Calculate Electric Field at the Center of Semicircular Loop

kjintonic
Messages
11
Reaction score
0
A semicircular loop of radius a carries positive charge Q distributed uniformly over its length.?
Find the electric field at the center of the loop (point P in the figure). Hint: Divide the loop into charge elements dq as shown in the figure, and write dq in terms of the angle d\theta. Then integrate over \theta to get the net field at P.


http://i533.photobucket.com/albums/ee336/shaneji_kotoba/RW-20-72.jpg

I don't know how to start this question...
 
kjintonic said:
A semicircular loop of radius a carries positive charge Q distributed uniformly over its length.?
Find the electric field at the center of the loop (point P in the figure). Hint: Divide the loop into charge elements dq as shown in the figure, and write dq in terms of the angle d\theta. Then integrate over \theta to get the net field at P.


http://i533.photobucket.com/albums/ee336/shaneji_kotoba/RW-20-72.jpg

I don't know how to start this question...
You must have some idea how to start, have you tried using the hint?
 
Probably I can use E= (S)dE= (S)kdq/r^2 inwhich r=a
(S) is integral sign
 
kjintonic said:
Probably I can use E= (S)dE= (S)kdq/r^2 inwhich r=a
(S) is integral sign
You're on the right lines, how about writing dq in terms of [itex]d\theta[/itex]?

HINT: Notice that the horizontal components will cancel so you need only consider the vertical components of the electric field.
 
hmmm... Sorry I kinda don't understant how to write dq in terms of d[itex]d\theta[itex]:([/itex][/itex]
 
bump?
 
First step you should note that by symmetry, one of field components in an axial direction is 0. Now you only have to find the other component. Set this problem up in the coordinate plane with point P at the origin. Draw a triangle for E in terms of E_x and E_y and theta. What can you say about how E_x is related to x? Write down the expression for dE, the differential electric field magnitude due to dq, then try to write dq in terms of [tex]\lambda dr[/tex], where lambda is Q/length.
 

Similar threads

Replies
8
Views
3K
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
1
Views
3K
  • · Replies 68 ·
3
Replies
68
Views
9K
Replies
3
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K