Dielectric boundary value problems

  • Thread starter Thread starter lys04
  • Start date Start date
Click For Summary
In solving dielectric boundary value problems, the introduction of an image charge in the z<0 region is necessary to accurately represent the electric potential. The image charge, while located at the same position as the original charge, has a different value, denoted as ##q''##, which is determined through analysis. The potential ##\phi_1## is calculated under the assumption that the dielectric material ##\varepsilon_1## extends throughout all space, affecting the image charge in the z>0 region. Similarly, the potential ##\phi_2## is derived with the assumption that the yellow dielectric ##\varepsilon_2## also permeates all space. This approach ensures a comprehensive understanding of the electric field behavior across the dielectric boundaries.
lys04
Messages
144
Reaction score
5
Homework Statement
Why in this problem when solving for the electric potential in the z<0 region we introduce another image charge that is the same as the original charge instead of just using the original charge? I don't really get the explanation they provide.
Relevant Equations
$$\vec{D}=\epsilon \vec{E}$$
1745389441297.png

1745389455722.png
 
Physics news on Phys.org
Hard to read.
The text should be fixed. It is not clear.
 
lys04 said:
Homework Statement: Why in this problem when solving for the electric potential in the z<0 region we introduce another image charge that is the same as the original charge instead of just using the original charge?
When finding the potential ##\phi_1## in region z<0, the image charge is not the same as the original charge. The image charge has the same location as the original charge, but the charge of the image is ##q''##, which differs from the original charge ##q##. The value of ##q''## will be determined in the analysis.

Also, when writing the potential ##\phi_1##, we imagine that the purple dielectric ##\varepsilon_1## extends throughout all of space: both z < 0 and z > 0. Thus, although ##q''## is located in the region ##z > 0##, ##q''## is taken to be immersed in dielectric material with ##\varepsilon_1##.
That's why you see this figure
1745430570324.png


Likewise, in writing the potential for ##\phi_2##, the yellow dielectric ##\varepsilon_2## is assumed to extend throughout all of space.
1745430531692.png
 
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
14
Views
2K
Replies
32
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
3K
  • · Replies 78 ·
3
Replies
78
Views
7K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K