What is the refraction index of a medium?

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

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Which of the following is true?

fenix001 [56]

Answer:

Its either A. Or C cause ive had a question like this before So Im sure But if not Then Im so sorry

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What best describes gravity?

DanielleElmas [232]

F = G*((m sub 1*m sub 2)/r^2)

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2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to

ArbitrLikvidat [17]

Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b) the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

F = m a

a = F / m

Let's use kinematics to find the velocity after x = 1 m

v² = v₀² + 2 a x

The initial speed is zero

v = √ (2 F/m x)

For the distance of x = 1 m

v₁ = √ (2 F / m)

Car 2

F = 2m a

a = F / 2m

v² = 2 a x

v = √ (F/m x)

For x = 1 m

v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m 2F / m

K₁ = F

Car 2

K₂ = ½ 2m v₂²

K₂ = ½ 2m F / m

K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m √ (2F / m)

Car 2

P₂ = 2m v²

P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

v = v₀ + a t

v = F / m t

v₁ = F / m 10

Car 2

v₂ = F / 2m 10

v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m (F / m 10)²

K₁ = 50 F² / m

Car2

K₂ = ½ 2m v₂²

K₂ = m (F / m 5)²

K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m F / m 10

P₁ = 10 F

Car 2

P₂ = 2m v₂

P₂ = 2m F / m 5

P₂ = 10 F

In this case the moment of the two cars is the same

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What are five atomic models that have been proposed over time ?

Art [367]

Air, water , atomic acid , and pocket rocket

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