Physics, 24.01.2020 21:31 cheergirl2133

Three point charges are located on the positive x-axis of a coordinate system. charge qı = 1.5 nc is 2.0 cm from the origin, charge q2 = -3.0 nc is 4.0 cm from the origin and charge 93 = 4.5 nc located at the origin. a. what is the magnitude of the net force exerted by the other two charges on charge q1q1q1 = 1.5 ncnc? b. what is the direction of the net force exerted by the other two charges on charge q1q1q1 = 1.5 ncnc?

Answers: 1

Show answers

Answers

Answer from: genyjoannerubiera

Solution: The magnitude of the net force acting on q₁ is : Fn = 25.312 [N]

and the direction is in the direction of the positive x axis

The electric force between two charges q₁ and q₂, is according to Coulombs´law as:

F₂₁ = K × q₁ × q₂ / d₁₂²

In that equation

F₂₁ is the force exerted by charge q₂ on the charge q₁

K = 9× 10⁹ [Nm²/C²] Coulomb contantq₁ is the charge upon which the force is actingq₂ is the force acting over q₁d₁₂ is the distance between the charges

It is pretty obvious that F₂₁ = F₁₂, and the force is of rejection if the charges are of the same sign or attraction if they are of opposite signs

Now in this case we calculate F₂₁

F₂₁ = [9×10⁹ × ( 1.5)×10⁻⁹ × (3)×10⁻⁹] /(0.02)²

F₂₁ = 40.5 × 10⁻⁹/ 4 × 10⁻⁴

F₂₁ = 10.125 × 10⁻⁵ [N]

F₂₁ is in the direction of the positive x ( attraction force)

And

F₃₁ = [9×10⁹ × ( 1.5)×10⁻⁹ × (4.5)×10⁻⁹] /(0.02)²

F₃₁ = 15.188× 10⁻⁵ [N]

F₃₁ is in the direction of the positive x ( rejection force)

Then the net force is Fn = 10.125 + 15.188

Fn = 25.312 [N] in direction of x positive

Three point charges are located on the positive x-axis of a coordinate system. charge qı = 1.5 nc is

Answer from: ramenbates81

$F_1=2.53\ 10^{-4} \ N$

The net force goes to the right

Explanation:

Electrostatic Force

Let's consider the situation where 2 point charges q1 and q2 are separated by a distance d. An electrostatic force appears between them whose magnitude can be computed by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

Where k is the constant of proportionality

$k=9.10^9\ Nw.m^2/c^2$

Two equally-signed charges repel each other, two opposite-signed charges attract each other.

We need to find the total net force exerted on q1 by q2 and q3. We're assuming the charges are placed to the right of the origin, so the distribution is shown in the figure below.

Since q3 repels q1, its force goes to the right, since q2 attracts q1, its force goes to the right also, thus the total force on q1 is :

$F_1=F_{31}+F_{21}$

It's directed to the right

Let's compute the individual forces. q3 is separated 2 cm from q1, so d=0.02 m

$\displaystyle F_{31}=\frac{9.10^9\ 4.5\ 10^{-9}\ 1.5\ 10^{-9}}{0.02^2}$

$F_{31}=0.000151875\ N$

$\displaystyle F_{21}=\frac{9.10^9\ 3\ 10^{-9}\ 1.5\ 10^{-9}}{0.02^2}$

$F_{21}=0.00010125\ N$

$F_1=F_{31}+F_{21}=0.000151875\ N+0.00010125\ N=0.000253125 \ N$

Expressing the force in scientific notation

$\boxed{F_1=2.53\ 10^{-4}\ N}$

The net force goes to the right

Answer from: Quest

Iwould believe out of the
For the car in the picture ,in which direction is the normal force? -into the screen -down -out of

Answer from: Quest

Perhaps seek some professional ? it isn't very asking some children who are just trying to find out answers to homework.also, it seems professional would actually be good to you. obviously it isn't going to magically fix your depression, but it'll do something. drugs, smoking weed, it won't do much, especially if your husband is against it. if you do it he'll dislike it. but talk to him, try and get him to listen to you, and figure everything out together, okay? there's a website- maybe it can offer some ?

Other questions on the subject: Physics

Physics, 22.06.2019 00:30, mme58

What is the acceleration of a spaceship that is initially at rest and then speeds up to 220 m/s in 20 seconds

Answers: 1

continue

Physics, 23.06.2019 00:00, cece3467

The horizontal blue line through the middle of the lens is called the optical axis. as the lamp is moved above or below the axis, keeping the horizontal distance to the center of the lens constant, how does the horizontal distance from the lens to the image change?

Answers: 2

continue

Physics, 23.06.2019 09:00, peternice2956

Choose the correct general statement concerning the total potential and kinetic energy at any point along the arc of swing. the total energy at any point along the arc of swing of the pendulum is always less than 12 joules. the total energy at any point along the arc of swing of the pendulum is always more than 12 joules. the total energy at any point along the arc of swing of the pendulum is always equal to 12 joules. not enough information to determine

Answers: 1

continue

Physics, 23.06.2019 12:10, smeeden

A47 kg person in a rollercoaster moving through the bottom of a curved track of radius 47.7 m feels a normal force of 895 n. what is the net centripedal force acting on the person

Answers: 1

continue

You know the right answer?

Three point charges are located on the positive x-axis of a coordinate system. charge qı = 1.5 nc is...