Help finding the Electric field at the center of charged arc

Click For Summary
The discussion centers on understanding the calculation of the electric field at the center of a charged arc. The user is confused about the length of the arc being pi/R/2, questioning if it results from dividing the circumference of a circle by four due to evaluating a quarter circle. Confirmation is provided that the charge +Q is indeed spread over this quarter arc length. Additionally, the user seeks a simpler approach to the problem, expressing difficulty in following the teacher's solution. Clarification is offered, inviting further questions on specific confusing parts of the solution.
guyvsdcsniper
Messages
264
Reaction score
37
Homework Statement
Determine electric field at center of curvature of arc.
Relevant Equations
E=KQ/r^2
I am having trouble understand where area circled in red.

I get that lamda is Q/L. The charge is +Q. Length is pi/R/2.

I am having trouble understanding why the length is pi/R/2? Is it because the circumference of a circle is 2*pi*R and since we have broken this problem down to just evaluating a quarter of the a circle, we divide 2*pi*R by 4 and get pi*R/2 which when we put that under Q we get 2Q/pi*R?

I wasnt sure until I typed out my question about and believe that may be the case, but just looking for some conformation.

Also, is there a simpler approach to this problem? My teacher wrote this out and I found it hard to follow.
 

Attachments

  • Screen Shot 2021-09-25 at 6.37.50 PM.png
    Screen Shot 2021-09-25 at 6.37.50 PM.png
    55.1 KB · Views: 156
Physics news on Phys.org
quittingthecult said:
I am having trouble understanding why the length is pi/R/2? Is it because the circumference of a circle is 2*pi*R and since we have broken this problem down to just evaluating a quarter of the a circle, we divide 2*pi*R by 4 and get pi*R/2 which when we put that under Q we get 2Q/pi*R?

I wasnt sure until I typed out my question about and believe that may be the case, but just looking for some conformation.
Yes, I think you have correctly answered your own question. The charge +Q is spread out over a length of a quarter of a circle of radius R.
quittingthecult said:
Also, is there a simpler approach to this problem? My teacher wrote this out and I found it hard to follow.
Your teacher's solution looks very good. If there is a specific place in the solution that you are confused about, we can help clarify.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
6
Views
1K
Replies
4
Views
3K
Replies
11
Views
1K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K