What is the electric potential at the center of the semicircle?

  • Thread starter physnoob
  • Start date
  • #1
15
0

Homework Statement


The wire in the following figure has linear charge density [tex]\lambda[/tex]. What is the electric potential at the center of the semicircle?


Homework Equations





The Attempt at a Solution


I can see a lot of integration going on here, but i am having trouble the setup the equation.
May i get some hints?

Thanks!
 

Attachments

  • 83.jpg
    83.jpg
    4.1 KB · Views: 1,751

Answers and Replies

  • #2
394
0
you can integrate from R to 3R and multiply it by 2 to account for the two straight lines. The easiest part, even though it may seem most difficult at first glance, is to account for the half circle. Just multiply lambda by piR and plug it into the equation for voltage of a point charge. This is because all the charge along that loop is exactly R away, so you don't need integration.
 
  • #3
15
0
you can integrate from R to 3R and multiply it by 2 to account for the two straight lines. The easiest part, even though it may seem most difficult at first glance, is to account for the half circle. Just multiply lambda by piR and plug it into the equation for voltage of a point charge. This is because all the charge along that loop is exactly R away, so you don't need integration.

Okay, I understand how to get the 2 lines now. But I am still a little bit confuse over the semicircle. How does lambda*pi*R relates to equation for voltage of a point charge?

Thanks for the help :)
 
  • #4
108
0
Okay, I understand how to get the 2 lines now. But I am still a little bit confuse over the semicircle. How does lambda*pi*R relates to equation for voltage of a point charge?

Thanks for the help :)

Components of electric potential don't cancel like they sometimes do when calculating electric field. Every charge chunk/point (dq) along the semicircle is contibuting a small amount of voltage (dV = (1/4[tex]\pi[/tex][tex]\epsilon[/tex]) * dq/r ). Since each chunk is the same radius R from the center, and the total charge of the semicircle is q = [tex]\lambda[/tex]*[tex]\pi[/tex]R, since there is constant charge density and the length is half a circle ( arc length is [tex]\pi[/tex]*R ), integrating dV along the semicircle will give V = (1/4[tex]\pi[/tex][tex]\epsilon[/tex]) [tex]\lambda[/tex]*[tex]\pi[/tex]R / R = (1/4[tex]\pi[/tex][tex]\epsilon[/tex]) [tex]\lambda[/tex]*[tex]\pi[/tex]. When first calculating electric potential, always start with the electric potential of a chunk of charge dV = (1/4[tex]\pi[/tex][tex]\epsilon[/tex]) dq / r , determine what dq is and the limits of integration, and integrate.
 
  • #5
15
0
Got it! Thank you so much guys!:tongue:
 
  • #6
15
0
Hmm, there is a little different with the answer i got and the answer in the book.
I have attached a pdf file with my work and the answer in the book, could you guys check and see what i did wrong? Thanks again!
 

Attachments

  • 1.pdf
    1.1 MB · Views: 2,071
Top