Electric field at origin from quarter-circle arc of charge

[Tapehead2]

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

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)

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)

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!

 

[Tapehead2]

Bump

 

[Tapehead2]

Bump

 

[TimmyJ1203]

I actually just did this problem, do you still need help?

 

[Tapehead2]

Fantastic, yes I do!

 

[TimmyJ1203]

Okay, have you already found the linear charge density lambda?

 

[Tapehead2]

The work above is as far as I have got, so no.

 

[TimmyJ1203]

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.

 

[Tapehead2]

I tried both positive and negative for both scientific notation and not. I don't think the program is wrong but it's possible?

 

[TimmyJ1203]

Okay, I really want to help you. Let me reference my answers.

 

[Tapehead2]

Thank you, I appreciate it.

 

[TimmyJ1203]

That has to be the answer; may I ask how you're inputting it in scientific notation?

 

[TimmyJ1203]

-1.3*10^-5C/m?

 

[Tapehead2]

Exactly, did not work.

 

[TimmyJ1203]

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.

 

[Tapehead2]

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.

 

[TimmyJ1203]

Sorry, it's probably something really simple!

 

[Tapehead2]

Thank you for all your help thus far!

 

[Tapehead2]

UPDATE:

Well I officially hate online physics homework.

 

[Feodalherren]

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