown in the first image) i.e. in horizontal direction, all of the electric field experienced by it will be in horizontal direction and the vertical component would be zero. Also the direction of Electric field will be towards the charge i.e. in left direction so the x-component of Electric field will be negative.

Thus, we can write the value of E in vector form to be:

E = - 67500i + 0j N/C

Part B)

We need to find the Electric Field at point (-5, 0)

Using the similar procedure as used in the previous step, first we find r:

$r=\sqrt{(-5-1)^{2}+(0-0)^{2}}=6$ cm = 0.06

Using the values in the formula of Electric field, we get:

$E = \frac{9 \times 10^{9} \times 12 \times 10^{-9}}{(0.06)^2}=30000$ N/C

In this case again, the point is located in a horizontal direction to the given charge, so all the Electric Field experienced by it will be in horizontal direction and the vertical component will be zero. The direction of Electric field will be towards the charge i.e towards Right, so the x-component will be positive in this case.

So, value of the electric field in component form would be:

E = 30000i + 0j N/C

Part C)

We need to find the value of electric field at the point (0, 5). First we find the value of r:

$r=\sqrt{(1-0)^2+(0-5)^2}=\sqrt{26}=5.1$ cm = 0.051 m

Using the values in the formula of E, we get:

$E=\frac{kQ}{r^{2}}=\frac{9 \times 10^{9} \times 12 \times 10^{-9}}{(0.051)^{2}}=41522$ N/C

The point (0, 5) is neither exactly to the left or exactly up. So, for this point we need to find both the horizontal and vertical components as shown in the 2nd figure below.

From the triangle, we have the opposite and adjacent side to the angle, so using the tangent we can find the value of angle theta.

$tan(\theta)=\frac{5}{1}\\\theta=tan^{-1}(5)=78.69$

The two angles shown in the figure will be equal as there are alternate interior angles. Now the angle which E will make with positive x-axis will lie in the 4th quadrant as it lies below the horizontal line. So, the angle with positive x-axis would be:

360 - 78.69 = 281.31 degrees

Ex = E cos(θ) = 41522 cos(281.31) = 8143.18 N/C

Ey = E sin(θ) = 41522 sin(281.31) = -40715.66 N/C

So, in component form the Electric field will be:

E = 8143.18i - -40715.66j

Part a what is the electric field at the position (x1,y1)=(5.0 cm , 0 cm) in component form? expres

Answer from: sosa1k18

The electric field at the position E (5.0 cm ,0 cm) is E - 67500i + 0j N/C.

The electric field at the position E (-5.0 cm ,0 cm) E is 30000i + 0j N/C.

The electric field at the position E (0 cm ,5 cm) E is 8143.18i - -40715.66j N/C

We have to determine, the electric field at the position (x1,y1) =( 5.0 cm , 0 cm) in component form? Express your answer in terms of the unit vectors i^ and j^.

According to the question,

Magnitude of the charge = Q = -12 nC = C

The negative charge, the direction of Electric Field lines will be directed toward the charge,

Location of the charge = (1, 0)

To the electric field at point (5, 0)

The formula for the magnitude of electric field due to a point charge is:

$E = \dfrac{Kq}{r^{2} }$

Where, k = Coulomb's Law Constant =

Q = Magnitude of the charge

r = Distance between the charge and the point

To find r by using the Distance Formula.

Therefore,

$r = \sqrt{(5-1)^{2} + (0-0)^2\\\\$

$r = \sqrt{16} \\\\r = 4cm = 0.04m$

By using these values in the formula,

$E = \dfrac{9\times 10^{9} \times 12 \times 10^{-12} }{(0.04)^{2}}$

$E = 67500\ Nc^{-1}$

Therefore,

Two point (5, 0) is to the right of the given charge (as shown in the first image) i.e. in horizontal direction, all of the electric field experienced by it will be in horizontal direction and the vertical component would be zero. Also the direction of Electric field will be towards the charge and in left direction so the x-component of Electric field will be negative.

Thus, the value of E in vector form to be:

E = - 67500i + 0j N/C

To find the Electric Field at point (-5, 0)

By using the similar procedure as used in the previous step, first we find r:

$r = \sqrt{(-5-1)^{2} + (0-0)^{2}}\\\\r = \sqrt{36} \\\\r = 6 cm\\\\r = 0.06m$

By using the values in the formula of Electric field,

$E = \dfrac{9 \times 10^{9} \times 12 \times 10^{-12}}{(0.06)^{2}} \\$

$E = 30000 \ NC^{-1}$

The point is located in a horizontal direction to the given charge, so all the Electric Field experienced by it will be in horizontal direction and the vertical component will be zero. The direction of Electric field will be towards the charge i.e towards Right, so the x-component will be positive in this case.

So, The value of the electric field in component form would be:

E = 30000i + 0j N/C

To find the value of electric field at the point (0, 5).

The value of r:

$r = \sqrt{(1-0)^{2} + (0-5)^2} \\\\r = \sqrt{25 +1 } \\\\r = \sqrt{26}\\\\r = 5.1cm \\\\r = 0.051m$

By using the values in the formula of E, we get:

$E = \dfrac{ 9 \times 10^{9} 12 \times 10^{-12}}{(0.051)^2}\\\\E = 41522\ NC{-1}$

The point (0, 5) is neither exactly to the left or exactly up. So, for this point To find both the horizontal and vertical components as shown in the 2nd figure below.

From the triangle, the opposite and adjacent side to the angle, so using the tangent ,

To find the value of angle theta.

$Tan\theta = \dfrac{5}{1}\\\\\theta = tan^{-1} \dfrac{5}{1}\\\\\theta = tan^{-1}(5)\\\\\theta = 78.69$

The two angles shown in the figure will be equal as there are alternate interior angles.

Now the angle which E will make with positive x-axis will lie in the 4th quadrant as it lies below the horizontal line. So, the angle with positive x-axis would be:

360 - 78.69 = 281.31 degrees

E_x = E cos(θ) = 41522 cos(281.31) = 8143.18 N/C

E_y = E sin(θ) = 41522 sin(281.31) = -40715.66 N/C

So, in component form the Electric field will be:

E = 8143.18i - -40715.66j

To know more about Electric field click the link given below.

link

Answer from: Quest

because its not ur topic and its not for is for homework not that.

Who is and what is this person i'm just being weird nobody answer this

Answer from: Quest

Okay i will add you

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Part a what is the electric field at the position (x1,y1)=(5.0 cm , 0 cm) in component form? expres...