Electric Field from Arc of Charge - Need Help

In summary: C)/(.09582m)λ=-1.25*10^-5 C/mIn summary, the problem involves finding the linear charge density along a quarter circle arc of radius 6.1 cm with a total charge of -1.2 μC. The solution is found by using the equation Q/ L, where L is the total length of the arc. The final answer is -1.25*10^-5 C/m.
  • #1
Tapehead2
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Electric Field from Arc of Charge - NEED HELP WITHIN AN HOUR

I have literally been working on this all day and I am finally turning it over to someone better at physics then myself. This is due within two hours and I'm starting to doubt my ability to finish this, any help will be beneficial. Thanks!


Homework Statement



23s6pts.png


A total charge Q = -1.2 μC is distributed uniformly over a quarter circle arc of radius a = 6.1 cm as shown.

1) What is λ the linear charge density along the arc?
C/m

2) What is Ex, the value of the x-component of the electric field at the origin (x,y) = (0,0) ?
N/C

3) What is Ey, the value of the y-component of the electric field at the origin (x,y) = (0,0) ?
N/C

4)

ic4w7k.png


How does the magnitude of the electric field at the origin for the quarter-circle arc you have just calculated comnpare to the electric field at the origin produced by a point charge Q = -1.2 μC located a distance a = 6.1 cm from the origin along a 45o line as shown in the figure?

a. The magnitude of the field from the point charge is less than that from the quarter-arc of charge.
b. The magnitude of the field from the point charge is equal to that from the quarter-arc of charge
c. The magnitude of the field from the point charge is greater than that from the quarter-arc of charge

5)

15wh3zq.png


What is the magnitude of the electric field at the origin produced by a semi-circular arc of charge = -2.4 μC, twice the charge of the quarter-circle arc?


The Attempt at a Solution



(2*pi*r)/4

(2*pi*.061m)/4=.0958m (the circumference of the circle we are concerned about)

(-1.2*10^-6)C/.0958m = -.000013 C/m

That is what I would say the density is but it is marked as incorrect. I cannot move on from this point. Any help will be great, thanks!
 
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  • #2
Bump
 
  • #3
Bump
 
  • #4
I actually just did this problem, do you still need help?
 
  • #5
Fantastic, yes I do!
 
  • #6
Okay, have you already found the linear charge density lambda?
 
  • #7
The work above is as far as I have got, so no.
 
  • #8
Q = λs
s = rθ

I always begin by changing my units to the base SI units:
Q = 1.2μC = 1.2*10-6C
r = 6.1cm = 0.061m

Q = λs = λrθ
1.2*10-6C = λ * (0.061m)(pi/2)
Solve for lambda and you should get the linear charge density.

I think the problem may have been with your sign, because I got to the same answer.
 
  • #9
I tried both positive and negative for both scientific notation and not. I don't think the program is wrong but it's possible?
 
  • #10
Okay, I really want to help you. Let me reference my answers.
 
  • #11
Thank you, I appreciate it.
 
  • #12
That has to be the answer; may I ask how you're inputting it in scientific notation?
 
  • #13
-1.3*10-5C/m?
 
  • #14
Exactly, did not work.
 
  • #15
Are you inputting the answers into Mastering Physics?

Perhaps, the asterisk is the issue. If you're entering -1.3x10^-5 C/m, it will not recognize that as a valid answer.
Also, if the units are already included on the side, you have to omit them and just submit the number: -1.3*10-5.
 
  • #16
Here is what I'm inputting and the programs response. Right need to get some shut eye but I will be on tomorrow and I'll see if I can ask my professor. The homework isn't due until a day after I thought so I'll have tomorrow to figure this out.

qrapvc.png
 
  • #17
Sorry, it's probably something really simple!
 
  • #18
Thank you for all your help thus far!
 
  • #19
UPDATE:

Well I officially hate online physics homework.

13z1jk0.png
 
  • #20
The circumference of a circle is 2πr

r=.061m

Since you have a quarter circle, the total length of your circle is L = (2∏(.061))/4 ≈ .09582m

You're given the charge and the problem states that the charge density is uniform.

λ=Q/L
 

FAQ: Electric Field from Arc of Charge - Need Help

What is an electric field?

An electric field is a physical phenomenon that is created by electrically charged particles. It is a vector field that describes the direction and magnitude of the force that a charged particle would experience at any given point in space.

What is an arc of charge?

An arc of charge is a portion of a charged object or particle that has a curved shape, similar to an arc. It can be created by a charged particle moving along a curved path, or by a charged object with a curved surface.

How is an electric field created by an arc of charge?

When a charged particle moves along a curved path, it creates an electric field that is tangential to the path at any given point. This electric field is strongest at the point where the particle is located and decreases as the distance from the particle increases. Similarly, a charged object with a curved surface will also create an electric field that is tangential to the surface at any given point.

What factors affect the strength of the electric field from an arc of charge?

The strength of the electric field from an arc of charge is affected by the magnitude of the charge, the distance from the arc, and the angle of the arc. The closer the distance and the larger the magnitude of the charge, the stronger the electric field will be. Additionally, the angle of the arc can affect the direction and shape of the electric field.

How is the electric field from an arc of charge calculated?

The electric field from an arc of charge can be calculated using Coulomb's law, which states that the electric field at a point is equal to the magnitude of the charge divided by the distance squared. This can be applied to each individual portion of the arc and then summed to find the overall electric field at a given point.

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