1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Scilla [17]
3 years ago
8

Two point charges (q1 = -4.8μC and q2 = 6.7 μC) are fixed along the x-axis, separated by a distance d = 7.6 cm. Point P is locat

ed at (x,y) = (d,d).
1)

What is Ex(P), the value of the x-component of the electric field produced by q1 and q2 at point P?N/C

You currently have 0 submissions for this question. Only 10 submission are allowed.
You can make 10 more submissions for this question.

2)

What is Ey(P), the value of the y-component of the electric field produced by q1 and q2 at point P?N/C

You currently have 0 submissions for this question. Only 10 submission are allowed.
You can make 10 more submissions for this question.

3)

h2_pointC

A third point charge q3 = 3.1 μC is now positioned along the y-axis at a distance d = 7.6 cm from q1 as shown. What is Ex(P), the x-component of the field produced by all 3 charges at point P?

N/C

You currently have 0 submissions for this question. Only 10 submission are allowed.
You can make 10 more submissions for this question.

4)

Suppose all charges are now doubled (i.e., q1 = -9.6 μC, q2 = 13.4 μC, q3 = 6.2 μC), how will the electric field at P change?

Its magnitude will increase by less than a factor of two and its direction will remain the same

Its magnitude will increase by less than a factor of two and its direction will change

Its magnitude will double and its direction will remain the same

Its magnitude will double and its direction will change

You currently have 0 submissions for this question. Only 10 submission are allowed.
You can make 10 more submissions for this question.

5)

How would you change q1 (keeping q2 and q3 fixed) in order to make the electric field at point P equal to zero?

Increase its magnitude and change its sign

Decrease its magnitude and change its sign

Increase its magnitude and keep its sign the same

Decrease its magnitude and keep its sign the same

There is no change you can make to q1 that will result in the electric field at point P being equal to zero.
Physics
1 answer:
amid [387]3 years ago
4 0

Answer:

1)  Eₓ = 8.99 10³3 (6.7 /(d-0.076)² - 4.8 / d²) ,

2)   E_{y}= 1,708 10⁴/d²      N /C

3)  Eₓ = 8.99 10³ (-0.7/d² + 6.7 / (d-0.076)²)

4) magnitud is double and maintains the same angle

5) There is no change that can be made.

Explanation:

The electric field is a vector quantity with formula

           E = k q / r²

Where k is the Coulomb constant with value 8.99 10⁹ N m² / C², q is the load and r is the distance from the load to the point of interest.

The total field is

              E = E1 + E2

             E = k q₁ / r₁² + k q₂ / r₂²

 

Since the field is a vector quantity we will look for each component

1) x axis

Let's look for the distance for simplicity, suppose the origin of the coordinate system is at point 1

          x₁ = x- x₀

          x = d - 0 = d

Point 2

          x₂ = d- 0.076

          Eₓ = E₁ₓ + E₂ₓ

          Eₓ = k q₁ / x₁² + k q₂ / x₂²

          Eₓ = 8.99 10⁹ (- 4.8 10⁻⁶/ d² + 6.7 10⁻⁶ / (d-0.076)² )

          Eₓ = 8.99 10³3 (6.7 /(d-0.076)² - 4.8 / d²)

.) Y axis

Distance

The points are on the x-axis so the y coordinate of the charges is zero

          y₁ = y-y₀

          y₁ = d-0

          y₂ = d-0

         

          E_{y} = E_{1y} + E_{2y}

          E_{y} = k q₁ / y₁² + k q₂ / y₂²

          E_{y} = 8.99 10⁹ (-4.8 10⁻⁶ / d² + 6.7 10⁻⁶ / d²)

           E_{y} = 8.99 10³ (1.9 / d²)

           E_{y}= 1,708 10⁴/d²      N /C

For explicit value of the electric field we must know the coordinates of point P

3) A third load is placed at a point d = 7.6 cm, on the y-axis, we calculate the field on the x-axis

Distance

           x₃ = d - 0 = d

           Eₓ = E₁ₓ + E₂ₓ + E₃ₓ

           Eₓ = k (q₁ / x₁² + q₂ / x₂² + q₃ / x₃²)

