The method of images

  • Thread starter M. next
  • Start date
  • #1
382
0

Main Question or Discussion Point

Hello :)
I would be very grateful if I get through with this bugging idea.
When I started reading about it, the first example was: A point charge in front of a finite plane conductor.
And I understood after some mathematical and geometrical procedure how it is okay to replace the finite plane with a charge of opposite sign.
Then I bumped into another problem:
Let us consider orthonormal system and point charge +q is placed on point (1,1) - it's image being in symmetry with respect to y-axis will be at (-1,1) and it's image abeing in symmetry with respect to x-axis will be at (1,-1). Until here I was okay, till I realized that another point charge was placed at (-1,-1). Why?
What's the whole point?

Thanks in advance.

M. next
 

Answers and Replies

  • #2
131
2
The net charge of the image charges should be -q shouldn't it? If you add the two charges that you made at (-1,1) and (1,-1), those would add up to be -2q. Thus, you need 1 more +q charge at (-1,-1).
 
  • #3
382
0
Who said that the net charge must be -q?
And another thing: if y-axis and x-axis are not perpendicular but they have an acute angle between them (say 60) and q is placed at (4,1) as in the old orthonormal system (am only placing coordinates to clear up the image). What will happen? What should we do then? How do we treat this case?
 
Last edited:
  • #4
131
2
If your angle is theta then the number of image charges you would need would be 360/theta -1. This number must be an integer so only angles that divide 360 are allowed for this method to work.
 
  • #5
382
0
nucl34rgg, I appreciate it a lot. But can you please tell me how do you know these equations? Because I read the whole course and didn't come across these equations.
And then you told me the nb of images but you didn't tell me where to locate them?
 
  • #6
131
2
Ok this picture is embarrassingly bad, but it will serve the purpose. Pretend each angle between the lines is 60 degrees, and pretend that the figure is radially symmetric. All you have to do is reflect around and go around in a circle. Treat the one +q in the top right as real and the rest are image charges. Then the lines along the boundary for that +q have V=0.
 

Attachments

  • #7
382
0
Thanks for your time, I really appreciate it. But what do you mean by "reflect around and go around in a circle"
You must be fed up by now, but I have to understand this by tomorrow.
Thanks again
 
  • #8
131
2
start with the positive q charge on the top right. reflect over the v=0 boundary line. add a -q charge there and reflect over the next line...add a +q charge there, etc
 

Attachments

Last edited:
  • #9
382
0
Oh now i get it! Finally!
Thank you loads nucl34rgg. I really appreciate your time, and patience. :)
 
  • #10
131
2
Glad to help! :)
 

Related Threads on The method of images

  • Last Post
Replies
3
Views
605
  • Last Post
Replies
3
Views
937
Replies
4
Views
1K
Replies
11
Views
743
Replies
4
Views
2K
Replies
6
Views
655
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
2
Views
13K
  • Last Post
Replies
2
Views
15K
  • Last Post
Replies
4
Views
957
Top