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
docker41 [41]
2 years ago
10

Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar

rangement is called an electric quadrupole. (a) Find the magnitude and direction of the electric field at points on the positive x-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for x>>a. Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole
Physics
1 answer:
den301095 [7]2 years ago
3 0

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

 a) Let's write the electric field for each charge and the total field

       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

You might be interested in
Describe why we cannot see details on the surface of Mars.​
bearhunter [10]

Answer: atmosphere

Explanation:

We cannot see the surface of Mars because of its thick atmosphere and of the iron dust that kicks up

4 0
3 years ago
Read 2 more answers
A -ion (negative ion, one with extra negative charge )
AlladinOne [14]
What is your question
8 0
3 years ago
Read 2 more answers
Astronomy
alina1380 [7]

Answer: B

Explanation:

X-rays coming into Earth's system from space are absorbed by the atmosphere. Telescopes that processes X-rays, therefore, are placed in orbit above the atmosphere

7 0
3 years ago
Read 2 more answers
How many times does the world go around
sattari [20]
I'm not sure what you're asking but the earth has the ability to infinitely continue to spin or the earth completes 365.25 rotations during a full cycle. 
6 0
3 years ago
The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
3 years ago
Other questions:
  • What is a Cartesian diver? What does it do?
    13·1 answer
  • A baseball player slides into third base with an initial speed of 4.0 m/s . if the coefficient of kinetic friction between the p
    12·1 answer
  • Which of these is an example of interactions between the atmosphere and biosphere?
    9·2 answers
  • A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car. Since a lot of damage is done on the car : Sel
    10·1 answer
  • PLLLLLLLLLLLZ HELP 15 POINTS
    8·1 answer
  • Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1450 N . Assume that the pla
    10·1 answer
  • On a coordinate plane, vertex A for triangle ABC is located at (6,4). Triangle ABC is dilated by a scale factor of 0.5 with the
    12·1 answer
  • A single, nonconstant force acts in the + x ‑direction on an object of mass M that is constrained to move along the x ‑axis. As
    13·1 answer
  • 2. Heather and Matthew walk with an average velocity of +0.87 m/s eastward.
    11·1 answer
  • 8. The mass of the sun is 1.99 x 1030 kilograms and its distance from Earth is 150 million kilometers (150 x 109 meters). What i
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!