           Eₓ = 8.99 10⁹ (- 4.8 / d² + 6.7 / (d-0.076)² + 3.1 / d²) 10⁻⁶

           Eₓ = 8.99 10³ (-0.7/d² + 6.7 / (d-0.076)²)

4) The electric field at point P is

               E₀ = RA (Eₓ² + E_{y}²)

        With the new charges

               Eₓ = 2 Eₓ₀

               E_{y} = 2 E_{yo}

               E = √ (4Exo + 4E_{yo})

               E = 2 E₀

As the two components increase the same amount the angle does not change

The answer:  magnitud is double and maintains the same angle

5) For the field to be zero at point P the two components must be zero

  Y Axis  

          Ey = 8.99 10⁹ (-4.8 10⁻⁶/d² + 6.7 10⁻⁶ / d²)

For this amount to be zero you must increase q₁ by keeping your sign

   X axis

         Ex = 8.99 10⁹ (- 4.8 / d² + 6.7 / (d-0.076)² + 3.1 / d²) 10⁻⁶

For this component you must also increase q1 without changing sign

But we have a problem, the magnitude of the change is different on each axis and, so there is no way to take the point / p to a zero field

The serious answer: There is no change that can be made.

You might be interested in
What does infrastructure refer to? process by which a nation improves the economic, political, and social well-being of its peop
Travka [436]

Answer:

Infrastructure is the term for the basic physical systems of a business or nation—transportation, communication, sewage, water, and electric systems are all examples of infrastructure

Explanation:

These systems tend to be high-cost investments and are vital to a country's economic development and prosperity.

7 0
2 years ago
How many joules of work are done on a box when a force of 25 N pushes it 3 m?
Anika [276]

Answer:

75joules

Explanation:

Workdone = force x distance

workdone = 25 x 3

workdone = 75joules

7 0
2 years ago
Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12
PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0
3 years ago
One difference between a solar flare and a CME is that a solar flare is composed of ___________, while a CME is composed of ____
JulsSmile [24]

Answer:

magnetic energy (proton) and magnetic plasma.

Explanation:

  • The solar fare consists of bright light that occurs in various wavelengths and is observed at the surface.
  • They are not as strong as compared to the coronal mass ejection or CME. The solar fares consist of 10²² joules, while the plasma is ejected from the solar corona and can be clearly seen from a distance.
  • The Solar flares represent an atmospheric disturbance and plasms are the medium for the growth and development of solar flare and lead to solar activity.
6 0
3 years ago
After a collision between two different massed objects; the larger objects accelerate at a faster rate than the smaller object?
Nitella [24]

Answer: Things continue doing what they are doing unless a force is applied to it. Objects have a natural tendency to resist change. This is INERTIA. Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects, so true

Explanation:

Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop. The Cadillac has more of a tendency to stay stationary (or continue moving), and resist a change in motion than a bicycle.

6 0
2 years ago
Other questions:
  • Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
    9·1 answer
  • An electric iron of resistance 20Ω takes a current of 5A. Calculate the heat developed in 30seconds?
    14·1 answer
  • How does the planet Venus look to a person standing on the Earth?
    14·1 answer
  • What is the de Broglie wavelength of an electron that strikes the back of the face of a TV screen at 1/10 the speed of light? __
    14·1 answer
  • Is energy released or absorbed during the formation of a solution?
    7·1 answer
  • Average speed is the total distance divided by the
    8·1 answer
  • Four mass–spring systems oscillate in simple harmonic motion. Rank the periods of oscillation for the mass–spring systems from l
    5·2 answers
  • Which practice is an unsustainable way of managing resources?<br> Please help
    12·1 answer
  • Caroline, a piano tuner, suspects that a piano's G4 key is out of tune. Normally, she would play the key along with her G4 tunin
    12·1 answer
  • a) CALCULATE the acceleration of a 300,000 kg jumbo jet, just before takeoff when the thrust of each of its 4 engines is 36,000
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!