Dielectric boundary value problems

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AI Thread 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
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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
 
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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
 
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